# Likelihood ratio test for exponential distribution with scale parameter

I need to find the exact test of level $\alpha$ for null hypothesis $H_0:\theta = \theta_0$ against the alternative hypothesis $H_0:\theta\neq\theta_0$ based on i.i.d data $y_1,\dots,y_n$ that follow the exponential distribution with scale parameter $\theta$.

I know that the likelihood function is $L(\theta)=\prod\frac{1}{\theta}e^{-\frac{y}{\theta}}=\frac{1}{\theta^n}e^{-\frac{\sum_y}{\theta}}$ and the relative MLE is $\hat \theta=\bar y$.

I use the definition of likelihood ratio:

$$\Lambda=\frac{L(\theta)}{\sup L(\theta)}=\frac{\theta^{-n}e^{-\frac{\sum y}{\theta}}}{\bar y^{-n}e^{-\frac{\sum y}{\bar y}}}=\left(\frac{\theta}{\bar y}\right)^{-n}e^{-\left(\frac{\sum y}{\theta}+\frac{\sum y}{\bar y}\right)}$$

But now I'm stuck on how to proceded. Any help or suggest?

Hint: the likelihood ratio test is not an exact test. The LRT is based upon the asymptotic distribution of the likelihood ratio statistic, i.e. $2 (\log L_1 - \log L_2) \rightarrow \chi^2_{p-q}$. With $L_1$ and $L_2$ arising from likelihoods in MLE parameters for $p$ and $q$ dimensional supports respectively.

An exact test is based upon the actual distribution of the sufficient statistic. What is the sufficient stat for an exponential distribution (how do you show that?) and what distribution does it take (under the null)?

• Hi, the exponential distribuion is a class of exponential family. I can write $p(y;\theta)$ like $c(\theta)^n \left[\prod h(y_i)\right]e^{\psi(\theta)\sum t(y_i)}$, so $\sum t(y_i)$ is a minimal sufficient statistic. Now, if $Y\sim Exp(\theta)$ then $T = \sum y\sim Ga(n, \theta)$.
– Paul
Aug 25, 2016 at 17:06
• @paul so calculate critical values from that distribution and call it a day. Aug 26, 2016 at 23:50
• If $T = \sum y\sim Ga(n, \theta)$ then $\frac{2n\bar y}{\theta}\sim \chi^2_{2n}$. Based on this assumption a exact test of level $\alpha$ will accept $H_0$ if $\frac{2n\bar y}{\theta} \lt c$ where $P(\chi_{2n}^2 \lt c)=\alpha$. I'm not sure that it is correct.
– Paul
Aug 27, 2016 at 15:59
• @Paul that's not correct. Where is the $\chi^2$ distribution coming from? That's not the small sample exact distribution of $\bar{Y}$. You can calculate critical values from anything, even a $Ga(n, \theta)$ distribution. Aug 29, 2016 at 15:25
• Let be $Y\sim Exp(\theta)$ with density $f(y)=\frac{1}{\theta}e^{\frac{y}{\theta}}$. Now set $X=\frac{2Y}{\theta} \Rightarrow Y=\frac{X\theta}{2}$. To find the new density I need $\frac{dY}{dX}=\frac{\theta}{2}$, so $f(x)=\frac{1}{\theta}e^{\frac{1}{\theta}\frac{x\theta}{2}} \cdot \frac{\theta}{2}=\frac{1}{2}e^{\frac{x}{2}}$, so $X\sim Exp(\frac{1}{2}) \sim \Gamma(1,\frac{1}{2}) \sim \chi^2_2$. Based on i.i.d. $\frac{2\sum_{1}^{n}Y}{\theta}=\frac{2n\bar Y}{\theta}\sim \chi^2_{2n}$.
– Paul
Aug 30, 2016 at 21:14