The problem involves Bayesian inference for a Poisson distribution with a Gamma prior.
Likelihood: $ Y | \theta \sim \text{Poisson}(\theta) $, where $ P(Y=y|\theta) = \frac{e^{-\theta} \theta^y}{y!} $ for $ y = 0, 1, 2, \dots $.
Prior: $ \tau(\theta) \propto e^{-\alpha \theta} \theta^{\beta - 1} $. This is the kernel of a Gamma distribution, specifically $ \text{Gamma}(\text{shape}=\beta, \text{rate}=\alpha) $.
1. Marginal Distribution Analysis
The marginal distribution of $Y$ is found by integrating the joint distribution $ P(Y=y, \theta) = P(Y=y|\theta) \tau(\theta) $ over $ \theta $.
The prior density is $ \tau(\theta) = \frac{\alpha^\beta}{\Gamma(\beta)} e^{-\alpha \theta} \theta^{\beta - 1} $.
The joint density is $ P(Y=y, \theta) \propto (e^{-\theta} \theta^y) (e^{-\alpha \theta} \theta^{\beta - 1}) = e^{-(\alpha+1)\theta} \theta^{y+\beta-1} $.
The marginal probability is $ P(Y=y) = \int_0^\infty P(Y=y|\theta) \tau(\theta) d\theta $.
$ P(Y=y) = \int_0^\infty \frac{e^{-\theta} \theta^y}{y!} \frac{\alpha^\beta}{\Gamma(\beta)} e^{-\alpha \theta} \theta^{\beta - 1} d\theta = \frac{\alpha^\beta}{y! \Gamma(\beta)} \int_0^\infty e^{-(\alpha+1)\theta} \theta^{y+\beta-1} d\theta $.
The integral evaluates to $ \frac{\Gamma(y+\beta)}{(\alpha+1)^{y+\beta}} $.
Therefore, $ P(Y=y) = \frac{\alpha^\beta \Gamma(y+\beta)}{y! \Gamma(\beta) (\alpha+1)^{y+\beta}} $. This is the probability mass function of a Negative Binomial distribution, not Hypergeometric.
Conclusion: Statement 1 is false.
2. Posterior Distribution Identification
The posterior distribution is proportional to the likelihood times the prior:
$ \tau(\theta|y) \propto P(Y=y|\theta) \tau(\theta) $
$ \tau(\theta|y) \propto (e^{-\theta} \theta^y) \times (e^{-\alpha \theta} \theta^{\beta - 1}) $
$ \tau(\theta|y) \propto e^{-(\alpha+1)\theta} \theta^{y+\beta-1} $
This form $ e^{-\text{rate} \cdot \theta} \theta^{\text{shape}-1} $ corresponds to the kernel of a Gamma distribution.
The posterior distribution is $ \text{Gamma}(\text{shape} = y + \beta, \text{rate} = \alpha + 1) $.
Conclusion: Statement 2 is true.
3. Conjugate Prior Check
A prior is conjugate if the posterior distribution belongs to the same family as the prior distribution.
- Prior distribution: $ \text{Gamma}(\text{shape}=\beta, \text{rate}=\alpha) $.
- Posterior distribution: $ \text{Gamma}(\text{shape}=y+\beta, \text{rate}=\alpha+1) $.
Since both the prior and the posterior are Gamma distributions, the Gamma prior is conjugate for the Poisson likelihood.
Conclusion: Statement 3 is true.
4. Bayes' Estimate Verification
For a squared error loss function, the Bayes' estimate of $ \theta $ is the posterior mean $ E[\theta|Y=y] $.
The posterior distribution is $ \text{Gamma}(\text{shape}=y+\beta, \text{rate}=\alpha+1) $.
The mean of a Gamma distribution with shape $ k $ and rate $ \lambda $ is $ k/\lambda $.
Therefore, the posterior mean is $ E[\theta|Y=y] = \frac{y+\beta}{\alpha+1} $.
This matches the estimate provided in the option.
Conclusion: Statement 4 is true.
Final Summary
Based on the analysis, statements 2, 3, and 4 are true.
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
