Difference between revisions of "Forest ErrAna StatDist"
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− | Here the above average of a parent distribution is defined in terms of an infinite sum of observations (x_i) of an observable x divided by the number of observations. | + | Here the above average of a parent distribution is defined in terms of an infinite sum of observations (<math>x_i</math>) of an observable x divided by the number of observations. |
<math>\bar{x}</math> is a calculation of the mean using a finite number of observations | <math>\bar{x}</math> is a calculation of the mean using a finite number of observations | ||
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Definition of the average | Definition of the average | ||
:<math>\mu \equiv \lim_{N\rightarrow \infty} \frac{\sum x_i}{N}</math> | :<math>\mu \equiv \lim_{N\rightarrow \infty} \frac{\sum x_i}{N}</math> | ||
+ | |||
+ | the deviation average would be given by | ||
:<math>\lim_{N\rightarrow \infty} \frac{\sum (x_i - \mu)}{N}</math> | :<math>\lim_{N\rightarrow \infty} \frac{\sum (x_i - \mu)}{N}</math> | ||
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The standard deviation is defined as the square root of the variance | The standard deviation is defined as the square root of the variance | ||
− | :S.D. = <math>\sqrt{\sigma}</math> | + | :S.D. = <math>\sqrt{\sigma^2}</math> |
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:<math>\frac{\sum(x_i-\mu)^2}{N} =\frac{\sum x_i^2}{N} - \mu^2</math> | :<math>\frac{\sum(x_i-\mu)^2}{N} =\frac{\sum x_i^2}{N} - \mu^2</math> | ||
+ | |||
+ | In the case where your observables are weighted by the probability of the observation one has an expectations value given by | ||
+ | |||
+ | <math>E[x] \equiv \sum x_i P_x(x)</math> | ||
+ | |||
+ | and a variance given by | ||
:<math>\Rightarrow \sigma^2 = E[(x-\mu)^2] = \sum_{x=0}^n (x_i - \mu)^2 P(x_i)</math> | :<math>\Rightarrow \sigma^2 = E[(x-\mu)^2] = \sum_{x=0}^n (x_i - \mu)^2 P(x_i)</math> | ||
: <math>= E[x^2] - \left ( E[x]\right )^2 = \sum_{x=0}^n x_i^2 P(x_i) - \left ( \sum_{x=0}^n x_i P(x_i)\right )^2</math> | : <math>= E[x^2] - \left ( E[x]\right )^2 = \sum_{x=0}^n x_i^2 P(x_i) - \left ( \sum_{x=0}^n x_i P(x_i)\right )^2</math> | ||
+ | |||
+ | to compare with the previous expression, let <math> P(x_i) = 1/N</math> | ||
== Average for an unknown probability distribution (parent population)== | == Average for an unknown probability distribution (parent population)== | ||
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===Arithmetic Mean and variance=== | ===Arithmetic Mean and variance=== | ||
− | If <math>n</math> observables are | + | If <math>n</math> observables are made in an experiment then the arithmetic mean of those observables is defined as |
:<math>\bar{x} = \frac{\sum_{i=1}^{i=N} x_i}{N}</math> | :<math>\bar{x} = \frac{\sum_{i=1}^{i=N} x_i}{N}</math> | ||
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:<math>\sigma^2 = \frac{\sum_{i=1}^{i=N} (x_i - \bar{x})^2}{N}</math> = RMS Error= Root Mean Squared Error | :<math>\sigma^2 = \frac{\sum_{i=1}^{i=N} (x_i - \bar{x})^2}{N}</math> = RMS Error= Root Mean Squared Error | ||
− | ;Note:RMS = Root Mean Square = <math>\frac{\sum_i^n x_i^2}{ | + | ;Note:RMS = Root Mean Square = <math>\frac{\sum_i^n x_i^2}{N}</math> = |
==== Statistical Variance decreases with N==== | ==== Statistical Variance decreases with N==== | ||
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:<math>\sigma^2_{\bar{x}} = \frac{\sum \left (\bar{x} -\overline{\left ( \bar{x} \right ) } \right )^2}{N} =\frac{\sum \bar{x_i}^2}{N} -\left( \overline{\left ( \bar{x} \right ) } \right )^2</math> | :<math>\sigma^2_{\bar{x}} = \frac{\sum \left (\bar{x} -\overline{\left ( \bar{x} \right ) } \right )^2}{N} =\frac{\sum \bar{x_i}^2}{N} -\left( \overline{\left ( \bar{x} \right ) } \right )^2</math> | ||
− | :<math>=\frac{\sum \bar{x_i}^2}{N} -\left( \bar{x} \right )^2</math> | + | ::<math>=\frac{\sum \bar{x_i}^2}{N} -\left( \bar{x} \right )^2</math> |
− | :<math>=\frac{\sum \left( \sum \frac{x_i}{N}\right)^2}{N} -\left( \bar{x} \right )^2</math> | + | ::<math>=\frac{\sum \left( \sum \frac{x_i}{N}\right)^2}{N} -\left( \bar{x} \right )^2</math> |
− | :<math>=\frac{1}{N^2}\frac{\sum \left( \sum x_i\right)^2}{N} -\left( \bar{x} \right )^2</math> | + | ::<math>=\frac{1}{N^2}\frac{\sum \left( \sum x_i\right)^2}{N} -\left( \bar{x} \right )^2</math> |
− | :<math>=\frac{1}{N^2}\frac{\sum \left (\sum x_i^2 + \sum_{i \ne j} x_ix_j \right )}{N} -\left( \bar{x} \right )^2</math> | + | ::<math>=\frac{1}{N^2}\frac{\sum \left (\sum x_i^2 + \sum_{i \ne j} x_ix_j \right )}{N} -\left( \bar{x} \right )^2</math> |
− | :<math>=\frac{1}{N^2} \left [ \frac{ | + | ::<math>=\frac{1}{N^2} \sum \left [ \frac{ \left(\sum x_i^2 \right)}{N} + \frac{ \left (\sum_{i \ne j} x_ix_j \right )}{N} \right ] -\left( \bar{x} \right )^2</math> |
− | + | ::<math>\sigma^2 =\frac{\sum x_i^2}{N} - \left ( \frac{\sum x_i}{N}\right )^2</math> | |
− | : | ||
− | |||
− | |||
− | :<math>\sigma^2_{\bar{x}}=\frac{1}{N^2} \left [ \frac{\sum \left(\sum x_i^2 \right)}{N} + \sum_{i \ne j} \bar{x}^2 \right ] -\left( \bar{x} \right )^2</math> | + | :<math>\sigma^2_{\bar{x}}=\frac{1}{N^2} \sum \left [ \sigma^2 + \left ( \frac{\sum x_i}{N}\right )^2 + \frac{ \left (\sum_{i \ne j} x_ix_j \right )}{N} \right ] -\left( \bar{x} \right )^2</math> |
+ | ::<math>= \frac{\sigma^2}{N} + \frac{1}{N^2} \sum \left [ \left ( \frac{\sum x_i}{N}\right )^2 + \frac{ \left (\sum_{i \ne j} x_ix_j \right )}{N} \right ] -\left( \bar{x} \right )^2</math> | ||
+ | ::<math>= \frac{\sigma^2}{N} + \frac{1}{N^2} \sum \left [ \left ( \bar x \right )^2 + \frac{ \left (\sum_{i \ne j} x_ix_j \right )}{N} \right ] -\left( \bar{x} \right )^2</math> | ||
