Suppose that $X_1,X_2,…,X_n$ is a random sample from the exponential distribution with rate $\lambda\gt0$.

Derive a hypothesis test of size $\alpha$ for

\[H_0:\lambda=\lambda_0\quad\text{vs.}\quad H_1:\lambda\gt\lambda_0\]

What statistic should we use?

Test1: Using the Sample Mean

Step One:

Choose a statistic.

\[\overline{X}\] Step Two:

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We have

\[f(x)=\lambda e^{-\lambda x}\\ E[X] = 1/\lambda\]

  • If we have a higher $\lambda$, we will see generally smaller values comming out of the distribution.

  • If we have a higher $\lambda$, the random variable has a smaller mean.

  • Overall, if we’re sampling values from the exponential, they’re going to be smaller as $\lambda$ gets larger.

Give the form of the test.

Reject $H_0$, in favor of $H_1$, if

\[\overline{X}\lt c\]

for some $c$ to be determined.

Step Three:

Find $c$.

\[\begin{align} \alpha &= P(\text{Type I Error})\\ &= P(\text{Reject }H_0;\lambda_0)\\ &= P(\overline{X}\lt c;\lambda_0)\\ &= P(W\lt2n\lambda_0c)\\ \end{align}\]

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We want

\[2n\lambda_0c=\chi^2_{1-\alpha,2n}\] Step Four:

Conclusion.

Reject $H_0$, in favor of $H_1$, if

\[\overline{X}\lt\frac{\chi^2_{1-\alpha,2n}}{2n\lambda_0}\]

Note:

To find $\chi^2_{\alpha,n}$.

In R, if we want to get $\chi^2_{0.10,6}$.

qchisq(0.90, 6)

Test1: Using the Sample Minimum

Step One:

Choose a statistic.

\[Y_n=\text{min}(X_1,X_2,...,X_n)\] Step Two:

png

We have

\[f(x)=\lambda e^{-\lambda x}\\ E[X] = 1/\lambda\]
  • If the alternate hypothesis is true, then $\lambda$ is large, that means that values coming out of the exponential distribution are actually smaller than in the null hypothesis.

Give the form of the test.

Reject $H_0$, in favor of $H_1$, if

\[Y_n\lt c\]

for some $c$ to be determined.

Step Three:

Find $c$.

\[\begin{align} \alpha &= P(\text{Type I Error})\\ &= P(\text{Reject }H_0;\lambda_0)\\ &= P(Y_n\lt c;\lambda_0)\\ &= P(n\lambda_0Y_n\lt cn\lambda_0;\lambda_0)\\ &= P(X\lt cn\lambda_0)\\ \end{align}\]

where $X\sim\text{exp}(\text{rate}=1)$.

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We want the probability that the random variable is less that or equal to some unknown number, and we want that to equal $\alpha$.

\[1 - e^{-?}=\alpha\\ \implies ? = -\ln(1-\alpha)\]

So

\[\alpha=P(X\lt cn\lambda_0)\\ \Downarrow\\ cn\lambda_0=-\ln(1-\alpha)\\ \Downarrow\\ c=\frac{-\ln(1-\alpha)}{n\lambda_0}\] Step Four:

Conclusion.

Reject $H_0$, in favor of $H_1$, if

\[Y_n=\text{min}(X_1,X_2,...,X_n)\lt\frac{-\ln(1-\alpha)}{n\lambda_0}\]

Compare the Tests

Test 1: based on $\overline{X}$

\[\begin{align} \gamma(\lambda)&=P(\text{Reject }H_0;\lambda)\\ &=P\Bigg(\overline{X}\lt\frac{\chi^2_{1-\alpha,2n}}{2n\lambda_0};\lambda\Bigg)\\ &=P\Bigg(2n\lambda\overline{X}\lt2n\lambda\frac{\chi^2_{1-\alpha,2n}}{2n\lambda_0};\lambda\Bigg)\\ &=P\Bigg(W\lt\frac{\lambda}{\lambda_0}\chi^2_{1-\alpha,2n}\Bigg)\\ \end{align}\]

So we have the power function for the first test

\[\gamma_1{\lambda}=P\Bigg(W\lt\frac{\lambda}{\lambda_0}\chi^2_{1-\alpha,2n}\Bigg)\\\]

Suppose that

\[n=10\quad\lambda_0=1\quad\alpha=0.05\\ \chi^2_{1-\alpha,2n}=\chi^2_{0.95,20}=10.851\] \[\begin{align} \gamma_1{\lambda}&=P\Bigg(W\lt\frac{\lambda}{\lambda_0}\chi^2_{1-\alpha,2n}\Bigg)\\ &=P(W\lt10.851\lambda) \end{align}\]

We plot in R:

lambda <- seq(0,6,0.01)
f<-pchisq(10.81*lambda,20)
plot(lambda, f, type='l')

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  • For the hypotheses, $\lambda \le \lambda_0$ which $\lambda_0=1$. So we want this power function, the probability we reject to be small when $\lambda < 1$, because if $\lambda$ is less then $1$ the null hypothesis is true and we don’t want to reject it when it’s true.

  • When $lambda \gt \lambda_0$, the region from $1$ and go the the right is where the alternate hypothesis is true, and we want that probability to be large.

Test 2: based on $Y_n=\text{min}(X_1,X_2,…,X_n)$

\[\begin{align} \gamma_2(\lambda)&=P(\text{Reject }H_0;\lambda)\\ &=P\Bigg(\underbrace{Y_n}_{\text{exp}(\text{rate}=n\lambda)}\lt\frac{-(\ln(1-\alpha)}{n\lambda_0};\lambda\Bigg)\\ &=1 - e^{-n\lambda(-\ln(1-\alpha)/n\lambda_0)}\\ &=1 - e^{\lambda(-\ln(1-\alpha)/\lambda_0}\\ &=1 - (1-\alpha)^{\lambda/\lambda_0}\\ \end{align}\]

png

  • We want the probability that reject the null hypothesis in the region $\lambda\gt\lambda_0$ which $\lambda_0=1$ is larger.

  • We can see in the gragh, the power function of test 1 shows higher probability in region that reject $H_0$, and lower in the region of $H_0$. So the test 1 is much better than the test 2. We should use statistic $\overline{X}$ instead of sample minimum.