Pivot Quantity for Exponential Distribution

Hi,

I am given two samples $\displaystyle \{x_1,x_2,...,x_n\} \sim \mbox{Exponential}(\lambda_1)$ and $\displaystyle \{y_1,y_2,...,y_m\} \sim \mbox{Exponential}(\lambda_2)$ and I wish to use a pivot quantity to test the hypothesis $\displaystyle H_0: \lambda_1=\lambda_2$ against $\displaystyle H_a: \lambda_1 \neq \lambda_2$ using a suitable pivot quantity.

I know that this means that I should find a statistics with a distribution that doesn't depend on $\displaystyle \lambda_1=\lambda_2$ under $\displaystyle H_0$. Since

$\displaystyle \sum_{i=0}^n X_i \sim Gamma(n,\lambda_1)$ this means that $\displaystyle \frac{\lambda_1}{n}\sum_{i=0}^n X_i \sim Gamma(n,n)$ and hence

T=$\displaystyle \frac{\lambda_1}{n}\sum_{i=0}^n X_i+\frac{\lambda_2}{m}\frac{m}{n}\sum_{i=0}^m Y_i \sim Gamma(m+n,n)$ is a quantity with a distribution that doesn't depend on $\displaystyle \lambda_1=\lambda_2$.

My questions are

1) is the above correct

2) How do I now proceed to compute a p-value. I see how I can simulate from an exponential distribution with any given parameter and compute the test statistics, but how do I evaluate the statistics for my given sample since I don't know $\displaystyle \lambda_1,\lambda_2$?

Thanks in advance,

David

Re: Pivot Quantity for Exponential Distribution

Hey DavidEriksson.

If you wish to use a pivotal quantity, then I would be looking at using the chi-square pivotal quantity for estimating whether the parameters are the same or not.

Recall that the pivotal quantity doesn't depend on the parameters or its distribution and what you are doing is the opposite where you are deriving specific sampling distributions to test your hypotheses: you can take this approach if you wish but its not the same as using pivotal quantities like the Normal Distribution or the chi-square distribution.

The reason also for using chi-square is that the exponential parameters must be greater than zero, so using a Normal pivotal quantity wouldn't work.