+ | ::<math>= \frac{\sigma^2}{N} + \frac{\left ( \bar x \right )^2}{N} + \frac{1}{N^2} \sum \left [ \frac{ \left (\sum_{i \ne j} x_ix_j \right )}{N} \right ] -\left( \bar{x} \right )^2</math> | ||
− | |||
− | + | ;If the measurements are all independent | |
− | + | ::<math> \sum \left [ \frac{ \left (\sum_{i \ne j} x_ix_j \right )}{N} \right ] = (N^2-N) \left ( \frac{\sum x_i}{N}\right )^2</math> | |
− | :<math>\sigma^2_{\bar{x}}=\frac{ | + | :<math>\sigma^2_{\bar{x}}= \frac{\sigma^2}{N} + \frac{\left ( \bar x \right )^2}{N} + (1-\frac{1}{N}) \left ( \frac{\sum x_i}{N}\right )^2 -\left( \bar{x} \right )^2</math> |
− | + | ::<math>= \frac{\sigma^2}{N} + \frac{\left ( \bar x \right )^2}{N} + \left( \bar{x} \right )^2 - \frac{\left ( \bar x \right )^2}{N} -\left( \bar{x} \right )^2</math> | |
− | :<math>=\frac{ | + | ::<math>= \frac{\sigma^2}{N} </math> |
− | |||
− | |||
− | :<math>= \frac{\sigma^2}{N} </math> | ||
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To determine the "true" variance consider taking average of several sample variances (this is the same argument used above which let to <math>\overline{(\bar{x})} = \bar{x}</math> ) | To determine the "true" variance consider taking average of several sample variances (this is the same argument used above which let to <math>\overline{(\bar{x})} = \bar{x}</math> ) | ||
− | :<math>\frac{\ | + | :<math>\frac{\sum_j \left [ \sigma_n^2 \right ]_j}{N} = \frac{ \sum_j \left [ \frac{\sum_i x_i^2}{n} -\left ( \bar{x} \right)^2 \right ]_j}{N}</math> |
:<math>= \frac{1}{n}\sum_i \left ( \frac{\sum_j x_j^2}{N} \right )_i - \frac {\sum_j \left ( \bar{x} \right)^2 }{N}</math> | :<math>= \frac{1}{n}\sum_i \left ( \frac{\sum_j x_j^2}{N} \right )_i - \frac {\sum_j \left ( \bar{x} \right)^2 }{N}</math> | ||
:<math>= \frac{1}{n}\sum_i \left ( \frac{\sum_j x_j^2}{N} \right )_i - \left [ \left ( \frac {\sum_j \bar{x}}{N} \right)^2 + \sigma_{\bar{x}}^2\right ]</math> : as shown previously <math>E[\bar{x}^2] = \left ( E[\bar{x}] \right )^2 + \sigma_{\bar{x}}^2</math> | :<math>= \frac{1}{n}\sum_i \left ( \frac{\sum_j x_j^2}{N} \right )_i - \left [ \left ( \frac {\sum_j \bar{x}}{N} \right)^2 + \sigma_{\bar{x}}^2\right ]</math> : as shown previously <math>E[\bar{x}^2] = \left ( E[\bar{x}] \right )^2 + \sigma_{\bar{x}}^2</math> | ||
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:<math>\frac{\sum \sigma_i^2}{N} =</math> an average of all possible sample variance which should be equivalent to the "true" population variance. | :<math>\frac{\sum \sigma_i^2}{N} =</math> an average of all possible sample variance which should be equivalent to the "true" population variance. | ||
− | :<math>\Rightarrow \frac{\sum \sigma_i^2}{N} \sim \frac{x_i-\bar{x}}{n}</math> | + | :<math>\Rightarrow \frac{\sum \sigma_i^2}{N} \sim \sum \frac{(x_i-\bar{x})^2}{n}</math> : if all the variances are the same this would be equivalent |
− | : <math>\sigma^2 = \frac{n}{n-1}\frac{\sum(x_i-\bar{x})}{n}</math> | + | : <math>\sigma^2 = \frac{n}{n-1}\frac{\sum(x_i-\bar{x})^2}{n}</math> |
− | : <math>= \frac{\sum(x_i-\bar{x})}{n-1} =</math> unbiased sample variance | + | : <math>= \frac{\sum(x_i-\bar{x})^2}{n-1} =</math> unbiased sample variance |
==Probability Distributions== | ==Probability Distributions== | ||
− | === Mean(Expectation value) and variance=== | + | === Mean(Expectation value) and variance for a Given Probability Distribution=== |
====Mean of Discrete Probability Distribution==== | ====Mean of Discrete Probability Distribution==== | ||
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<math>x_i =</math> ith observable quantity | <math>x_i =</math> ith observable quantity | ||
− | <math>P(x_i) =</math> probability of observing <math>x_i</math> = Probability | + | <math>P(x_i) =</math> probability of observing <math>x_i</math> = Probability Distribution for a discrete probability distribution |
====Mean of a continuous probability distibution==== | ====Mean of a continuous probability distibution==== | ||
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− | This means the probability of getting exactly | + | This means the probability of getting exactly <math>x</math> successes after <math>n</math> trials is |
:<math>P_B(x) = {n\choose x}p^{x}q^{n-x} </math> | :<math>P_B(x) = {n\choose x}p^{x}q^{n-x} </math> | ||
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:<math>P = \frac{\mbox{number of desired outcomes}}{\mbox{number of possible outcomes}} = \frac{1}{2}</math> = Uniform distribution with a=0 (tails) b=1 (heads). | :<math>P = \frac{\mbox{number of desired outcomes}}{\mbox{number of possible outcomes}} = \frac{1}{2}</math> = Uniform distribution with a=0 (tails) b=1 (heads). | ||
− | Suppose you toss a coin 4 times. Here are the possible outcomes | + | Suppose you toss a coin 4 times. |
+ | |||
+ | |||
+ | |||
+ | |||
+ | Here are the possible outcomes | ||
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|16||h|| h || h|| h||4 | |16||h|| h || h|| h||4 | ||
|} | |} | ||
+ | |||
+ | |||
+ | How many possible ways can you get all tails (x=4). | ||
+ | |||
+ | |||
+ | <math>{n\choose x}=\frac{n!}{x!(n-x)!}=\frac{4!}{4!(4-4)!} =1</math> 0! = 1 | ||
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===Poisson Distribution=== | ===Poisson Distribution=== | ||
− | The Poisson distribution is an approximation to the binomial distribution in the event | + | The Poisson distribution is an approximation to the binomial distribution in the event that the probability of a success is quite small <math>(p \ll 1)</math>. As the number of repeated observations (n) gets large, the binomial distribution becomes more difficult to evaluate because of the leading term |
:<math>\frac{n !}{ \left [ (n-x) ! x !\right ]}</math> | :<math>\frac{n !}{ \left [ (n-x) ! x !\right ]}</math> | ||
