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
Random Variable Hypothesis Test Analysis
We analyze the given test $\phi$ for the hypotheses $H_0: \theta = \theta_0$ versus $H_1: \theta = \theta_1$. The test function is defined as $\phi(x) = 1$ if $x \in \{0, 1\}$ and $\phi(x) = 0$ if $x \in \{2, 3\}$.
Probability Distributions Table
| x | $p_{\theta_0}(x)$ | $p_{\theta_1}(x)$ |
|---|---|---|
| 0 | 0.01 | 0.02 |
| 1 | 0.04 | 0.08 |
| 2 | 0.50 | 0.40 |
| 3 | 0.45 | 0.50 |
Test Size Calculation (Option D)
The size of a test (or significance level $\alpha$) is the maximum probability of rejecting the null hypothesis ($H_0$) when it is true. Here, $\Theta_0 = \{\theta_0\}$.
$ \alpha = \sup_{\theta \in \Theta_0} E_\theta[\phi(X)] = E_{\theta_0}[\phi(X)] $
The expectation is calculated as:
$ E_{\theta_0}[\phi(X)] = \sum_{x=0}^{3} \phi(x) p_{\theta_0}(x) $
$ E_{\theta_0}[\phi(X)] = \phi(0) p_{\theta_0}(0) + \phi(1) p_{\theta_0}(1) + \phi(2) p_{\theta_0}(2) + \phi(3) p_{\theta_0}(3) $
Substituting the values:
$ \alpha = (1)(0.01) + (1)(0.04) + (0)(0.50) + (0)(0.45) = 0.01 + 0.04 = 0.05 $
The calculated size is exactly 0.05. Thus, the test $\phi$ is a test of size 0.05. Option D is correct.
Likelihood Ratio Calculation
To evaluate options A and B, we compute the likelihood ratio (LR) $\frac{p_{\theta_1}(x)}{p_{\theta_0}(x)}$ for each $x$:
- For $x=0$: $LR = \frac{0.02}{0.01} = 2$
- For $x=1$: $LR = \frac{0.08}{0.04} = 2$
- For $x=2$: $LR = \frac{0.40}{0.50} = 0.8$
- For $x=3$: $LR = \frac{0.50}{0.45} = \frac{10}{9} \approx 1.11$
The test $\phi$ rejects $H_0$ (sets $\phi(x)=1$) for $x=0, 1$. These are the values of $x$ with the highest likelihood ratios.
Most Powerful Test Analysis (Option A)
The Neyman-Pearson Lemma states that the most powerful test at a given level $\alpha$ rejects $H_0$ if the likelihood ratio exceeds a threshold $k$, chosen such that the size constraint is met. Our test rejects for $x=0, 1$ (where LR = 2). We found the size is exactly 0.05 for this rejection region $\{0, 1\}$. Since these $x$ values yield the highest LR values, this test is indeed the most powerful test for level $\alpha = 0.05$. Option A is correct.
Likelihood Ratio Test Analysis (Option B)
A likelihood ratio test is a test that rejects the null hypothesis when the likelihood ratio $\frac{p_{\theta_1}(x)}{p_{\theta_0}(x)}$ is greater than some constant $k$. Since the test $\phi$ rejects $H_0$ precisely for the values $x=0, 1$ which correspond to the largest likelihood ratios, it fits the definition of a likelihood ratio test. As it achieves the level of 0.05, it is a likelihood ratio test at level 0.05. Option B is correct.
Unbiased Test Analysis (Option C)
A test is unbiased if its power (the probability of rejecting $H_0$ when $H_1$ is true) is greater than its size (the probability of rejecting $H_0$ when $H_0$ is true), i.e., $E_{\theta_1}[\phi(X)] > E_{\theta_0}[\phi(X)]$.
We already know the size: $E_{\theta_0}[\phi(X)] = 0.05$.
Now, we calculate the power:
$ \text{Power} = E_{\theta_1}[\phi(X)] = \sum_{x=0}^{3} \phi(x) p_{\theta_1}(x) $
$ \text{Power} = \phi(0) p_{\theta_1}(0) + \phi(1) p_{\theta_1}(1) + \phi(2) p_{\theta_1}(2) + \phi(3) p_{\theta_1}(3) $
$ \text{Power} = (1)(0.02) + (1)(0.08) + (0)(0.40) + (0)(0.50) = 0.02 + 0.08 = 0.10 $
Since Power (0.10) > Size (0.05), the test $\phi$ is unbiased. Option C is correct.
Summary
The test $\phi$ is confirmed to be:
- A test of size 0.05 (Option D).
- The most powerful test at level 0.05 (Option A).
- A likelihood ratio test at level 0.05 (Option B).
- An unbiased test (Option C).
Therefore, all options A, B, C, and D correctly describe the test $\phi$.
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 $X_2$?
