$Y = 1$ if all the three tosses result in heads
$Y = 2$ if all the three tosses result in tails
$Y = 3$ otherwise
If the prior density of $\theta$ is Beta $(\alpha, \beta)$, and $\hat{\theta}_i$ is the posterior mean of $\theta$ given $Y = i$, for $i = 1, 2$, then
Posterior Means Comparison ($\hat{\theta}_1$ vs $\hat{\theta}_2$)
The prior distribution for the probability of heads, $\theta$, is Beta($\alpha$, $\beta$). The posterior mean of a Beta distribution is given by the ratio of its shape parameters: $\frac{\text{shape}_1}{\text{shape}_1 + \text{shape}_2}$.
Case Y = 1 (All Heads - HHH):
- The outcome is three heads ($n_H=3, n_T=0$).
- Given the Beta($\alpha$, $\beta$) prior, the posterior distribution becomes Beta($\alpha+3$, $\beta$).
- The posterior mean is calculated as $\hat{\theta}_1 = \frac{\alpha+3}{(\alpha+3)+\beta} = \frac{\alpha+3}{\alpha+\beta+3}$.
Case Y = 2 (All Tails - TTT):
- The outcome is three tails ($n_H=0, n_T=3$).
- The posterior distribution becomes Beta($\alpha$, $\beta+3$).
- The posterior mean is calculated as $\hat{\theta}_2 = \frac{\alpha}{\alpha+(\beta+3)} = \frac{\alpha}{\alpha+\beta+3}$.
Comparison:
- To compare $\hat{\theta}_1$ and $\hat{\theta}_2$, we look at their numerators since the denominators ($\alpha+\beta+3$) are identical and positive (assuming $\alpha, \beta > 0$).
- The numerator of $\hat{\theta}_1$ is $\alpha+3$, and the numerator of $\hat{\theta}_2$ is $\alpha$.
- Since $\alpha+3$ is always greater than $\alpha$, it follows that $\hat{\theta}_1 > \hat{\theta}_2$.
- Therefore, statement A ($\hat{\theta}_1 > \hat{\theta}_2$) is correct.
Posterior Density Analysis for Y=3
Case Y = 3 (Mixed Outcomes):
- This event occurs if the sequence of three tosses is neither all heads (HHH) nor all tails (TTT).
- The probability of event Y=3, given $\theta$, is $P(Y=3|\theta) = 1 - P(\text{HHH}|\theta) - P(\text{TTT}|\theta)$.
- Substituting the probabilities: $P(Y=3|\theta) = 1 - \theta^3 - (1-\theta)^3$.
- Expanding $(1-\theta)^3 = 1 - 3\theta + 3\theta^2 - \theta^3$.
- Simplifying $P(Y=3|\theta)$: $1 - \theta^3 - (1 - 3\theta + 3\theta^2 - \theta^3) = 1 - \theta^3 - 1 + 3\theta - 3\theta^2 + \theta^3 = 3\theta - 3\theta^2 = 3\theta(1-\theta)$.
Posterior Form Derivation:
- The posterior density function is proportional to the product of the likelihood and the prior probability density function: $f(\theta|Y=3) \propto P(Y=3|\theta) \times f(\theta)$.
- The prior PDF is $f(\theta) \propto \theta^{\alpha-1}(1-\theta)^{\beta-1}$ (for a Beta($\alpha$, $\beta$) distribution).
- So, $f(\theta|Y=3) \propto [3\theta(1-\theta)] \times [\theta^{\alpha-1}(1-\theta)^{\beta-1}]$.
- Combining the terms involving $\theta$ and $(1-\theta)$: $f(\theta|Y=3) \propto 3 \cdot \theta^{1+\alpha-1} (1-\theta)^{1+\beta-1}$.
- This simplifies to $f(\theta|Y=3) \propto \theta^{\alpha} (1-\theta)^{\beta}$.
- Rewriting in the standard Beta form: $f(\theta|Y=3) \propto \theta^{(\alpha+1)-1} (1-\theta)^{(\beta+1)-1}$.
Conclusion on Posterior Density:
- The derived form $\theta^{(\alpha+1)-1} (1-\theta)^{(\beta+1)-1}$ is the kernel of a Beta distribution with parameters $\alpha+1$ and $\beta+1$.
- Therefore, the posterior density of $\theta$ given $Y=3$ is indeed a Beta density. Statement C is correct.
Final Answer Determination
Based on the detailed analysis, both statement A and statement C are mathematically 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 $X_2$?
