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I've been trying to solve the following problem but am thrown off by the "guessing in case she doesn't know the answer". The problem is as follows:

A student takes a test with n questions of equal difficulty. For each question, the student has a probability $\theta$ of knowing the answer and in that case she solves the question perfectly. If she does not know the answer, she guesses and then she has a probability of 0.5 of solving it correctly. The responses to all n questions can be considered independent.

Based on this information, the following is asked:

Derive analytically the likelihood function for $\theta$ given y correctly solved items.

The "likelihood given y correctly solved items" confuses me. I've seen many examples of solving $P(\theta|Y=1)$ using Bayes' theorem (e.g. see here). But how can this be generalized to a likelihood for $\theta$ given y correctly solved items? Do we even need Bayes' theorem or can this just be answered with a Binomial distribution corrected for the guessing given she doesn't know the answer?

Furthermore, a follow-up question is asked:

Assuming a beta prior with constants a and b, derive an expression for the marginal probability $p(y)$. Simplify the expression, but leave the integral if there is no closed-form solution.

This makes me believe Bayes' theorem would indeed be used for the first question. Any help on any part of this problem is greatly appreciated.

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    $\begingroup$ May I confirm that the required likelihood function $L(\theta\mid Y=y) = P(Y=y\mid \theta)$, as defined in likelihood function on Wikipedia? $\endgroup$
    – peterwhy
    Commented Aug 13, 2022 at 0:36
  • $\begingroup$ The phrasing is a bit awkward in my opinion. Usually I wouldnt say "likelihood given $Y$" since "given" implies that you are conditioning on $Y$, but that is not typically how the likelihood is defined. See my answer below. $\endgroup$
    – dmh
    Commented Aug 13, 2022 at 1:50

1 Answer 1

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The likelihood of $\theta$ is $\mathbb P(Y=y | \theta)$ (the probability of the data given the parameter). But $Y$ is just a binomial random variable given $\theta$, where the success probability is given by the probability of getting a question correct. So all we need to do is calculate the probability of getting any one question correct.

$$p_\theta:=\mathbb P (\text{correct}) = \mathbb P(\text{correct} | \text{known})\mathbb P(\text{known}) + \mathbb P(\text{correct} | \text{unknown})\mathbb P(\text{unknown}) = \theta + (1-\theta)\cdot0.5$$

Then the likelihood is: $$\sum_i^n \binom{n}{i}{p_\theta}^i(1-p_\theta)^{n-i}$$

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