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− | :<math>P(0,x, \lambda) = | + | :<math>P(0,x, \lambda) = 1 - \frac{x}{\lambda} \approx e^{\frac{-x}{\lambda}}</math> |
Now we wish to find the probability of finding <math>N</math> events over a distance <math>x</math> given the mean free path. | Now we wish to find the probability of finding <math>N</math> events over a distance <math>x</math> given the mean free path. | ||
− | This is calculated as a joint probability. If it were the case that we wanted to know the probability of only one interaction over a distance <math>L</math> | + | This is calculated as a joint probability. If it were the case that we wanted to know the probability of only one interaction over a distance <math>L</math>, then we would want to multiply the probability that an interaction happened after a distance <math>dx</math> by the probability that no more interactions happen by the time the particle reaches the distance <math>L</math>. |
For the case of <math>N</math> interactions, we have a series of <math>N</math> interactions happening over N intervals of <math>dx</math> with the probability <math>dx/\lambda</math> | For the case of <math>N</math> interactions, we have a series of <math>N</math> interactions happening over N intervals of <math>dx</math> with the probability <math>dx/\lambda</math> | ||
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− | The N! accounts for the degeneracy in which for every N! permutations there is really | + | The N! accounts for the degeneracy in which for every N! permutations there is really only one new combination. ie we are double counting when we integrate. |
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====Variance of Poisson Dist==== | ====Variance of Poisson Dist==== | ||
− | For [http://wiki.iac.isu.edu/index.php/TF_ErrAna_Homework#Poisson_Prob_Dist Homework] you | + | For [http://wiki.iac.isu.edu/index.php/TF_ErrAna_Homework#Poisson_Prob_Dist Homework] you will show, in a manner similar to the above mean calculation, that the variance of the Poisson distribution is |
:<math>\sigma^2 = \mu</math> | :<math>\sigma^2 = \mu</math> | ||
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The Gaussian distribution is accepted as one of the most likely distributions to describe measurements. | The Gaussian distribution is accepted as one of the most likely distributions to describe measurements. | ||
− | A Gaussian distribution which is normalized such that its integral is unity is | + | A Gaussian distribution which is normalized such that its integral is unity is refered to as the Normal distribution. You could mathematically construct a Gaussian distribution that is not normalized to unity (this is often done when fitting experimental data). |
:<math>P_G(x,\mu, \sigma) = \frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{1}{2} \left( \frac{x -\mu}{\sigma} \right) ^2}</math> = probability of observing <math>x</math> from a Gaussian parent distribution with a mean <math>\mu</math> and standard deviation <math>\sigma</math>. | :<math>P_G(x,\mu, \sigma) = \frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{1}{2} \left( \frac{x -\mu}{\sigma} \right) ^2}</math> = probability of observing <math>x</math> from a Gaussian parent distribution with a mean <math>\mu</math> and standard deviation <math>\sigma</math>. | ||
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==== Half-Width <math>\Gamma</math> (a.k.a. Full Width as Half Max)==== | ==== Half-Width <math>\Gamma</math> (a.k.a. Full Width as Half Max)==== | ||
− | The half width <math>\Gamma</math> is used to describe the range of <math>x</math> through which the | + | The half width <math>\Gamma</math> is used to describe the range of <math>x</math> through which the distribution's amplitude decreases to half of its maximum value. |
;ie: <math>P_G(\mu \pm \frac{\Gamma}{2}, \mu, \sigma) = \frac{P_G(\mu,\mu,\sigma)}{2}</math> | ;ie: <math>P_G(\mu \pm \frac{\Gamma}{2}, \mu, \sigma) = \frac{P_G(\mu,\mu,\sigma)}{2}</math> | ||
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==== Binomial with Large N becomes Gaussian==== | ==== Binomial with Large N becomes Gaussian==== | ||
− | Consider the binomial distribution in which a fair coin is tossed a large number of times (N is very large and | + | Consider the binomial distribution in which a fair coin is tossed a large number of times (N is very large and an EVEN number N=2n) |
What is the probability you get exactly <math>\frac{1}{2}N -s</math> heads and <math>\frac{1}{2}N +s</math> tails where s is an integer? | What is the probability you get exactly <math>\frac{1}{2}N -s</math> heads and <math>\frac{1}{2}N +s</math> tails where s is an integer? | ||
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similarly | similarly | ||
− | :<math>\ln\left [(n-s)! \right ] \sim \ln \left [ \left (2 \pi \right)^{1/2} \right ] +\ln \left [ (n- | + | :<math>\ln\left [(n-s)! \right ] \sim \ln \left [ \left (2 \pi \right)^{1/2} \right ] +\ln \left [ (n-s)^{1/2} \right ] + (n-s)(\ln\left [ (n-s) \right ] -1 )</math> |
− | :<math>\ln\left [(n+s)! \right ] \sim \ln \left [ \left (2 \pi \right)^{1/2} \right ] +\ln \left [ (n+ | + | :<math>\ln\left [(n+s)! \right ] \sim \ln \left [ \left (2 \pi \right)^{1/2} \right ] +\ln \left [ (n+s)^{1/2} \right ] + (n+s)(\ln\left [ (n+s) \right ] -1 )</math> |
:<math>\Rightarrow \ln (R) = 2 \times \left (\ln \left [ \left (2 \pi \right)^{1/2} \right ] +\ln \left [ n^{1/2} \right ] +n(\ln\left [ n \right ] -1 ) \right ) </math> | :<math>\Rightarrow \ln (R) = 2 \times \left (\ln \left [ \left (2 \pi \right)^{1/2} \right ] +\ln \left [ n^{1/2} \right ] +n(\ln\left [ n \right ] -1 ) \right ) </math> | ||
− | :<math>- \left ( \ln \left [ \left (2 \pi \right)^{1/2} \right ] +\ln \left [ (n- | + | :<math>- \left ( \ln \left [ \left (2 \pi \right)^{1/2} \right ] +\ln \left [ (n-s)^{1/2} \right ] + (n-s)(\ln\left [ (n-s) \right ] -1 )\right )</math> |
