Suppose $X$ has density $f(x|\theta)$ where $\theta$ is $0$ or $1$. Also,
$f(x|0) = 1$ if $0 < x < 1$, and $0$ otherwise,
$f(x|1) = \frac{1}{2\sqrt{x}}$ if $0 < x < 1$ and $0$ otherwise.
To test $H_0: \theta = 0$ versus $H_1: \theta = 1$ at level $\alpha, 0 < \alpha < 1$, the Most Powerful test
Most Powerful Test Derivation
We need to find the Most Powerful (MP) test for $H_0: \theta = 0$ versus $H_1: \theta = 1$. The MP test is determined using the Neyman-Pearson lemma, which involves the likelihood ratio.
Likelihood Ratio Analysis
The densities under the two hypotheses are:
- \(f(x|0) = 1\) for \(0 < x < 1\)
- \(f(x|1) = \frac{1}{2\sqrt{x}}\) for \(0 < x < 1\)
The likelihood ratio $\text{LR}(x)$ is calculated as:
$ \text{LR}(x) = \frac{f(x|1)}{f(x|0)} = \frac{1 / (2\sqrt{x})}{1} = \frac{1}{2\sqrt{x}} \quad \text{for } 0 < x < 1 $
The Neyman-Pearson lemma states that the MP test rejects $H_0$ if $\text{LR}(x) > k$ for some constant $k$.
Rejection Region Calculation
Setting up the inequality for rejection:
$ \frac{1}{2\sqrt{x}} > k $
Solving for $x$:
$ 1 > 2k\sqrt{x} \implies \sqrt{x} < \frac{1}{2k} \implies x < \left(\frac{1}{2k}\right)^2 $
Let the critical value be $c = \left(\frac{1}{2k}\right)^2$. The rejection region is $X < c$. To find $c$ at significance level $\alpha$, we use $P(\text{Reject } H_0 | H_0 \text{ is true}) = \alpha$. Under $H_0$, $X \sim \text{Uniform}(0, 1)$, so its CDF is $F(x|0) = x$ for \(0 < x < 1\). $ P(X < c | \theta = 0) = F(c|0) = c $ Setting $c = \alpha$, the rejection region is $X < \alpha$. This corresponds to option B.
Power of the Test Calculation
The power is the probability of rejecting $H_0$ when $H_1$ is true: Power = \( P(\text{Reject } H_0 | \theta = 1) \).
Using the rejection region $X < \alpha$:
$ \text{Power} = P(X < \alpha | \theta = 1) = \int_{0}^{\alpha} f(x|1) dx $
$ \text{Power} = \int_{0}^{\alpha} \frac{1}{2\sqrt{x}} dx = \frac{1}{2} \int_{0}^{\alpha} x^{-1/2} dx $
$ \text{Power} = \frac{1}{2} \left[ \frac{x^{1/2}}{1/2} \right]_{0}^{\alpha} = [\sqrt{x}]_{0}^{\alpha} = \sqrt{\alpha} - 0 = \sqrt{\alpha} $
The power of this MP test is $\sqrt{\alpha}$. This corresponds to option D.
Summary of MP Test Characteristics
The Most Powerful test is characterized by its rejection region $X < \alpha$ (Option B) and its power $\sqrt{\alpha}$ (Option D).
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 $X_1, X_2, \dots, X_7$ 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
Consider the pdf
$f(x;\theta,\sigma) = \frac{0.9}{\sigma}\phi\left(\frac{x-\theta}{\sigma}\right) + 0.1\ \phi(x-\theta)$, where $-\infty < \theta < \infty$ and $\sigma > 0$ are unknown parameters and $\phi$ denotes the pdf of $N(0,1)$. Let $X_1, X_2, \cdots, X_n$ be a random sample from this probability distribution. Then which of the following is (are) correct?
