# Conditional Probability

Tags:
1. Oct 27, 2014

### joshthekid

Basically I am wondering how you deal with a conditional cdf and turning that into a conditional pdf when the random variables are independent. I know that f(X|Y) =f(X)f(Y)/f(Y)=f(X)

I tried to derive this in a nice attached laTex document but it does not seem right to me.

Note(this is for a homework problem but this is only a derivation I am trying to use to solve it so I decided to post it here because it is not a textbook problem)

#### Attached Files:

• ###### ECE514hw5.pdf
File size:
54.8 KB
Views:
91
2. Oct 28, 2014

### RUber

I have only seen this explained as
$f(X|Y) = \frac{f(X \cap Y)}{f(Y)}$ where $f(X \cap Y) = f(X)f(Y)$ by the definition of independence.
In your work, it seems like in part (6) you were taking the integral with respect to y, where you should be considering a fixed y and taking the integral with respect to x.
I have not put pen to paper, but it looks like that could get you something in a more recognizable form.

3. Oct 29, 2014

### Stephen Tashi

The inequality $g(x,y) < z$ can't necessarily be rewritten in the form $x < h(y,z)$.

For example, the solution $x^2 + y < z$ might require that $x$ be in an interval of the form $-a < x < a$ rather than in an interval of the form $x < a$.