# Probability Questions

• Sep 16th 2009, 05:48 PM
Raji
Probability Questions
Hi,

I have been trying to work this question for two days, but I could not figure it out. If any one help me to direct me that would be much appreciated. please!!!

Suppose X follows a Normal distribution with mean (mu) (with mu>0) and a variance that is some function of the mean, (ie) (sigma^2) = h(u). Find h(u) such that P(X less than or equal to 0) does not depend on mean (mu).
• Sep 18th 2009, 07:57 AM
Moo
Hello,

Well, I'm not too sure about my method, but I get a result...

$\displaystyle \mathbb{P}(X\leq 0)=\int_{-\infty}^0 \frac{1}{\sqrt{2\pi} \cdot \sqrt{h(u)}} \cdot \exp\left(-\frac{(x-u)^2}{2h(u)}\right) ~dx$

Now substitute $\displaystyle t=\frac{x-u}{\sqrt{h(u)}}$

You'll get :

$\displaystyle \mathbb{P}(X\leq 0)=\int_{-\infty}^{-u/\sqrt{h(u)}} \frac{1}{\sqrt{2\pi}} \cdot \exp\left(-\frac{t^2}{2}\right) ~dt$

So the only place left where there's u is in the boundary of the integral.
Find any function h such that $\displaystyle -\frac{u}{\sqrt{h(u)}}=\text{constant}$ (the only way not to have u)

Spoiler:
$\displaystyle h(u)=au^2$, for any positive constant a
• Sep 18th 2009, 08:05 PM
Raji
Many Thanks
Many thanks for the help.