# Pivot Quantity for Exponential Distribution

• Oct 24th 2012, 12:12 AM
DavidEriksson
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 $\displaystyle \lambda_1,\lambda_2$?