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.
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:Originally Posted by jedoob
$\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