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
- 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
- Suppose $X|\theta \sim \text{Binomial}(7,\theta)$, $0 < \theta < 1$, and the prior distribution of $\theta$ is $\text{Beta}(\alpha, \beta)$ where $\alpha > 0$ and $\beta > 0$ are known. Then which of the following statements MAY NOT be true?
Let $X$ be a random sample from an exponential distribution with mean $1/\lambda$. If $\lambda$ has a prior distribution with probability density function
$g(\lambda) = \begin{cases} \lambda e^{-\lambda} & ; \quad \lambda > 0 \\ 0 & ; \quad \lambda \leq 0 \end{cases}$
then the Bayes estimator of $1/\lambda$ with respect to the squared error loss function is
Let $X_1, X_2, \dots, X_7$ be a random sample from $N(\mu, \sigma^2)$ where $\mu$ and $\sigma^2$ are unknown. Consider the problem of testing $H_0: \mu = 2$ against $H_1: \mu > 2$. Suppose the observed values of $x_1, x_2, \dots, x_7$ are $1.2, 1.3, 1.7, 1.8, 2.1, 2.3, 2.7$. If we use the Uniformly Most Powerful test, which of the following is true?
Suppose $X_i \mid \theta_i \sim N(\theta_i, \sigma^2), i = 1, 2$ are independently distributed. Under the prior distribution, $\theta_1$ and $\theta_2$ are i.i.d $N(\mu, \tau^2)$, where $\sigma^2, \mu$ and $\tau^2$ are known. Then which of the following is true about the marginal distributions of $X_1$ and $X_2$?
