# Thread: Another Joint pdf question

1. ## Another Joint pdf question

Hi,

Iam trying to find the conditional pdf f(x│1/2) and have the joint pdf of

f(x,y) = 3.3 - y , 0 < y < < 1

Iam not sure how to set up the limits of integration to get f(x│y) because of the $x^2$.

Any help how be apprectiated

2. Originally Posted by Number Cruncher 20
Hi,

Iam trying to find the conditional pdf f(x│1/2) and have the joint pdf of

f(x,y) = 3.3 - y , 0 < y < < 1

Iam not sure how to set up the limits of integration to get f(x│y) because of the $x^2$.

Any help how be apprectiated
$f(x | y) = \frac{f(x, y)}{f_Y(y)}$.

The joint pdf is non-zero for values of X and Y lying in the region of the XY-plane enclosed by x = 0, x = 1, y = 0 and y = x^2. Therefore the marginal density function of Y is

$f_Y(y) = \int_{x = +\sqrt{y}}^{x=1} f(x, y) \, dx$.

3. Originally Posted by mr fantastic
$f(x | y) = \frac{f(x, y)}{f_Y(y)}$.

The joint pdf is non-zero for values of X and Y lying in the region of the XY-plane enclosed by x = 0, x = 1, y = 0 and y = x^2. Therefore the marginal density function of Y is

$f_Y(y) = \int_{x = +\sqrt{y}}^{x=1} f(x, y) \, dx$.
My mistake. It should be

$f_Y(y) = \int_{x = +\sqrt{y}}^{x=1} f(x, y) \, dx + \int_{x=-1}^{x = -\sqrt{y}} f(x, y) \, dx = 2 \int_{x = +\sqrt{y}}^{x=1} f(x, y) \, dx$ (since f(x, y) is symmetric in x).

And by now you've realised that it's 1.8 and not 3.3 (see my other mistake in the other post)