# Conditional pmf

1. Oct 26, 2012

### mrkb80

1. The problem statement, all variables and given/known data

let $f_{X,Y}(x,y)=2e^{-(x+y)}$ for $0 \le x \le y$ and $y \ge 0 \\$find $P(Y<1 | X < 1)$

2. Relevant equations
$f(X=x | y=y) = \dfrac{f_{X,Y}(x,y)}{f_y(y)}$

3. The attempt at a solution
$P(Y<1 | X<1) = \int_0^1 \dfrac{f_{X,Y}(x,y)}{\int_0^1 f_X(x) dx} dy$

before I integrate, I want to make sure I understand the concept, which I don't think I do.

2. Oct 26, 2012

### Ray Vickson

Use the formula
$$P(A|B) = \frac{P(A\, \& \, B)}{P(B)}.$$
with appropriately-defined A and B.

RGV

3. Oct 26, 2012

### mrkb80

I'm thinking that would be something like this:

P(A) = P(Y<1)
P(B) = P(X<1)=$\int_0^1 f_X(x) dx$
so then $P(A \cap B) =P(Y<1 \cap X<1) = \int_0^1 \int_0^y f_{X,Y}(x,y) dxdy$

and then $\dfrac{\int_0^1 \int_0^y f_{X,Y}(x,y) dxdy}{\int_0^1 f_X(x) dx}$

or am I still not understanding?

4. Oct 27, 2012

### Ray Vickson

This is OK. Now all you need do is find fX, and do the integrations in the numerator and the denominator.

RGV

5. Oct 27, 2012

### mrkb80

Great. Thanks for the help again.

6. Oct 29, 2012

### mrkb80

I thought I understood it, but something is not correct. just working on the denominator, I get $f_X(x)=-2(e^{-(x+y)}-e^{-x})$ and then if I try to integrate that I get $\int_0^1 f_X(x) dx = 2e^{-y-1}+2e^{-y}+e^{-1}+1$ What am I missing here?

7. Oct 29, 2012

### Ray Vickson

You are missing the fact that f_X(x) cannot have y in it. You need to start again.

RGV

8. Oct 29, 2012

### mrkb80

You're right. I see my mistake: $f_X(x)=2e^{-x}$ makes much more sense.