# construct an interval estimator (exponential)

• Feb 13th 2011, 08:38 PM
Volga
construct an interval estimator (exponential)
I think I solved it but I don't have the answer in my book to check against.

Question.

Let $\displaystyle Y_1, ... Y_n$ be a random sample from an exponential distribution with density function

$\displaystyle f_Y(y)={\theta}e^{-{\theta}y}, y>0$

where $\displaystyle \theta>0$ is an unknown parameter. Define $\displaystyle X=Y_1+...+Y_n$. Find the moment generating function of X. Show that $\displaystyle 2X\theta$ is a pivotal function, and hence construct an interval estimator for $\displaystyle \theta$ with confidence coefficient 0.95. (Note. You may use the fact that the $\displaystyle \chi^2$ distribution with k degrees of freedom has moment generating function $\displaystyle (1-2t)^{-k/2}$.)

$\displaystyle M_X(t)=E(e^{Xt})=E(e^{Y_1t}...e^{Y_nt})=(M_Y(t))^n =(\frac{\theta}{\theta-t})^n=(\frac{1}{(1-t/n)})^n$ which, I think, accidentally means $\displaystyle X{\sim}Gamma(n,\theta)$

Now we show that the distribution of $\displaystyle 2X\theta$ does not depend on $\displaystyle \theta$, so that it can be a pivotal function

Let $\displaystyle U=2X\theta$

$\displaystyle M_U(t)=E(e^{Ut})=E(e^{X2{\theta}t})=M_X(2{\theta}t )=(\frac{1}{(1-\frac{2{\theta}t}{\theta}})^n=(\frac{1}{1-2t})^n$ which is mgf for a chi-square distribution with 2n degrees of freedom.

So $\displaystyle U=2X\theta\sim\chi^2(2n)$ and the distribution of U does not depend on $\displaystyle \theta$.

Now we construct an interval estimator for $\displaystyle \theta$ with confidence coefficient 0.95. Using the distribution of U,

$\displaystyle P(t_{0.025}(2n){\leq}2X{\theta}{\leq}t_{0.975}(2n) )=0.95$

I finally got the interval estimator $\displaystyle [\frac{t_{0.025}(2n)}{2X}; \frac{t_{0.975}(2n)}{2X}]$.