Understanding the Problem Parameters

This problem involves Bayesian inference, where we update our beliefs about a parameter ($\theta$) based on observed data ($X$). Let's break down the given information:

• Distribution of X given $\theta$: The variable $X$ follows a normal distribution with mean $\theta$ and variance $15$. Mathematically, this is written as $X | \theta \sim N(\theta, 15)$. This tells us that if we knew $\theta$, the distribution of $X$ would be centered at $\theta$ and have a spread (variance) of $15$.
• Prior Distribution of $\theta$: Before observing any data, our initial belief about $\theta$ is represented by its prior distribution. Here, it's an improper prior, meaning it doesn't integrate to $1$. It's stated as being proportional to $1$ for all possible values of $\theta$ ($-\infty < \theta < \infty$). This is often called a flat prior, indicating no strong initial belief about $\theta$'s value. We write this as $\pi(\theta) \propto 1$.
• Observed Data: The value of $X$ that was actually observed is $13$. So, $x = 13$.

Calculating the Maximum Likelihood Estimate (MLE)

The Maximum Likelihood Estimate (MLE) is the value of the parameter $\theta$ that makes the observed data most probable. We find it by maximizing the likelihood function.

• The likelihood function, $L(\theta | x)$, is the probability density function of $X$ given $\theta$, evaluated at the observed $x$: $ L(\theta | x) = \frac{1}{\sqrt{2\pi \cdot 15}} \exp\left(-\frac{(x - \theta)^2}{2 \cdot 15}\right) $
• To find the MLE, we typically maximize the logarithm of the likelihood function, or in this simple case, we can minimize the term $(x - \theta)^2$ within the exponent. This term is smallest when $\theta$ is exactly equal to $x$.
• Given the observed value $x = 13$, the MLE of $\theta$ is $\hat{\theta}_{MLE} = 13$.

Deriving the Posterior Distribution

Using Bayes' theorem, we combine the likelihood of the data with the prior belief to get the posterior distribution, which represents our updated belief about $\theta$ after seeing the data.

• The relationship is: Posterior $\propto$ Likelihood $\times$ Prior. $ p(\theta | x) \propto L(x | \theta) \pi(\theta) $
• Substituting the known functions: $ p(\theta | x) \propto \left( \frac{1}{\sqrt{2\pi \cdot 15}} \exp\left(-\frac{(x - \theta)^2}{2 \cdot 15}\right) \right) \times 1 $
• Since the constant term $\frac{1}{\sqrt{2\pi \cdot 15}}$ doesn't affect the shape of the distribution with respect to $\theta$, we focus on the exponential part: $ p(\theta | x) \propto \exp\left(-\frac{(\theta - x)^2}{2 \cdot 15}\right) $
• This form is the kernel of a normal distribution. The standard normal distribution's kernel is proportional to $\exp\left(-\frac{(\theta - \mu)^2}{2\sigma^2}\right)$. By comparing these, we can identify the parameters of our posterior distribution:
  • Posterior mean ($\mu$) = $x$
  • Posterior variance ($\sigma^2$) = $15$
• Since the observed value is $x = 13$, the posterior distribution is: $ \theta | x \sim N(13, 15) $

Determining Posterior Mean and Variance

From the posterior distribution $\theta | x \sim N(13, 15)$, we can directly identify its mean and variance.

• Posterior mean = $13$.
• Posterior variance = $15$.

Comparing Results and Verifying the Statement

Now, let's compare our calculated values with the elements mentioned in the question's options.

• Comparison 1: Posterior Mean vs. MLE
  • Posterior mean = $13$.
  • Maximum likelihood estimate of $\theta$ ($\hat{\theta}_{MLE}$) = $13$.
  • Conclusion: Posterior mean = Maximum likelihood estimate of $\theta$. This part of the statement is true.
• Comparison 2: Posterior Variance vs. Var($X|\theta$)
  • Posterior variance = $15$.
  • The problem states that the variance of $X$ given $\theta$, Var($X|\theta$), is $15$.
  • Conclusion: Posterior variance = Var($X|\theta$). This part of the statement is also true.

Since both conditions stated in the first option are true (Posterior mean equals the MLE and Posterior variance equals Var($X|\theta$)), the statement is verified.