Find k for this probability density function to be valid.

[TelusPig]

Homework Statement

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.

Homework 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

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.

 

[Dick]

Homework Statement

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.

Homework 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

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.

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.

 

[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

 

[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