1. ## Need Help With Joint Density Problem Please!

I got the problem

2. OK, to find (a) you'll need to calculate the marginal densities of Y1 and Y2. You have the joint density, but you'll need the density of Y1 and the density of Y2 so you can calculate E(Y1), the expected value of Y1, and E(Y2), the expected value of Y2. Are you familiar with this?

Note, you'll also use the marginal density of Y2 to solve for your conditional distribution in part (c).

Hope that helps. Good luck!

3. I have but its been a while can you explain or set up the equation?

4. OK, to find the marginal density of Y1, for example, you integrate the joint density function with respect to Y2, so dy2.

And for the marginal density of Y2, you integrate the joint density with respect to Y1.

I don't like using Y1 and Y2 because I'm used to using X and Y, but that's just me.

Now, you have to be careful on your limits when integrating. Fortunately, the region of positive density (given by your problem's joint density: 0 ≤ y1 ≤ 1 , 0 ≤ y2 ≤ 1) is an easy region to integrate. If you draw the graph you'll find this region is a square. So, you'll integrate from 0 to 1 for dy1 and dy2 (dy1 is like saying dx and dy2 is like dy if that helps you).

For part (b) check in your notes or book about variances. Don't hold me to this because I'm going off memory, but I think V(3Y1 - 2Y2) = 3^2Var(Y1) - 2^2Var(Y2). I think you square the constant term in front of your variance. Again, check this, but maybe the hint helps.

5. ......................

6. OK, to solve for (a) you need to find your marginal densities. For the sake of typing and for clarity, let me rewrite the problem with easier variables:

let Y1 = X
let Y2 = Y

You don't have to do this, but I find it easier to follow because I'm used to the xy-plane.

Your joint density is f(x, y) = X + Y, in the region 0 < x < 1, 0 < y < 1.

The marginal density of X, fx(x) = ∫x + y dy, integrate from 0 to 1. When you perform the integration with respect to y, your function becomes xy + y^2/2. Solving from 0 to 1 and you're left with x + 1/2.

The marginal density of Y, fy(y) = ∫x + y dx, integrate from 0 to 1. Note, you'll get the same answer as the marginal density of x, but with a y in its place: y + 1/2.

Now, to solve for E(X), the equation is ∫x*fx(x) dx --> ∫x^2 + x/2 dx, from to 1.

To solve for E(Y), the equation is ∫y*fy(y) dy. Of course, go back and replace x with Y1 and y with Y2.

Did that help? Please let me know if you have any questions.

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts