# construct an interval estimator (exponential)

• February 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 $Y_1, ... Y_n$ be a random sample from an exponential distribution with density function

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

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

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

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

Let $U=2X\theta$

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

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

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

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

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