− | :<math> -\left ( \ln \left [ \left (2 \pi \right)^{1/2} \right ] +\ln \left [ (n+ | + | :<math> -\left ( \ln \left [ \left (2 \pi \right)^{1/2} \right ] +\ln \left [ (n+s)^{1/2} \right ] + (n+s)(\ln\left [ (n+s) \right ] -1 )\right ) </math> |
− | : <math>= 2 \ln \left [ n^{1/2} \right ] +2 n(\ln\left [ n \right ] -1 ) - \ln \left [ (n- | + | : <math>= 2 \ln \left [ n^{1/2} \right ] +2 n(\ln\left [ n \right ] -1 ) - \ln \left [ (n-s)^{1/2} \right ] - (n-s)(\ln\left [ (n-s) \right ] -1 ) -\ln \left [ (n+s)^{1/2} \right ] - (n+s)(\ln\left [ (n+s) \right ] -1 )</math> |
− | : <math>\ln \left [ n^{1/2} \right ] = \ln \left [ (n- | + | : <math>\ln \left [ n^{1/2} \right ] = \ln \left [ (n-s)^{1/2} \right ] = \ln \left [ (n+s)^{1/2} \right ]</math> For Large <math>n \gg s</math> |
:<math> \ln (R) = 2 n(\ln\left [ n \right ] -1 ) - (n-s)(\ln\left [ (n-s) \right ] -1 ) - (n+s)(\ln\left [ (n+s) \right ] -1 )</math> | :<math> \ln (R) = 2 n(\ln\left [ n \right ] -1 ) - (n-s)(\ln\left [ (n-s) \right ] -1 ) - (n+s)(\ln\left [ (n+s) \right ] -1 )</math> | ||
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:<math>\sigma^2 = \mu</math> | :<math>\sigma^2 = \mu</math> | ||
− | replacing dummy variable x with r - \mu | + | replacing dummy variable x with r - <math>\mu</math> |
:<math>P_P(r) = \frac{1}{\sqrt{2 \pi \sigma^2}} e^{\frac{-(r - \mu)^2}{2\sigma^2} } =\frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{1}{2} \left( \frac{r -\mu}{\sigma} \right) ^2}</math> = Gaussian distribution when <math>\mu \gg 1</math> | :<math>P_P(r) = \frac{1}{\sqrt{2 \pi \sigma^2}} e^{\frac{-(r - \mu)^2}{2\sigma^2} } =\frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{1}{2} \left( \frac{r -\mu}{\sigma} \right) ^2}</math> = Gaussian distribution when <math>\mu \gg 1</math> | ||
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:<math>A_G(\Delta x, \mu, \sigma) = \frac{1}{\sigma \sqrt{2 \pi}} \int_{\mu - \Delta x}^{\mu + \Delta x} e^{- \frac{1}{2} \left ( \frac{x - \mu}{\sigma}\right )^2} dx</math> | :<math>A_G(\Delta x, \mu, \sigma) = \frac{1}{\sigma \sqrt{2 \pi}} \int_{\mu - \Delta x}^{\mu + \Delta x} e^{- \frac{1}{2} \left ( \frac{x - \mu}{\sigma}\right )^2} dx</math> | ||
− | The advantage of this definition becomes apparent when you are | + | The advantage of this definition becomes apparent when you are interested in quantifying the probability that a measurement would fall outside a range <math>\Delta x</math>. |
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− | There is an 8% probability that 5 or more | + | There is an 8% probability that 5 or more students will fail the class |
===== 2 SD rule of thumb for Gaussian PDF ===== | ===== 2 SD rule of thumb for Gaussian PDF ===== | ||
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− | The above integral can | + | The above integral can be done numerically by expanding the exponential in a power series |
:<math>e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}</math> | :<math>e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}</math> | ||
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: <math>P_G(X \le \mu - 2 \sigma, \mu , \sigma ) = \frac{1}{ \sqrt{2 \pi}} \int_{-\infty}^{\mu - 2\sigma} \left ( 1 -\frac{z^2}{2}+ \frac{z^4}{8} - \frac{z^6}{48} \cdots \right)dz</math> | : <math>P_G(X \le \mu - 2 \sigma, \mu , \sigma ) = \frac{1}{ \sqrt{2 \pi}} \int_{-\infty}^{\mu - 2\sigma} \left ( 1 -\frac{z^2}{2}+ \frac{z^4}{8} - \frac{z^6}{48} \cdots \right)dz</math> | ||
:<math> = \left . \frac{1}{ \sqrt{2 \pi}} \left ( z -\frac{z^3}{6}+ \frac{z^5}{40} - \frac{z^7}{48 \times 7} \cdots \right ) \right |_{-\infty}^{\mu - 2\sigma}</math> | :<math> = \left . \frac{1}{ \sqrt{2 \pi}} \left ( z -\frac{z^3}{6}+ \frac{z^5}{40} - \frac{z^7}{48 \times 7} \cdots \right ) \right |_{-\infty}^{\mu - 2\sigma}</math> | ||
− | : <math>= \frac{1}{\pi} \sum_{j=0 | + | : <math>=\left . \frac{1}{\pi} \sum_{j=0}^{\infty} \frac{(-1)^j \left (\frac{x}{\sqrt{2}} \right)^{2j+1}}{j! (2j+1)} \right |_{x=\mu - 2\sigma}</math> |
+ | |||
+ | The probability is one which you can compute. | ||
+ | |||
+ | Below is a table representing the cumulative probability <math>P_G(x< \mu - \delta \mbox{ and } x> \mu + \delta , \mu, \sigma)</math> for events to occur outside and interval of <math>\pm \delta</math> in a Gaussian distribution | ||
+ | |||
+ | |||
+ | |||
+ | {| border="1" |cellpadding="20" cellspacing="0 | ||
+ | |- | ||
+ | |<math>P_G(x< \mu - \delta \mbox{ and } x> \mu + \delta , \mu, \sigma)</math> || <math>\delta</math> | ||
+ | |- | ||
+ | |<math>3.2 \times 10^{-1}</math> ||1<math>\sigma</math> | ||
+ | |- | ||
+ | |<math>4.4 \times 10^{-2}</math> ||2<math>\sigma</math> | ||
+ | |- | ||
+ | |<math>2.7 \times 10^{-3}</math> ||3<math>\sigma</math> | ||
+ | |- | ||
+ | |<math>6.3 \times 10^{-5}</math> ||4<math>\sigma</math> | ||
+ | |} | ||
+ | |||
+ | |||
+ | [[File:TF_Error_CDF_Gauss.png| 400 px]] | ||
===Cauchy/Lorentzian/Breit-Wigner Distribution=== | ===Cauchy/Lorentzian/Breit-Wigner Distribution=== | ||
+ | In Mathematics, the Cauchy distribution is written as | ||
+ | :<math>P_{CL}(x, x_0, \Gamma) = \frac{1}{\pi} \frac{\Gamma/2}{(x -x_0)^2 + (\Gamma/2)^2}</math> = Cauchy-Lorentian Distribution | ||
− | : | + | :Note; The probability does not fall as rapidly to zero as the Gaussian. As a result, the Gaussian's central peak contributes more to the area than the Lorentzian's. |
− | :<math>P_{ | + | This distribution happens to be a solution to physics problems involving forced resonances (spring systems driven by a source, or a nuclear interaction which induces a metastable state). |
+ | |||
+ | :<math>P_{BW} = \sigma(E)= \frac{1}{2\pi}\frac{\Gamma}{(E-E_0)^2 + (\Gamma/2)^2}</math> = Breit-Wigner distribution | ||
