# Thread: Cumulative distribution function help!

1. ## Cumulative distribution function help!

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.

2. Originally Posted by knighty
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

3. Originally Posted by CaptainBlack
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?

4. Originally Posted by knighty
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)$

but $p(Z

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

CB