1. Identify Distributional Properties:
The prior distribution is $\theta \sim \text{Beta}(\alpha, \beta)$. Its mean is given by:
Prior Mean ($E[\theta]$) = $\frac{\alpha}{\alpha + \beta}$
The likelihood is $X|\theta \sim \text{Binomial}(n, \theta)$ with $n = 7$. Since Beta is the conjugate prior for the Binomial distribution, the posterior distribution after observing $X = x$ is:
$\theta|X = x \sim \text{Beta}(\alpha + x, \beta + 7 - x)$
The posterior mean is: $E[\theta|X = x] = \frac{\alpha + x}{\alpha + \beta + 7}$
2. Comparing Prior and Posterior Means:
Let $P$ be the prior mean and $M$ be the posterior mean. The posterior mean can be written as a weighted average of the prior mean and the sample mean ($x/7$):
$M = \left( \frac{\alpha + \beta}{\alpha + \beta + 7} \right) \cdot P + \left( \frac{7}{\alpha + \beta + 7} \right) \cdot \frac{x}{7}$
From this relation, we can conclude:
- If $x/7 > P$, then $M > P$ (Posterior mean increases).
- If $x/7 < P$, then $M < P$ (Posterior mean decreases).
- If $x/7 = P$, then $M = P$ (Posterior mean remains unchanged).
3. Evaluating the Options (Which "MAY NOT" be true):
The truth of each statement depends entirely on the unknown values of $\alpha$ and $\beta$.
- Option 1: If the prior mean is already very small (e.g., $0.1$), and we observe $X=2$ (sample mean $\approx 0.28$), then $M > P$. Thus, Option 1 is False in some cases.
- Option 2: If the prior mean is very large (e.g., $0.9$), and we observe $X=3$ (sample mean $\approx 0.43$), then $M < P$. Thus, Option 2 is False in some cases.
- Option 3: If the prior parameters are chosen such that $\alpha/(\alpha+\beta) = 4/7$, then $M = P$. Thus, the statement "not equal" is False in some cases.
- Option 4: Unless the prior mean happens to be exactly $5/7$, $M$ will not equal $P$. Thus, Option 4 is False in most cases.
Final Answer:
All four provided statements are dependent on the specific values of $\alpha$ and $\beta$. Therefore, Option 1, Option 2, Option 3, and Option 4 all represent statements that MAY NOT be 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
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 $X_2$?
Let $(X,Y)$ have the joint discrete distribution such that $X \mid Y=y \sim \text{Binomial} \quad (y, 0.5)$ and $Y \sim \text{Poisson}(\lambda), \lambda > 0$, where $\lambda$ is an unknown parameter. Let $T = T(X,Y)$ be any unbiased estimator of $\lambda$. Then
