# Thread: Conditional distribution of continuous random variables

1. ## Conditional distribution of continuous random variables

Hello, I have a question about conditional distributions in the continuous case.

We are given that the joint distribution function of two continuous random variables X and Y is

$\displaystyle f(x,y) = \frac{3}{2}(x^2+y^2)$ for $\displaystyle \;\;0 \leq x \leq 1 , 0 \leq y \leq 1$

The marginal distribution of Y can be found by integrating with respect to x, the marginal distribution of Y is

$\displaystyle f_Y(y) = \frac{3}{2}y^2+\frac{1}{2}$ for $\displaystyle 0 \leq y \leq 1$

The book asks to find the conditional density of X given that Y =.3

This is the answer that is listed in the book

$\displaystyle f_{X|Y}(x|Y =.3) = \frac{f(x,.3)}{f_Y(.3)} = \frac{\frac{3}{2}(x^2+.3^2)}{\frac{3}{2}*.3^2+.5}= \frac{\frac{3}{2}(x^2+.09)}{.635}$

Why is the book just plugging in the value .3 for y? Shouldn't we be integrating since this is continous?

2. ## Re: Conditional distribution of continuous random variables

Hey Jame.

You can think of the conditional distribution as the slice (or cross sectional) distribution where the Y value is fixed. Since you knowe the value of y, you treat it like a constant and substitute it in to get the distribution for the random variable X.

If you think of it as a constant and X being the random variable, then you get a PDF in terms of X only with all the properties (integrates to 1, greater than zero, and so on).

3. ## Re: Conditional distribution of continuous random variables

That makes sense.

I guess I was thrown off by the single value and the fact that the rvs were continous.

If I was given that Y was between 0 and .3, then would I integrate?

Thank you for assistance.

4. ## Re: Conditional distribution of continuous random variables

Yes you would integrate in that situation over the appropriate region.