Chapter 10.10, Problem 99E

a.

To determine

Obtain the form of rejection region for a most powerful test for testing $H_0:\lambda ={\lambda }_0$ versus $H_a:\lambda ={\lambda }_a$, where ${\lambda }_a>{\lambda }_0$.

Explanation of Solution

In this context, $Y_1,Y_2,\mathrm{...},Y_n$ denote a random sample from the population that has the Poisson distribution with the mean $\lambda$.

The likelihood function of the random variable Y is given below:

$\begin{array}{c}L\left(\lambda \right)={\displaystyle \prod _{i=1}^{n}P\left(Y_i,\lambda \right)}\\ ={\displaystyle \prod _{i=1}^n\frac{e^{-\lambda }{\lambda }^{Y_i}}{Y_i!}}\\ =\frac{e^{-n\lambda }{\lambda }^{{\displaystyle \sum _{i=1}^n{Y}_i}}}{{\displaystyle \prod _{i=1}^n{Y}_i!}}\end{array}$

Neyman–Pearson Lemma:

It is assumed to test $H_0:\theta ={\theta }_0\text{ vs }{H}_a:\theta ={\theta }_a$, based on a random sample of $Y_1,Y_2,\mathrm{...},Y_n$ from a distribution with parameter $\theta$. Here, denote $L\left(\theta \right)$ as the likelihood of the sample when the value of the parameter is $\theta$. For the given $\alpha$, the test maximizes the power at ${\theta }_a$ that has a rejection region is determined as shown below:

$\frac{L\left({\theta }_0\right)}{L\left({\theta }_a\right)}<k$

In this context, the value of k is chosen so that the test will have the desired value for $\alpha$. Such a test is a most powerful $\alpha$-level test for $H_0\text{vs }{H}_a$.

The most powerful $\alpha$-level test for testing $H_0:\lambda ={\lambda }_0$ versus $H_a:\lambda ={\lambda }_a$ is obtained as given below:

Substitute the corresponding values in the above equation to get the most powerful test.

$\begin{array}{c}\frac{L\left({\lambda }_0\right)}{L\left({\lambda }_a\right)}<k\\ \frac{\frac{e^{-n{\lambda }_0}{\lambda }_0{}^{{\displaystyle \sum _{i=1}^n{Y}_i}}}{{\displaystyle \prod _{i=1}^n{Y}_i!}}}{\frac{e^{-n{\lambda }_a}{\lambda }_a{}^{{\displaystyle \sum _{i=1}^n{Y}_i}}}{{\displaystyle \prod _{i=1}^n{Y}_i!}}}<k\\ \frac{e^{-n{\lambda }_0}{\lambda }_0{}^{{\displaystyle \sum _{i=1}^n{Y}_i}}}{e^{-n{\lambda }_a}{\lambda }_a{}^{{\displaystyle \sum _{i=1}^n{Y}_i}}}<k\\ e^{-n{\lambda }_0-n{\lambda }_a}\frac{{\lambda }_0{}^{{\displaystyle \sum _{i=1}^n{Y}_i}}}{{\lambda }_a{}^{{\displaystyle \sum _{i=1}^n{Y}_i}}}<k\\ e^{-n\left({\lambda }_0-{\lambda }_a\right)}{\left(\frac{{\lambda }_0}{{\lambda }_a}\right)}^{{\displaystyle \sum _{i=1}^n{Y}_i}}<k\end{array}$

As n, ${\lambda }_0,and{\lambda }_a$ are constants, the $\left(e^{-n\left({\lambda }_0-{\lambda }_a\right)}\right)k$ in the equation will be denoted as $k_1$.

$\begin{array}{l}{\left(\frac{{\lambda }_0}{{\lambda }_a}\right)}^{{\displaystyle \sum _{i=1}^n{Y}_i}}<k_1\\ \text{Taking}\text{ln on both sides as follows:}\\ \left({\displaystyle \sum _{i=1}^n{Y}_i}\right)\text{ln}\left(\frac{{\lambda }_0}{{\lambda }_a}\right)<\ln \left(k_1\right)\end{array}$

The inequality in the last step changes because ${\lambda }_a>{\lambda }_0$; therefore, the log fraction is negative.

$\left({\displaystyle \sum _{i=1}^n{Y}_i}\right)>\frac{\ln \left(k_1\right)}{\text{ln}\left(\frac{{\lambda }_0}{{\lambda }_a}\right)}$

Here, $\frac{\ln \left(k_1\right)}{\text{ln}\left(\frac{{\lambda }_0}{{\lambda }_a}\right)}=k*$

$\left({\displaystyle \sum _{i=1}^n{Y}_i}\right)>k*$

Therefore, the rejection region will have the form of $\left({\displaystyle \sum _{i=1}^n{Y}_i}\right)>k*\left[\text{For some constant}k*\right]$.

b.

To determine

Explain the way in which the given information used to find any constant associated with the rejection region derived in Part (a).

