TF ErrorAna StatInference

Statistical Inference

Frequentist -vs- Bayesian Inference

When it comes to testing a hypothesis, there are two dominant philosophies known as a Frequentist or a Bayesian perspective.

The dominant discussion for this class will be from the Frequentist perspective.

frequentist statistical inference

Statistical inference is made using a null-hypothesis test; that is, ones that answer the question Assuming that the null hypothesis is true, what is the probability of observing a value for the test statistic that is at least as extreme as the value that was actually observed?

The relative frequency of occurrence of an event, in a number of repetitions of the experiment, is a measure of the probability of that event. Thus, if nt is the total number of trials and nx is the number of trials where the event x occurred, the probability P(x) of the event occurring will be approximated by the relative frequency as follows:

[math]P(x) \approx \frac{n_x}{n_t}.[/math]

Bayesian inference.

Statistical inference is made by using evidence or observations to update or to newly infer the probability that a hypothesis may be true. The name "Bayesian" comes from the frequent use of Bayes' theorem in the inference process.

Bayes' theorem relates the conditional probability|conditional and marginal probability|marginal probabilities of events A and B, where B has a non-vanishing probability:

[math]P(A|B) = \frac{P(B | A)\, P(A)}{P(B)}\,\! [/math].

Each term in Bayes' theorem has a conventional name:

• P(A) is the prior probability or marginal probability of A. It is "prior" in the sense that it does not take into account any information about B.
• P(B) is the prior or marginal probability of B, and acts as a normalizing constant.
• P(A|B) is the conditional probability of A, given B. It is also called the posterior probability because it is derived from or depends upon the specified value of B.
• P(B|A) is the conditional probability of B given A.

Bayes' theorem in this form gives a mathematical representation of how the conditional probabability of event A given B is related to the converse conditional probabablity of B given A.

Example

Suppose there is a school having 60% boys and 40% girls as students.

The female students wear trousers or skirts in equal numbers; the boys all wear trousers.

An observer sees a (random) student from a distance; all the observer can see is that this student is wearing trousers.

What is the probability this student is a girl?

The correct answer can be computed using Bayes' theorem.

[math] P(A) \equiv[/math] probability that the student observed is a girl = 0.4

[math]P(B) \equiv[/math] probability that the student observed is wearing trousers = 60+20/100 = 0.8

[math]P(B|A) \equiv[/math] probability the student is wearing trousers given that the student is a girl

[math]P(A|B) \equiv[/math] probability the student is a girl given that the student is wearing trousers

[math]P(B|A) =0.5[/math]

[math]P(A|B) = \frac{P(B|A) P(A)}{P(B)} = \frac{0.5 \times 0.4}{0.8} = 0.25.[/math]

To compute P(A|B), we first need to know: P(A), or the probability that the student is a girl regardless of any other information. Since the observers sees a random student, meaning that all students have the same probability of being observed, and the fraction of girls among the students is 40%, this probability equals 0.4. P(B|A), or the probability of the student wearing trousers given that the student is a girl. As they are as likely to wear skirts as trousers, this is 0.5. P(B), or the probability of a (randomly selected) student wearing trousers regardless of any other information. Since P(B) = P(B|A)P(A) + P(B|A')P(A'), this is 0.5×0.4 + 1×0.6 = 0.8. Given all this information, the probability of the observer having spotted a girl given that the observed student is wearing trousers can be computed by substituting these values in the formula:

[math]P(A|B) = \frac{P(B|A) P(A)}{P(B)} = \frac{0.5 \times 0.4}{0.8} = 0.25.[/math]

Another, essentially equivalent way of obtaining the same result is as follows. Assume, for concreteness, that there are 100 students, 60 boys and 40 girls. Among these, 60 boys and 20 girls wear trousers. All together there are 80 trouser-wearers, of which 20 are girls. Therefore the chance that a random trouser-wearer is a girl equals 20/80 = 0.25. Put in terms of Bayes´ theorem, the probability of a student being a girl is 40/100, the probability that any given girl will wear trousers is 1/2. The product of two is 20/100, but you know the student is wearing trousers, so you remove the 20 non trouser wearing students and receive a probability of (20/100)/(80/100), or 20/80.

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