:<math>E_0 =</math> mass resonance | :<math>E_0 =</math> mass resonance | ||
Line 856: | Line 899: | ||
− | ==== | + | A Beit-Wigner function fit to cross section measured as a function of energy will allow one to evaluate the rate increases that are produced when the probing energy excites a resonant state that has a mass <math>E_0</math> and lasts for the time <math>\tau</math> derived from the Half Width <math>\Gamma</math>. |
+ | |||
+ | ==== mean==== | ||
+ | |||
+ | Mean is not defined | ||
+ | |||
+ | Mode = Median = <math>x_0</math> or <math>E_0</math> | ||
+ | |||
+ | ==== Variance ==== | ||
+ | |||
+ | |||
+ | The variance is also not defined but rather the distribution is parameterized in terms of the Half Width <math>\Gamma</math> | ||
+ | |||
Let | Let | ||
Line 866: | Line 921: | ||
− | The above integral does not converge. The width of the distribution is instead characterized by <math>\Gamma</math> = FWHM | + | The above integral does not converge for large z [large deviations <math>(x -\mu)</math>] . The width of the distribution is instead characterized by <math>\Gamma</math> = FWHM |
===Landau=== | ===Landau=== | ||
Line 873: | Line 928: | ||
where <math>c</math> is any positive real number. | where <math>c</math> is any positive real number. | ||
− | To simplify computation | + | To simplify computation it is more convenient to use the equivalent expression |
:<math>P_L(x) = \frac{1}{\pi} \int_0^\infty\! e^{-t \log t - x t} \sin(\pi t)\, dt.</math> | :<math>P_L(x) = \frac{1}{\pi} \int_0^\infty\! e^{-t \log t - x t} \sin(\pi t)\, dt.</math> | ||
+ | |||
+ | |||
+ | The above distribution was derived by Landau (L. Landau, "On the Energy Loss of Fast Particles by Ionization", J. Phys., vol 8 (1944), pg 201 ) to describe the energy loss by particles traveling through thin material ( materials with a thickness on the order of a few radiation lengths). | ||
+ | |||
+ | Bethe-Bloch derived an expression to determine the AVERAGE amount of energy lost by a particle traversing a given material <math>(\frac{dE}{dx})</math> assuming several collisions which span the physical limits of the interaction. | ||
+ | |||
+ | For the case of thin absorbers, the number of collisions is so small that the central limit theorem used to average over several collision doesn't apply and there is a finite possibility of observing large energy losses. | ||
+ | |||
+ | As a result one would expect a distribution which is Gaussian like but with a "tail" on the <math>\mu + \sigma</math> side of the distribution. | ||
===Gamma=== | ===Gamma=== | ||
− | :<math> P_{\gamma}(x | + | :<math> P_{\gamma}(x,k,\theta) = x^{k-1} \frac{e^{-x/\theta}}{\theta^k \, \Gamma(k)}\text{ for } x > 0\text{ and }k, \theta > 0.\,</math> |
where | where | ||
::<math> \Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\,dt\;</math> | ::<math> \Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\,dt\;</math> | ||
+ | |||
+ | |||
+ | The distribution is used for "waiting time" models. How long do you need to wait for a rain storm, how long do you need to wait to die,... | ||
+ | |||
+ | Climatologists use this for predicting how rain fluctuates from season to season. | ||
+ | |||
+ | If <math>k =</math> integer then the above distribution is a sum of <math>k</math> independent exponential distributions | ||
+ | |||
+ | :<math> P_{\gamma}(x,k,\theta) = 1 - e^{-x/\theta} \sum_{j=0}^{k-1}\frac{1}{j!}\left ( \frac{x}{\theta}\right)^j </math> | ||
+ | |||
+ | ==== Mean==== | ||
+ | |||
+ | :<math>\mu = k \theta</math> | ||
+ | |||
+ | ====Variance==== | ||
+ | |||
+ | :<math>\sigma^2 = k \theta^2</math> | ||
+ | |||
+ | ====Properties==== | ||
+ | |||
+ | :<math>\lim_{X \rightarrow \infty} P_{\gamma}(x,k,\theta) = \left \{ {\infty \;\;\;\; k <1 \atop 0 \;\;\;\; k>1} \right .</math> | ||
+ | : <math>= \frac{1}{\theta} \;\; \mbox{if} k=1</math> | ||
===Beta=== | ===Beta=== | ||
Line 893: | Line 979: | ||
^{\alpha-1}(1-x)^{\beta-1}\!</math> | ^{\alpha-1}(1-x)^{\beta-1}\!</math> | ||
− | === | + | ====Mean==== |
+ | :<math>\mu = \frac{\alpha}{\alpha + \beta}</math> | ||
− | === | + | ==== Variance==== |
− | = | + | :<math>\sigma^2 = \frac{\alpha \beta }{(\alpha + \beta)^2 (\alpha + \beta + 1)}</math> |
− | |||
− | === | + | ===Exponential=== |
− | The | + | The exponential distribution may be used to describe the processes that are in between Binomial and Poisson (exponential decay) |
− | :<math> | + | :<math> P_{e}(x,\lambda) = \left \{ {\lambda e^{-\lambda x} \;\;\;\; x \ge 0\atop 0 \;\;\;\; x<0} \right .</math> |
− | |||
+ | :<math> CDF_{e}(x,\lambda) = \left \{ {\lambda 1-e^{-\lambda x} \;\;\;\; x \ge 0\atop 0 \;\;\;\; x<0} \right .</math> | ||
− | ==== | + | ==== Mean==== |
+ | :<math>\mu = \frac{1}{\lambda}</math> | ||
− | + | ==== Variance ==== | |
− | == Skewness and | + | :<math>\sigma = \frac{1}{\lambda^2}</math> |
+ | |||
+ | == Skewness and Kurtosis== | ||
+ | |||
+ | Distributions may also be characterized by how they look in terms of Skewness and Kurtosis | ||
=== Skewness=== | === Skewness=== | ||
+ | |||
+ | Measures the symmetry of the distribution | ||
− | Skewness = <math>\frac{\sum (x_i - \bar{x})^3}{(N-1)s^3}</math> | + | Skewness = <math>\frac{\sum (x_i - \bar{x})^3}{(N-1)s^3} = \frac{\mbox{3rd moment}}{\mbox{2nd moment}}</math> |
where | where | ||
:<math>s^2 = \frac{\sum (x_i - \bar{x})^2}{N-1}</math> | :<math>s^2 = \frac{\sum (x_i - \bar{x})^2}{N-1}</math> | ||
+ | |||
+ | ;The higher the number the more asymmetric (or skewed) the distribution is. The closer to zero the more symmetric. | ||
− | + | A negative skewness indicates a tail on the left side of the distribution. | |
+ | Positive skewness indicates a tail on the right. | ||
===Kurtosis=== | ===Kurtosis=== | ||
