Find k for this probability density function to be valid.

1. Jan 18, 2013

TelusPig

1. The problem statement, all variables and given/known data

Find k such that the function $f(x)=ke^{-\frac{x-\mu}{\theta}}$ is a probability density function (pdf), for $x > \mu, \mu$ and $\theta$ are constant.

2. Relevant equations

The property of a pdf says that the integral of f(x) from $-\infty$ to $\infty$ equals 1, that is $\int\limits_{-\infty}^\infty f(x)dx=1$

3. The attempt at a solution

$\int\limits_{-\infty}^\infty ke^{-\frac{x-\mu}{\theta}}dx$
$=k\int\limits_{-\infty}^\infty e^{-\frac{x-\mu}{\theta}}dx$
Let $t=-\frac{x-\mu}{\theta} => -\theta dt=dx$

$=> -k\int\limits_\infty^{-\infty} e^t(-\theta)dt$

$=>k\theta\int\limits_{-\infty}^\infty e^tdt$

But $e^t$ would diverge going towards infinity? How could this integral be equal to 1. I am not sure what I'm doing wrong.

2. Jan 18, 2013

Dick

It's supposed to be a density function only for x>μ. You don't want to integrate from -infinity to +infinity. You want to integrate from μ to +infinity. The pdf will be zero for x<μ.

3. Jan 18, 2013

TelusPig

Ok I see what you mean. But even if I do that, the upper bound of infinity is giving me problems because e^(infinity) is infinity :S

4. Jan 18, 2013

TelusPig

Never mind!! I finally figured it out... I had a -infinity instead. All is good :) I got k = 1/theta if anyone was interested