Consider the case of a binary parameter \theta \in \{ 0, 1\}. We have the prior distribution \pi (\theta ) that satisfies \pi (0)=p and \pi (1)=1-p, for some 0 \leq p \leq 1. Then, we observe X_1, \ldots, X_ n with corresponding conditional likelihood function L(X_1, \ldots , X_ n | \theta ) that is positive for both \theta =0 and \theta =1. Which of the following statements about the posterior distribution \pi (\theta |X_1, \ldots , X_ n) is true? (Choose all that apply.)
If p=0, then the posterior distribution will be identical to the prior distribution.
If p=1, then the posterior distribution will be identical to the prior distribution.
If p < \frac{1}{2}, then in the posterior distribution, \pi (\theta =0|X_1, \ldots , X_ n) \leq \pi (\theta =1|X_1, \ldots , X_ n) will necessarily be true.
If p > \frac{1}{2}, then in the posterior distribution, \pi (\theta =0|X_1, \ldots , X_ n) \geq \pi (\theta =1|X_1, \ldots , X_ n) will necessarily be true.
The posterior distribution $\pi(\theta|X_1,\ldots,X_n)$ is proportional to the product of the prior distribution $\pi(\theta)$ and the likelihood function $L(X_1,\ldots,X_n|\theta)$. Let's consider each statement separately:
1) If $p=0$, then the prior distribution is $\pi(0)=0$ and $\pi(1)=1$. In this case, the likelihood function $L(X_1,\ldots,X_n|\theta)$ will not change the posterior, as any non-zero probability multiplied by 0 will still be 0. Therefore, the posterior distribution will be identical to the prior distribution.
2) If $p=1$, then the prior distribution is $\pi(0)=1$ and $\pi(1)=0$. Again, the likelihood function will not change the posterior, as any non-zero probability multiplied by 0 will still be 0. So the posterior distribution will be identical to the prior distribution.
3) If $p<\frac{1}{2}$, then $\pi(\theta=0|X_1,\ldots,X_n) \leq \pi(\theta=1|X_1,\ldots,X_n)$ is not necessarily true. The posterior distribution depends on both the prior and the likelihood function, so it is possible for the likelihood to favor $\theta=0$ over $\theta=1$ even if $p<\frac{1}{2}$.
4) If $p>\frac{1}{2}$, then $\pi(\theta=0|X_1,\ldots,X_n) \geq \pi(\theta=1|X_1,\ldots,X_n)$ is not necessarily true. The posterior distribution depends on both the prior and the likelihood function, so it is possible for the likelihood to favor $\theta=1$ over $\theta=0$ even if $p>\frac{1}{2}$.
Therefore, the correct statements are:
- If $p=0$, then the posterior distribution will be identical to the prior distribution.
- If $p=1$, then the posterior distribution will be identical to the prior distribution.
To understand which statements about the posterior distribution are true, we need to use Bayes' theorem. Bayes' theorem states that the posterior distribution is proportional to the product of the prior distribution and the likelihood function. In this case, we have a binary parameter, so our posterior distribution will have two components, one for each possible value of \theta.
Let's break down the statements one by one:
1. If p=0, then the prior distribution is \pi (\theta =0) = 0 and \pi (\theta =1) = 1. In this case, the prior distribution assigns all probability mass to \theta = 1. When we apply Bayes' theorem, the likelihood function will determine the posterior distribution, so the statement is false. The posterior distribution will not be identical to the prior distribution.
2. If p=1, then the prior distribution is \pi (\theta =0) = 1 and \pi (\theta =1) = 0. Similar to the previous case, the prior distribution assigns all probability mass to \theta = 0. When we apply Bayes' theorem, the likelihood function will determine the posterior distribution, so the statement is false. The posterior distribution will not be identical to the prior distribution.
3. If p < \frac{1}{2}, then the prior distribution assigns a higher probability to \theta = 1 compared to \theta = 0. When we apply Bayes' theorem, the likelihood function will contribute more to the posterior distribution, especially if it strongly supports \theta = 1. Therefore, it is possible that \pi (\theta =0|X_1, \ldots , X_ n) \leq \pi (\theta =1|X_1, \ldots , X_ n), but it is not necessarily true. The statement is not always true.
4. If p > \frac{1}{2}, then the prior distribution assigns a higher probability to \theta = 0 compared to \theta = 1. When we apply Bayes' theorem, the likelihood function will contribute more to the posterior distribution, especially if it strongly supports \theta = 0. Therefore, it is possible that \pi (\theta =0|X_1, \ldots , X_ n) \geq \pi (\theta =1|X_1, \ldots , X_ n), but it is not necessarily true. The statement is not always true.
To summarize, statements 1 and 2 are false, while statements 3 and 4 are not necessarily true.