+ | Measures the "pointyness" of the distribution | ||
Kurtosis = <math>\frac{\sum (x_i - \bar{x})^4}{(N-1)s^4}</math> | Kurtosis = <math>\frac{\sum (x_i - \bar{x})^4}{(N-1)s^4}</math> | ||
Line 938: | Line 1,035: | ||
:<math>s^2 = \frac{\sum (x_i - \bar{x})^2}{N-1}</math> | :<math>s^2 = \frac{\sum (x_i - \bar{x})^2}{N-1}</math> | ||
− | + | ||
K=3 for Normal Distribution | K=3 for Normal Distribution | ||
+ | |||
+ | In ROOT the Kurtosis entry in the statistics box is really the "excess kurtosis" which is the subtraction of the kurtosis by 3 | ||
+ | |||
+ | Excess Kurtosis = <math>\frac{\sum (x_i - \bar{x})^4}{(N-1)s^4} - 3</math> | ||
+ | |||
+ | In this case a Positive excess Kurtosis will indicate a peak that is sharper than a gaussian while a negative value will indicate a peak that is flatter than a comparable Gaussian distribution. | ||
+ | |||
+ | |||
+ | [[File:ForeErrAna_Gaus-Cauchy_SkeKurt.gif|200 px]][[File:ForeErrAna_Gaus-Landau_SkeKurt.gif|200 px]][[File:ForeErrAna_Gaus-gamma_SkeKurt.gif|200 px]] | ||
[http://wiki.iac.isu.edu/index.php/Forest_Error_Analysis_for_the_Physical_Sciences] [[Forest_Error_Analysis_for_the_Physical_Sciences]] | [http://wiki.iac.isu.edu/index.php/Forest_Error_Analysis_for_the_Physical_Sciences] [[Forest_Error_Analysis_for_the_Physical_Sciences]] |
Latest revision as of 21:49, 31 January 2020
Parent Distribution
Let
represent our ith attempt to measurement the quantityDue to the random errors present in any experiment we should not expect
.If we neglect systematic errors, then we should expect
to, on average, follow some probability distribution around the correct value .This probability distribution can be referred to as the "parent population".
Average and Variance
Average
The word "average" is used to describe a property of a "parent" probability distribution or a set of observations/measurements made in an experiment which gives an indication of a likely outcome of an experiment.
The symbol
is usually used to represent the "mean" of a known probability (parent) distribution (parent mean) while the "average" of a set of observations/measurements is denoted as
and is commonly referred to as the "sample" average or "sample mean".
Definition of the mean
Here the above average of a parent distribution is defined in terms of an infinite sum of observations ( ) of an observable x divided by the number of observations.
is a calculation of the mean using a finite number of observations
This definition uses the assumption that the result of an experiment, measuring a sample average of , asymptotically approaches the "true" average of the parent distribution .
Variance
The word "variance" is used to describe a property of a probability distribution or a set of observations/measurements made in an experiment which gives an indication how much an observation will deviate from and average value.
A deviation
of any measurement from a parent distribution with a mean can be defined asthe deviations should average to ZERO for an infinite number of observations by definition of the mean.
Definition of the average
the deviation average would be given by
But the AVERAGE DEVIATION is given by an average of the magnitude of the deviations given by
- = a measure of the dispersion of the expected observations about the mean
Taking the absolute value though is cumbersome when performing a statistical analysis so one may express this dispersion in terms of the variance
A typical variable used to denote the variance is
and is defined as
Standard Deviation
The standard deviation is defined as the square root of the variance
- S.D. =
The mean should be thought of as a parameter which characterizes the observations we are making in an experiment. In general the mean specifies the probability distribution that is representative of the observable we are trying to measure through experimentation.
The variance characterizes the uncertainty associated with our experimental attempts to determine the "true" value. Although the mean and true value may not be equal, their difference should be less than the uncertainty given by the governing probability distribution.
Another Expression for Variance
Using the definition of variance (omitting the limit as
)- Evaluating the definition of variance
In the case where your observables are weighted by the probability of the observation one has an expectations value given by
and a variance given by
to compare with the previous expression, let
Average for an unknown probability distribution (parent population)
If the "Parent Population" is not known, you are just given a list of numbers with no indication of the probability distribution that they were drawn from, then the average and variance may be calculate as shown below.
Arithmetic Mean and variance
If
observables are made in an experiment then the arithmetic mean of those observables is defined as
The "unbiased" variance of the above sample is defined as
- If you were told that the average is then you can calculate the
"true" variance of the above sample as
- = RMS Error= Root Mean Squared Error
- Note
- RMS = Root Mean Square = =
Statistical Variance decreases with N
The repetition of an experiment can decrease the STATISTICAL error of the experiment
Consider the following:
The average value of the mean of a sample of n observations drawn from the parent population is the same as the average value of each observation. (The average of the averages is the same as one of the averages)
- sample mean
- if all means are the same
This is the reason why the sample mean is a measure of the population average (
)Now consider the variance of the average of the averages (this is not the variance of the individual measurements but the variance of their means)
- If the measurements are all independent
The above is the essence of counting statistics.
