Discussion
Suppose given \(\theta\), the random variable X follows a normal distribution with mean \(\theta\) and variance 1. Let the prior density of \(\theta\) be Bernoulli with parameter 1/2. Then the posterior distribution of \(\theta\) is :
Answer: C
The posterior is \(\dfrac{e^{x\theta - \theta^{2}/2}}{1 + e^{x - 1/2}}\), for \(\theta \in \{0, 1\}\).
The prior puts mass 1/2 on each of \(\theta = 0\) and \(\theta = 1\), so the posterior is proportional to the likelihood:
\[ \begin{aligned} \pi(\theta \mid x) &\propto e^{-(x - \theta)^{2}/2} \propto e^{x\theta - \theta^{2}/2} \\ \text{at } \theta = 0 &: 1,\quad \text{at } \theta = 1 : e^{x - 1/2} \end{aligned} \]
after dropping the factor \(e^{-x^{2}/2}\), which does not depend on \(\theta\). Normalising by the sum \(1 + e^{x - 1/2}\) gives the third option.
The first two options are unnormalised likelihoods that do not sum to 1 over \(\theta\). The last keeps an \(e^{x^{2}}\) factor in the numerator that should have cancelled, so its two values do not add to 1.
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