Cumulative distribution function help!

knighty · #1 $Old$ March 16th, 2009, 10:46 AM

Hey, I'm new here. I'm stuck with this problem. Will post what I've tried to do. Hopefully someone can provide hints to help me out. Thanks.

If Z is a standard normal random variable, and Y = -ln (1-cdf(Z)), what is the distribution of the random variable Y. (note: cdf is cumulative distribution function)

Here's what I've done:
cdf(y) = P ( Y < y) = P (-ln(1-cdf(Z))<y) = ... = P ( cdf (Z) < 1 - e^-y ) -> stuck..

my approach is to find cdf of y and then differentiate it to get pdf of y, which is what's required.

CaptainBlack · #2 $Old$ March 17th, 2009, 01:08 AM

Quote:

Originally Posted by knighty $View Post$

Hey, I'm new here. I'm stuck with this problem. Will post what I've tried to do. Hopefully someone can provide hints to help me out. Thanks.

If Z is a standard normal random variable, and Y = -ln (1-cdf(Z)), what is the distribution of the random variable Y. (note: cdf is cumulative distribution function)

Here's what I've done:
cdf(y) = P ( Y < y) = P (-ln(1-cdf(Z))<y) = ... = P ( cdf (Z) < 1 - e^-y ) -> stuck..

my approach is to find cdf of y and then differentiate it to get pdf of y, which is what's required.

If you don't know this prove it, otherwise just use it:

$\text{cdf}(Z) \sim U(0,1)$

so:

$1-\text{cdf}(Z) \sim U(0,1)$

CB

knighty · #3 $Old$ March 17th, 2009, 08:33 AM

Quote:

Originally Posted by CaptainBlack $View Post$

If you don't know this prove it, otherwise just use it:

$\text{cdf}(Z) \sim U(0,1)$

so:

$1-\text{cdf}(Z) \sim U(0,1)$

CB

why is the cdf of Z a uniform distribution?

CaptainBlack · #4 $Old$ March 17th, 2009, 09:45 AM

Quote:

Originally Posted by knighty $View Post$

why is the cdf of Z a uniform distribution?

let $f(Z)=cdf(Z)$ then as f is strictly increasing it is invertable and so:

$p(cdf(Z)<a)=p(f(Z)<a)=p(Z<f^{-1}(a))$

but $p(Z<f^{-1}(a))=f(f^{-1}(a))=a$

So if $X=cdf(Z)$ then $p(X<a)=a$ which is the cdf of the uniform distribution $U(0,1).$

CB