Let X be a random sample from a Poisson distribution with parameter $\lambda$. The parameter $\lambda$ has a prior distribution $f(z)$; where $f(z) = \begin{cases} e^{-z}; & z > 0 \\ 0, & \text{otherwise}. \end{cases}$ Under the squared error loss function, which of the following statements are correct?
Problem Analysis
The problem involves a random sample \(X\) drawn from a Poisson distribution with parameter \(\lambda\). The parameter \(\lambda\) has a prior distribution, and we need to find the correct statements regarding the posterior distribution and Bayes' estimators under squared error loss.
- Likelihood: \( P(X|\lambda) \propto \frac{\lambda^X e^{-\lambda}}{X!} \propto \lambda^X e^{-\lambda} \) (since \(X!\) is constant with respect to \(\lambda\)).
- Prior: \( f(\lambda) = e^{-\lambda} \) for \( \lambda > 0 \).
- Loss Function: Squared error loss.
Posterior Distribution Derivation
The posterior density function is proportional to the product of the likelihood and the prior:
\( f(\lambda|X) \propto P(X|\lambda) f(\lambda) \)
\( f(\lambda|X) \propto (\lambda^X e^{-\lambda}) (e^{-\lambda}) \)
\( f(\lambda|X) \propto \lambda^X e^{-2\lambda} \)
This is the kernel of a Gamma distribution. A standard Gamma distribution with shape \(\alpha\) and rate \(\beta\) has the PDF kernel \( \lambda^{\alpha-1} e^{-\beta\lambda} \). Comparing this with our posterior kernel:
- \( \alpha - 1 = X \implies \alpha = X+1 \)
- \( \beta = 2 \)
Therefore, the posterior distribution of \(\lambda\) given \(X\) is a Gamma distribution: \(\lambda|X \sim \text{Gamma}(X+1, 2)\). This confirms statement C.
Posterior Mean Calculation
For a Gamma distribution \(\text{Gamma}(\alpha, \beta)\), the mean is given by \( E[\lambda] = \frac{\alpha}{\beta} \).
Using the parameters derived for the posterior distribution (\(\alpha = X+1, \beta = 2\)):
\( E[\lambda|X] = \frac{X+1}{2} \)
The posterior mean of \(\lambda\) is \( \frac{X+1}{2} \). This confirms statement B.
Bayes' Estimator for \( e^\lambda \)
Under the squared error loss function, the Bayes' estimator for a function \( g(\lambda) \) is the posterior expectation \( E[g(\lambda)|X] \). We need the estimator for \( g(\lambda) = e^\lambda \).
We need to compute \( E[e^\lambda|X] \) where \(\lambda \sim \text{Gamma}(X+1, 2)\).
The moment generating function (MGF) of a Gamma\((\alpha, \beta)\) distribution is \( M_Y(t) = E[e^{tY}] = \left(\frac{\beta}{\beta-t}\right)^\alpha \).
For our posterior distribution \(\lambda \sim \text{Gamma}(X+1, 2)\), the MGF is:
\( E[e^{t\lambda}|X] = \left(\frac{2}{2-t}\right)^{X+1} \)
To find the Bayes' estimator of \( e^\lambda \), we set \( t=1 \):
\( E[e^{\lambda}|X] = \left(\frac{2}{2-1}\right)^{X+1} = \left(\frac{2}{1}\right)^{X+1} = 2^{X+1} \)
The Bayes' estimator of \( e^\lambda \) is \( 2^{X+1} \). This confirms statement A.
Conclusion on Correct Statements
Based on the derivations:
- Statement A is correct.
- Statement B is correct.
- Statement C is correct.
Statements A, B, and C are correct.
Important Questions from Elementary Bayesian Inference
- Suppose the distribution of $X$ given $\theta$ is normal with mean $\theta$ and variance $15$. Further, let the prior (improper) distribution of $\theta$ be proportional to $1, \ -\infty<\theta<\infty$. If the observed value of $X$ is $13$, then which of the following statements is true?
- Let $X_1, X_2, . . ., X_n$ be a random sample from $N(\theta, 1)$, $\theta \in R$. If $\hat{\theta}$ is the Bayes estimator of $\theta$ with respect to some prior $\pi(\theta)$ and loss function $L(\theta, d)$. Then, which of the following statements are true?
- Let $X|\theta \sim \text{Uniform}[0, \theta]$ and $\theta$ has an Exponential distribution with mean $\lambda$, where $\lambda > 3$ is known. If the realized value of $X$ is $2025$, then the posterior mode equals
- $X_1, X_2, \cdots, X_n$ are independent and identically distributed $N(\theta, 1)$ random variables, where $\theta$ takes only integer values i.e.
$\theta \in \{\cdots, -2, -1, 0, 1, 2, \cdots\}$.
Which of the following is the maximum likelihood estimator of $\theta$? - Suppose the probability mass function of a random variable X under the parameter $\theta = \theta_0$ and $\theta = \theta_1 (\ne \theta_0)$ are given by
x 0 1 2 3 $p_{\theta_0}(x)$ 0.01 0.04 0.5 0.45 $p_{\theta_1}(x)$ 0.02 0.08 0.4 0.5
Define a test $\phi$ such that $\phi(x) = 1$ if $x = 0, 1$, and $0$ if $x = 2, 3$.
For testing $H_0: \theta = \theta_0$ against $H_1: \theta = \theta_1$, the test $\phi$ is
