I have the following question: Find the best size $\alpha$ test of $H_0:\theta=\theta_0$ vs $H_1:\theta =\theta_1$, with $\theta_1>\theta_0$, write down the expression for the power function when $X_1,\dots,X_n$ are IID exponential random variables, with parameter $\theta$.

Use the $\chi^2$ tables to find the least sample size which will allow us to test $H_0:\theta=1$ against $H_1:\theta=3$ with $\alpha=5\%$ and $\beta\leq 10\%$. Describe the appropiate critical region numerically.

I have managed to find the best test size in the following way:

First of all, finding the likelihood ratio:

$\begin{gather}\frac{f_1(x|\theta_1)}{f_0(x|\theta_0)}= \left(\frac{\theta_1}{\theta_0}\right)^n \cdot e^{-(\sum x_i)(\theta_1-\theta_0)} \end{gather}$.

Then, for the sufficient statistic $T = \sum X_i$, we condition that the likelihood ratio be bigger than $k(\alpha)$. Computing the inequality, we get that we need to have $T<c_s$, where $c_s=\frac{n(ln\theta_1 - ln\theta_0)-lnk(\alpha)}{\theta_1-\theta_0}$.

Now, we know further that $T\sim Gamma(n,\theta)$. Hence, by Neyman-Pearson Lemma we have that :

$\begin{gather}\alpha = \int_{c_s}^{\infty}\frac{1}{\Gamma(n)}t^{n-1}e^{-t}dt \end{gather}$

We can compute $c_s$. However, I do not know what to do for the next part.

  • $\begingroup$ Rejection region is $T<c$ but not sure how you got the $c$. It is found subject to the size restriction $P_{\theta=\theta_0}(T<c)=\alpha$, where you get $c$ in terms of chi-square quantiles. Note that if $X_i$s are iid Exp with mean $1/\theta$ then $2\theta X_i$s are iid $\chi^2_2$, so that $2\theta T\sim \chi^2_{2n}$. The power function is given by $P_{\theta}(T<c)$. $\endgroup$ Jul 1, 2020 at 6:48

1 Answer 1


As you calculated, we Reject $\mathcal{H}_0$ if $\sum_i X_i <c^*$

where $c^*$ is calculated by

$$\int_0^{c^*}\frac{\theta_0^n}{\Gamma(n)}t^{n-1}e^{-\theta_0 t}dt=\alpha$$

Then the power of the test is, by definition,

$$\mathbb{P}[\text{Reject }\mathcal{H}_0|\mathcal{H}_1]$$ that is

$$\gamma=\int_0^{c^*}\frac{\theta_1^n}{\Gamma(n)}t^{n-1}e^{-\theta_1 t}dt$$

Now, considering the second part, we have the following system to verify

$$ \begin{cases} \mathcal{H}_0: \theta=1 \\ \mathcal{H}_1: \theta=3 \end{cases}$$

We know that $\sum_i X_i\sim Gamma(n;\theta)$ that means $2\theta\sum_i X_i\sim \chi_{(2n)}^2$

Considering the text request, we must search the minimum $n$ such that $\alpha=5\%$ with a power $\gamma=1-\beta \geq 90\%$

In other words we are looking for the following condition to be satisfied:

$$ \begin{cases} \mathbb{P}[2\sum_i X_i<c_1]=0.05 \\ \mathbb{P}[6\sum_i X_i<c_2] > 0.90 \end{cases}$$

By simply scrolling the $\chi^2$ table, we have to find the first row where the 90° percentile is more than 3 times the 5° percentile. This row is the one corresponding to the 16 degrees of freedom:



and $7.96\times3=23.88$

So the minimum sample size is 8 which gets a test with $\alpha=5\%$ and a power $\gamma \approx 91\%$


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