Question
$\theta$ is the probability of obtaining a head in the toss of a coin. The coin is tossed three times and we record
$Y = 1$ 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.
Was this answer helpful?
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
Need Expert Advice?
