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.
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: $$\sum_i^n \binom{n}{i}{p_\theta}^i(1-p_\theta)^{n-i}$$