# 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.
• 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 '16 at 17:06
• 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 '16 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. – AdamO Aug 29 '16 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 '16 at 21:14