It says that the STATISTICAL error in an experiment decreases as a function of
Biased and Unbiased variance
Where does this idea of an unbiased variance come from?
Using the same procedure as the previous section let's look at the average variance of the variances.
A sample variance of
measurements of is
To determine the "true" variance consider taking average of several sample variances (this is the same argument used above which let to )
- : as shown previously
- : also shown previously the universe average is the same as the sample average
Here
- the sample variance
- an average of all possible sample variance which should be equivalent to the "true" population variance.
- : if all the variances are the same this would be equivalent
- unbiased sample variance
Probability Distributions
Mean(Expectation value) and variance for a Given Probability Distribution
Mean of Discrete Probability Distribution
In the case that you know the probability distribution you can calculate the mean
or expectation value E(x) and standard deviation asFor a Discrete probability distribution
where
number of observations
number of different possible observable variables
ith observable quantity
probability of observing = Probability Distribution for a discrete probability distribution
Mean of a continuous probability distibution
The average (mean) of a sample drawn from any probability distribution is defined in terms of the expectation value E(x) such that
The expectation value for a continuous probability distribution is calculated as
Variance
Variance of a discrete PDF
Variance of a Continuous PDF
Expectation of Arbitrary function
If
is an arbitrary function of a variable governed by a probability distributionthen the expectation value of
is
or if a continuous distribtion
Uniform
The Uniform probability distribution function is a continuous probability function over a specified interval in which any value within the interval has the same probability of occurring.
Mathematically the uniform distribution over an interval from a to b is given by
Mean of Uniform PDF
Variance of Uniform PDF
Now use ROOT to generate uniform distributions.
http://wiki.iac.isu.edu/index.php/TF_ErrAna_InClassLab#Day_3
Binomial Distribution
Binomial random variable describes experiments in which the outcome has only 2 possibilities. The two possible outcomes can be labeled as "success" or "failure". The probabilities may be defined as
- p
- the probability of a success
and
- q
- the probability of a failure.
If we let
represent the number of successes after repeating the experiment timesExperiments with
are also known as Bernoulli trails.Then
is the Binomial random variable with parameters and .The number of ways in which the
successful outcomes can be organized in repeated trials is- where the denotes a factorial such that .
The expression is known as the binomial coefficient and is represented as
The probability of any one ordering of the success and failures is given by
This means the probability of getting exactly successes after trials is
Mean
It can be shown that the Expectation Value of the distribution is
- :summation starts from x=1 and not x=0 now
- :factor out : replace n-1 with m everywhere and it looks like binomial distribution
- :change summation index so y=x-1, now n become n-1
- :
- :definition of binomial expansion
- :q+p =1
variance
- Remember
To calculate the variance of the Binomial distribution I will just calculate and then subtract off .
- : x=0 term is zero so no contribution
Let m=n-1 and y=x-1
Examples
The number of times a coin toss is heads.
The probability of a coin landing with the head of the coin facing up is
- = Uniform distribution with a=0 (tails) b=1 (heads).
Suppose you toss a coin 4 times.
Here are the possible outcomes
order Number | Trial # | # of Heads | |||
1 | 2 | 3 | 4 | ||
1 | t | t | t | t | 0 |
2 | h | t | t | t | 1 |
3 | t | h | t | t | 1 |
4 | t | t | h | t | 1 |
5 | t | t | t | h | 1 |
6 | h | h | t | t | 2 |
7 | h | t | h | t | 2 |
8 | h | t | t | h | 2 |
9 | t | h | h | t | 2 |
10 | t | h | t | h | 2 |
11 | t | t | h | h | 2 |
12 | t | h | h | h | 3 |
13 | h | t | h | h | 3 |
14 | h | h | t | h | 3 |
15 | h | h | h | t | 3 |
16 | h | h | h | h | 4 |
How many possible ways can you get all tails (x=4).
0! = 1
The probability of order #1 happening is
P( order #1) =
P( order #2) =
The probability of observing the coin land on heads 3 times out of 4 trials is.
A 6 sided die
A die is a 6 sided cube with dots on each side. Each side has a unique number of dots with at most 6 dots on any one side.
P=1/6 = probability of landing on any side of the cube.
Expectation value :
- The expected (average) value for rolling a single die.
The variance:
If we roll the die 10 times what is the probability that X dice will show a 6?
A success will be that the die landed with 6 dots face up.
So the probability of this is 1/6 (p=1/6) , we toss it 10 times (n=10) so the binomial distribution function for a success/fail experiment says
So the probability the die will have 6 dots face up in 4/10 rolls is
Mean = np =
Variance =Poisson Distribution
The Poisson distribution is an approximation to the binomial distribution in the event that the probability of a success is quite small
. As the number of repeated observations (n) gets large, the binomial distribution becomes more difficult to evaluate because of the leading term
The poisson distribution overcomes this problem by defining the probability in terms of the average .
Poisson as approximation to Binomial
To drive home the idea that the Poisson distribution approximates a Binomial distribution at small p and large n consider the following derivation
The Binomial Probability Distriubtions is
The term
- IFF we have x terms above
- then
- example:
This leave us with
- For
Derivation of Poisson Distribution
The mean free path of a particle traversing a volume of material is a common problem in nuclear and particle physics. If you want to shield your apparatus or yourself from radiation you want to know how far the radiation travels through material.
The mean free path is the average distance a particle travels through a material before interacting with the material.
- If we let represent the mean free path
- Then the probability of having an interaction after a distance x is
as a result
- = probability of getting no events after a length dx
When we consider
( we are looking for small distances such that the probability of no interactions is high)
Now we wish to find the probability of finding
events over a distance given the mean free path.This is calculated as a joint probability. If it were the case that we wanted to know the probability of only one interaction over a distance
, then we would want to multiply the probability that an interaction happened after a distance by the probability that no more interactions happen by the time the particle reaches the distance .For the case of
interactions, we have a series of interactions happening over N intervals of with the probability
- = probability of finding events within the length
The above expression represents the probability for a particular sequence of events in which an interaction occurs after a distance then a interaction after ,
So in essence the above expression is a "probability element" where another probability element may be
where the first interaction occurs after the distance
.
