Poisson Mean Hypothesis Test Overview

The problem involves hypothesis testing for the mean ($\theta$) of a Poisson distribution using a random sample $X, Y$. The hypotheses are $H_0 : \theta = 1$ versus $H_1 : \theta = \frac{1}{2}$. The test rejects $H_0$ if $X = 0$ or if ($X = 1$ and $X + Y \leq 2$).

Type I Error Calculation for Poisson Test

A Type I error occurs when $H_0$ is rejected, but $H_0$ is true. We calculate this probability under $H_0$, where $\theta = 1$. The rejection region is $R = \{X=0\} \cup \{X=1 \text{ and } X+Y \leq 2\}$.

• Probability of $X=0$:

  For $X \sim \text{Poisson}(1)$, the probability mass function is $P(X=k | \theta) = \frac{e^{-\theta} \theta^k}{k!}$. Under $H_0$, $\theta=1$.

  $ P(X=0 | \theta=1) = \frac{e^{-1} (1)^0}{0!} = e^{-1} $

• Probability of $X=1$ and $X+Y \leq 2$:

  If $X=1$, the condition $X+Y \leq 2$ becomes $1+Y \leq 2$, which means $Y \leq 1$ (i.e., $Y=0$ or $Y=1$). Since $X$ and $Y$ are independent Poisson(1) variables:

  $ P(X=1, Y=0 | \theta=1) = P(X=1 | \theta=1) P(Y=0 | \theta=1) = \left(\frac{e^{-1} 1^1}{1!}\right) \left(\frac{e^{-1} 1^0}{0!}\right) = e^{-1} \cdot e^{-1} = e^{-2} $

  $ P(X=1, Y=1 | \theta=1) = P(X=1 | \theta=1) P(Y=1 | \theta=1) = \left(\frac{e^{-1} 1^1}{1!}\right) \left(\frac{e^{-1} 1^1}{1!}\right) = e^{-1} \cdot e^{-1} = e^{-2} $

  The probability for this case ($X=1$ and $Y \leq 1$) is the sum: $e^{-2} + e^{-2} = 2e^{-2}$.

The events ($X=0$) and ($X=1$) are mutually exclusive. Therefore, the total probability of Type I error is the sum of the probabilities calculated above:

$ P(\text{Type I error}) = P(X=0 | \theta=1) + P(X=1, Y \leq 1 | \theta=1) = e^{-1} + 2e^{-2} $

This result matches Option A. The size of the test is defined as the probability of Type I error, which is $e^{-1} + 2e^{-2}$. Option 3 states the size is $e^{-1} + e^{-2}$, which is incorrect.

Test Power and Type II Error Calculation

The power of the test is the probability of rejecting $H_0$ when $H_1$ is true. We calculate the probability of the rejection region $R$ under $H_1$, where $\theta = \frac{1}{2}$. Assume $X, Y \sim \text{Poisson}(\frac{1}{2})$.

• Probability of $X=0$:

  $ P(X=0 | \theta=\frac{1}{2}) = \frac{e^{-1/2} (\frac{1}{2})^0}{0!} = e^{-1/2} $

• Probability of $X=1$ and $X+Y \leq 2$:

  This implies $X=1$ and $Y \leq 1$. We need the probabilities under $\theta = \frac{1}{2}$:

  $ P(X=1 | \theta=\frac{1}{2}) = \frac{e^{-1/2} (\frac{1}{2})^1}{1!} = \frac{1}{2}e^{-1/2} $

  $ P(Y=0 | \theta=\frac{1}{2}) = \frac{e^{-1/2} (\frac{1}{2})^0}{0!} = e^{-1/2} $

  $ P(Y=1 | \theta=\frac{1}{2}) = \frac{e^{-1/2} (\frac{1}{2})^1}{1!} = \frac{1}{2}e^{-1/2} $

  Using independence:

  $ P(X=1, Y=0 | \theta=\frac{1}{2}) = P(X=1 | \theta=\frac{1}{2}) P(Y=0 | \theta=\frac{1}{2}) = (\frac{1}{2}e^{-1/2}) (e^{-1/2}) = \frac{1}{2}e^{-1} $

  $ P(X=1, Y=1 | \theta=\frac{1}{2}) = P(X=1 | \theta=\frac{1}{2}) P(Y=1 | \theta=\frac{1}{2}) = (\frac{1}{2}e^{-1/2}) (\frac{1}{2}e^{-1/2}) = \frac{1}{4}e^{-1} $

  The probability for this case ($X=1$ and $Y \leq 1$) is the sum: $\frac{1}{2}e^{-1} + \frac{1}{4}e^{-1} = \frac{3}{4}e^{-1}$.

The total probability of rejecting $H_0$ under $H_1$ (Power) is:

$ \text{Power} = P(X=0 | \theta=\frac{1}{2}) + P(X=1, Y \leq 1 | \theta=\frac{1}{2}) = e^{-1/2} + \frac{3}{4}e^{-1} $

This result matches Option D. Thus, Option 4 is true.

Type II error occurs when $H_0$ is accepted, but $H_1$ is true. The probability is $P(\text{Type II error}) = 1 - \text{Power}$.

$ P(\text{Type II error} | \theta=\frac{1}{2}) = 1 - \left(e^{-1/2} + \frac{3}{4}e^{-1}\right) $

Option 2 gives $1 - \frac{1}{2}e^{-1} - e^{-1/2}$, which does not match our calculation and is incorrect.

Conclusion

The calculations show that Option A (P[type I error] = $e^{-1} + 2e^{-2}$) and Option D (Power of the test = $\frac{3}{4}e^{-1} + e^{-1/2}$) are true statements.