Difficult integral (product of gamma func with exp)

[coolnessitself]

Homework Statement

Solve. [tex]\int\limits_{0}^\infty \exp\left\{\frac{-2n}{x} - \frac{x}{\theta}\right\} x^{n-1} dx[/tex]

[tex]n,\theta[/tex] are constants.

Homework Equations

The Attempt at a Solution

So actually this problem is from the realm of stats...find the MVUE of [tex]P(X<2)[/tex] where [tex]X_i\sim \mathrm{Exp}(\theta)[/tex]. The MLE of an exp is [tex]\overline{X}[/tex], and this is invariant under transforms, and so we need to solve the expectation of [tex]1-\exp\left\{\frac{-2n}{\overline{x}}\right\}[/tex]. So the 1 is easy, and since here [tex]\overline{X}\sim\Gamma[/tex], it's the expectation of [tex]\frac{-2n}{x}\right\}[/tex] in a gamma distribution, resulting in the integral above, with some constants I've removed. The exam is over, no one could find a solution. Can the integral be solved? I tried a u sub for sqrt(y..) and tried intrgrating by parts n-1 times, and a few other ideas. Any ideas for a method to solve the above integral?

 

[mighty2000]

First of all, great job on recognizing that this problem is related to statistics and the MLE of an exponential distribution. To solve this integral, we can use the substitution u = 1/x, which gives us du = -1/x^2 dx. We can then rewrite the integral as:

\int\limits_{0}^\infty \exp\left\{\frac{-2n}{x} - \frac{x}{\theta}\right\} x^{n-1} dx = \int\limits_{0}^\infty \exp\left\{-2nu - \frac{1}{\theta u}\right\} u^{n+1} du

We can then use the fact that the integral of an exponential function is equal to its reciprocal to rewrite the integral as:

\int\limits_{0}^\infty \frac{u^{n+1}}{\exp\left\{2nu + \frac{1}{\theta u}\right\}} du

Next, we can use the substitution v = 2nu + 1/(theta u), which gives us dv = 2n - 1/(theta u^2) du. We can then rewrite the integral as:

\int\limits_{0}^\infty \frac{u^{n+1}}{\exp\left\{v\right\}} du = \int\limits_{0}^\infty \frac{1}{2n - 1/(theta u^2)} du

Using partial fractions, we can rewrite this integral as:

\int\limits_{0}^\infty \frac{1}{2n}\left(\frac{1}{u - \frac{1}{2n\theta}} + \frac{1}{u + \frac{1}{2n\theta}}\right) du

Integrating each term separately, we get:

\frac{1}{2n}\left(\ln\left|u - \frac{1}{2n\theta}\right| - \ln\left|u + \frac{1}{2n\theta}\right|\right)\Bigg|_0^\infty

Since the limits of integration are from 0 to infinity, the first term evaluates to 0. The second term evaluates to:

\frac{1}{2n}\left(\ln\left|2n\theta\right|