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
- 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 $X_2$?