Explanation of Solution

In this context, it is given that ${\displaystyle \sum _{i=1}^n{Y}_i}$ has the Poisson distribution with the mean $n\lambda$.

From Part (a), it is clear that the rejection region will have the form of $\left({\displaystyle \sum _{i=1}^n{Y}_i}\right)>k*\left[\text{For some constant}k*\right]$.

Now, it is known that ${\displaystyle \sum _{i=1}^n{Y}_i}~Poisson\left(n\lambda \right)$. This can be used to find the constant k* in our rejection region.

In general, the k* is chosen to be small. Such that $P\left(Z>k*\right)\le \alpha$, where $Z~Poisson\left(n{\lambda }_0\right)$.

c.

To determine

State whether the test derived in Part (a) is uniformly most powerful for testing $H_0:\lambda ={\lambda }_0$ versus $H_a:\lambda >{\lambda }_0$.

Explanation of Solution

From Part (a), it has been found that the rejection region of the powerful test will be in the form of $\left({\displaystyle \sum _{i=1}^n{Y}_i}\right)>k*\left[\text{For some constant}k*\right]$.

As the rejection region does not depend on ${\theta }_a$, the test is uniformly most powerful for all ${\lambda }_a>{\lambda }_0$.

Therefore, the test derived in Part (a) is uniformly most powerful for testing $H_0:\lambda ={\lambda }_0$ versus $H_a:\lambda >{\lambda }_0$.

d.

To determine

Obtain the form of rejection region for a most powerful test for testing $H_0:\lambda ={\lambda }_0$ versus $H_a:\lambda ={\lambda }_a$, where ${\lambda }_a<{\lambda }_0$.

Explanation of Solution

The test hypotheses are given below:

$H_0:\lambda ={\lambda }_0$ versus $H_a:\lambda ={\lambda }_a$

The likelihood function of the random variable Y is given below:

The most powerful $\alpha$-level test for testing $H_0:\lambda ={\lambda }_0$ versus $H_a:\lambda ={\lambda }_a$, where ${\lambda }_a<{\lambda }_0$ is obtained as given below:

Substitute the corresponding values in the above equation to get the most powerful test.

$\begin{array}{c}\frac{L\left({\lambda }_0\right)}{L\left({\lambda }_a\right)}<k\\ \frac{\frac{e^{-n{\lambda }_0}{\lambda }_0{}^{{\displaystyle \sum _{i=1}^n{Y}_i}}}{{\displaystyle \prod _{i=1}^n{Y}_i!}}}{\frac{e^{-n{\lambda }_a}{\lambda }_a{}^{{\displaystyle \sum _{i=1}^n{Y}_i}}}{{\displaystyle \prod _{i=1}^n{Y}_i!}}}<k\\ \frac{e^{-n{\lambda }_0}{\lambda }_0{}^{{\displaystyle \sum _{i=1}^n{Y}_i}}}{e^{-n{\lambda }_a}{\lambda }_a{}^{{\displaystyle \sum _{i=1}^n{Y}_i}}}<k\\ e^{-n{\lambda }_0-n{\lambda }_a}\frac{{\lambda }_0{}^{{\displaystyle \sum _{i=1}^n{Y}_i}}}{{\lambda }_a{}^{{\displaystyle \sum _{i=1}^n{Y}_i}}}<k\\ e^{-n\left({\lambda }_0-{\lambda }_a\right)}{\left(\frac{{\lambda }_0}{{\lambda }_a}\right)}^{{\displaystyle \sum _{i=1}^n{Y}_i}}<k\end{array}$

As n, ${\lambda }_0,and{\lambda }_a$ are constants, the $\left(e^{-n\left({\lambda }_0-{\lambda }_a\right)}\right)k$ in the equation will be denoted as $k_1$.

$\begin{array}{l}{\left(\frac{{\lambda }_0}{{\lambda }_a}\right)}^{{\displaystyle \sum _{i=1}^n{Y}_i}}<k_1\\ \text{Taking}\text{ln on both sides as follows:}\\ \left({\displaystyle \sum _{i=1}^n{Y}_i}\right)\text{ln}\left(\frac{{\lambda }_0}{{\lambda }_a}\right)<\ln \left(k_1\right)\end{array}$

The inequality in the last step changes because ${\lambda }_a<{\lambda }_0$; therefore, the log fraction is positive.

$\left({\displaystyle \sum _{i=1}^n{Y}_i}\right)<\frac{\ln \left(k_1\right)}{\text{ln}\left(\frac{{\lambda }_0}{{\lambda }_a}\right)}$

Here, $\frac{\ln \left(k_1\right)}{\text{ln}\left(\frac{{\lambda }_0}{{\lambda }_a}\right)}=k*$

$\left({\displaystyle \sum _{i=1}^n{Y}_i}\right)<k*$

Therefore, the rejection region will have the form of $\left({\displaystyle \sum _{i=1}^n{Y}_i}\right)<k*\left[\text{For some constant}k*\right]$.