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Setting up a likelihood ratio test where for the exponential distribution, with pdf:

$$f(x;\lambda)=\begin{cases}\lambda e^{-\lambda x}&,\,x\ge0\\0&,\,x<0\end{cases}$$

And we are looking to test:

$$H_0:\lambda=\lambda_0 \quad\text{ against }\quad H_1:\lambda\ne \lambda_0$$

Now, when $H_1$ is true we need to maximise its likelihood, so I note that in that case the parameter $\lambda$ would merely be the maximum likelihood estimator, in this case, the sample mean.

Some algebra yields a likelihood ratio of:

$$\left(\frac{\frac{1}{n}\sum_{i=1}^n X_i}{\lambda_0}\right)^n \exp\left(\frac{\lambda_0-n\sum_{i=1}^nX_i}{n\lambda_0}\right)$$

And if I let the statistic:

$$Y=\sum_{i=1}^n X_i$$

I can rewrite the above equation as:

$$\left(\frac{\frac{1}{n}Y}{\lambda_0}\right)^n \exp\left(\frac{\lambda_0-nY}{n\lambda_0}\right)$$

When the null hypothesis is true, what would be the distribution of $Y$?

And if I were to be given values of $n$ and $\lambda_0$ (e.g. $n=50$ and $\lambda_0=3/2$ , how would I go about determining a test based on $Y$ at the $1\%$ level of significance?

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  • $\begingroup$ Please use MathJax for typesetting math on this site. $\endgroup$ – StubbornAtom Oct 5 '18 at 14:11
  • $\begingroup$ Do you see why the likelihood ratio you found is not correct? $\endgroup$ – StubbornAtom Oct 7 '18 at 5:50
  • $\begingroup$ I do! Mea culpa...I was mixing the differing parameterisations of the exponential distribution. Understand now! My thanks. $\endgroup$ – Sloth Oct 7 '18 at 5:59
  • $\begingroup$ I see you have not voted or accepted most of your questions so far. If your queries have been answered sufficiently, you might consider upvoting and/or accepting those answers. $\endgroup$ – StubbornAtom Oct 7 '18 at 6:06
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Assuming you are working with a sample of size $n$, the likelihood function given the sample $(x_1,\ldots,x_n)$ is of the form

$$L(\lambda)=\lambda^n\exp\left(-\lambda\sum_{i=1}^n x_i\right)\mathbf1_{x_1,\ldots,x_n>0}\quad,\,\lambda>0$$

The LR test criterion for testing $H_0:\lambda=\lambda_0$ against $H_1:\lambda\ne \lambda_0$ is given by

$$\Lambda(x_1,\ldots,x_n)=\frac{\sup\limits_{\lambda=\lambda_0}L(\lambda)}{\sup\limits_{\lambda}L(\lambda)}=\frac{L(\lambda_0)}{L(\hat\lambda)}$$

, where $\hat\lambda$ is the unrestricted MLE of $\lambda$.

A routine calculation gives $$\hat\lambda=\frac{n}{\sum_{i=1}^n x_i}=\frac{1}{\bar x}$$

Then we have

$$\Lambda(x_1,\ldots,x_n)=\lambda_0^n\,\bar x^n \exp(n(1-\lambda_0\bar x))=g(\bar x)\quad,\text{ say }$$

Now study the function $g$ to justify that $$g(\bar x)<c \iff \bar x<c_1\quad\text{ or }\quad \bar x>c_2$$

, for some constants $c_1,c_2$ determined from the level $\alpha$ restriction

$$P_{H_0}(\overline X<c_1)+P_{H_0}(\overline X>c_2)\leqslant \alpha$$

You are given an exponential population with mean $1/\lambda$. So we can multiply each $X_i$ by a suitable scalar to make it an exponential distribution with mean $2$, or equivalently a chi-square distribution with $2$ degrees of freedom. Note the transformation

\begin{align} X_i\stackrel{\text{ i.i.d }}{\sim}\text{Exp}(\lambda)&\implies 2\lambda X_i\stackrel{\text{ i.i.d }}{\sim}\chi^2_2 \\&\implies 2\lambda \sum_{i=1}^n X_i\sim \chi^2_{2n} \end{align}

That is, we can find $c_1,c_2$ keeping in mind that under $H_0$, $$2n\lambda_0 \overline X\sim \chi^2_{2n}$$

We use this particular transformation to find the cutoff points $c_1,c_2$ in terms of the fractiles of some common distribution, in this case a chi-square distribution.

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