So we can write a differential probability element which we need to add up as
The N! accounts for the degeneracy in which for every N! permutations there is really only one new combination. ie we are double counting when we integrate.
Using the integral formula
we end up with
Mean of Poisson Dist
Variance of Poisson Dist
For Homework you will show, in a manner similar to the above mean calculation, that the variance of the Poisson distribution is
Gaussian
The Gaussian (Normal) distribution is an approximation of the Binomial distribution for the case of a large number of possible different observations. Poisson approximated the binomial distribution for the case when p<<1 ( the average number of successes is a lot smaller than the number of trials
).The Gaussian distribution is accepted as one of the most likely distributions to describe measurements.
A Gaussian distribution which is normalized such that its integral is unity is refered to as the Normal distribution. You could mathematically construct a Gaussian distribution that is not normalized to unity (this is often done when fitting experimental data).
- = probability of observing from a Gaussian parent distribution with a mean and standard deviation .
Half-Width (a.k.a. Full Width as Half Max)
The half width
is used to describe the range of through which the distribution's amplitude decreases to half of its maximum value.- ie
- Side note
- the point of steepest descent is located at such that
Probable Error (P.E.)
The probable error is the range of
in which half of the observations (values of ) are expected to fall.Binomial with Large N becomes Gaussian
Consider the binomial distribution in which a fair coin is tossed a large number of times (N is very large and an EVEN number N=2n)
What is the probability you get exactly
heads and tails where s is an integer?The Binomial Probability distribution is given as
p = probability of success= 1/2
q= 1-p = 1/2
N = number of trials =2n
x= number of successes=n-s
Now let's cast this probability with respect to the probability that we get an even number of heads and tails by defining the following ratio R such that
Take the natural logarithm of both sides
Stirling's Approximation says
similarly
- For Large
If
Then
or
as a result
In binomial distributions
for this problem
or
= probability of exactly heads AND tails after flipping the coin N times (N is and even number and s is an integer).
If we let
and realize that for a binomial distributions
Then
- So when N gets big the Gaussian distribution is a good approximation to the Binomianl
Gaussian approximation to Poisson when
- = Poisson probability distribution
substitute
- Stirling's Approximation when
- : if Note:
another mathematical identity
if
thenIn the Poisson distribution
replacing dummy variable x with r -
- = Gaussian distribution when
Integral Probability (Cumulative Distribution Function)
The Poisson and Binomial distributions are discrete probability distributions (integers).
The Gaussian distribution is our first continuous distribution as the variables are real numbers. It is not very meaningful to speak of the probability that the variate (x) assumes a specific value.
One could consider defining a probability element
which is really an integral over a finite region such thatThe advantage of this definition becomes apparent when you are interested in quantifying the probability that a measurement would fall outside a range
.
The Cumulative Distribution Function (CDF), however, is defined in terms of the integral from the variates min value
- Probability that you measure a value less than or equal to
discrete CDF example
The probability that a student fails this class is 7.3%.
What is the probability more than 5 student will fail in a class of 32 students?
Answ:
- = 8%
There is an 8% probability that 5 or more students will fail the class
2 SD rule of thumb for Gaussian PDF
In the above example you calculated the probability that more than 5 student will fail a class. You can extend this principle to calculate the probability of taking a measurement which exceeds the expected mean value.
One of the more common consistency checks you can make on a sample data set which you expect to be from a Gaussian distribution is to ask how many data points appear more than 2 S.D. (
) from the mean value.
The CDF for this is
Let
The above integral can be done numerically by expanding the exponential in a power series
The probability is one which you can compute.
Below is a table representing the cumulative probability
for events to occur outside and interval of in a Gaussian distribution
1 | |
2 | |
3 | |
4 |
Cauchy/Lorentzian/Breit-Wigner Distribution
In Mathematics, the Cauchy distribution is written as
- = Cauchy-Lorentian Distribution
- Note; The probability does not fall as rapidly to zero as the Gaussian. As a result, the Gaussian's central peak contributes more to the area than the Lorentzian's.
This distribution happens to be a solution to physics problems involving forced resonances (spring systems driven by a source, or a nuclear interaction which induces a metastable state).
- = Breit-Wigner distribution
- mass resonance
- FWHM
- = uncertainty principle
- lifetime of resonance/intermediate state particle
A Beit-Wigner function fit to cross section measured as a function of energy will allow one to evaluate the rate increases that are produced when the probing energy excites a resonant state that has a mass and lasts for the time derived from the Half Width .
mean
Mean is not defined
Mode = Median =
orVariance
The variance is also not defined but rather the distribution is parameterized in terms of the Half Width
Let
Then
The above integral does not converge for large z [large deviations ] . The width of the distribution is instead characterized by = FWHM
Landau
where
is any positive real number.To simplify computation it is more convenient to use the equivalent expression
The above distribution was derived by Landau (L. Landau, "On the Energy Loss of Fast Particles by Ionization", J. Phys., vol 8 (1944), pg 201 ) to describe the energy loss by particles traveling through thin material ( materials with a thickness on the order of a few radiation lengths).
Bethe-Bloch derived an expression to determine the AVERAGE amount of energy lost by a particle traversing a given material
assuming several collisions which span the physical limits of the interaction.For the case of thin absorbers, the number of collisions is so small that the central limit theorem used to average over several collision doesn't apply and there is a finite possibility of observing large energy losses.
As a result one would expect a distribution which is Gaussian like but with a "tail" on the
side of the distribution.Gamma
where
The distribution is used for "waiting time" models. How long do you need to wait for a rain storm, how long do you need to wait to die,...
Climatologists use this for predicting how rain fluctuates from season to season.
If
integer then the above distribution is a sum of independent exponential distributionsMean
Variance
Properties
Beta
Mean
Variance
Exponential
The exponential distribution may be used to describe the processes that are in between Binomial and Poisson (exponential decay)
Mean
Variance
Skewness and Kurtosis
Distributions may also be characterized by how they look in terms of Skewness and Kurtosis
Skewness
Measures the symmetry of the distribution
Skewness =
where
- The higher the number the more asymmetric (or skewed) the distribution is. The closer to zero the more symmetric.
A negative skewness indicates a tail on the left side of the distribution.
Positive skewness indicates a tail on the right.
Kurtosis
Measures the "pointyness" of the distribution
Kurtosis =
where
K=3 for Normal Distribution
In ROOT the Kurtosis entry in the statistics box is really the "excess kurtosis" which is the subtraction of the kurtosis by 3
Excess Kurtosis =
In this case a Positive excess Kurtosis will indicate a peak that is sharper than a gaussian while a negative value will indicate a peak that is flatter than a comparable Gaussian distribution.