# probability density function

• Feb 23rd 2006, 08:37 AM
jedoob
probability density function
Hey,

Can anyone help me with the following question:

f(x) = K(1+x)^n such that x>0
and
f(x) = 0 otherwise.

for some constants K and n.

For what values of n can f be made into a valid probability density?

Any help would be greatly appreciated.
• Feb 23rd 2006, 09:26 AM
CaptainBlack
Quote:

Originally Posted by jedoob
Hey,

Can anyone help me with the following question:

f(x) = K(1+x)^n such that x>0
and
f(x) = 0 otherwise.

for some constants K and n.

For what values of n can f be made into a valid probability density?

Any help would be greatly appreciated.

A function is defined as follows:

$\displaystyle f(x)=\left{ \begin{array}{cc} K(1+x)^n & \mbox{,\ if}\ x>0\\ 0 & \mbox{,\ if}\ x \le 0 \end{array}\right$,

for some $\displaystyle K$ and $\displaystyle n$.

For what values of $\displaystyle n$ can $\displaystyle f$ be made into a valid probability density.

Now $\displaystyle f$ is a probability density if and only if $\displaystyle f(x)\ge\ 0$ for all $\displaystyle x$, and

$\displaystyle \int_{-\infty}^{\infty}f(x) dx = 1$.

Now the integral is infinite when $\displaystyle n \ge \ 0$, and if $\displaystyle n < \ 0$:

$\displaystyle \int_{-\infty}^{\infty}f(x) dx = \frac{-K}{n+1}=1$.

Now $\displaystyle K > \ 0$ as otherwise $\displaystyle f(x)$ would not be be
greater than or equal to zero. Then $\displaystyle n+1=-K$, which can only
be made to work if $\displaystyle n <0$.

RonL
• Feb 23rd 2006, 09:46 AM
jedoob
Thanks
Thanks a million.