1. ## Joint p.d.f

Can anyone help with the following question please?

The continuous random variables X and Y have the joint probability density function,

f XY (x,y) = 6x, 0 < x < 1, 0 < y < 1-x

Derive the marginal p.d.f for X and Y, and find the marginal expected values for both X and Y.

For the marginal of X, I got 3x(1-x) using the limits of 0 to 1-x
and for Y I got 3, using the limits of 0 to 1.

And using the limits from 0 to 1, I got E(X) = 1/4

My question is what limits, would I used for E(Y), as when I use 0 to 1-x I get E(Y) = (1-x)^3 - 3(1-x)^2

2. Firstly the marginal pdf of X that you have written is incorrect.

You can check if your odf of X is correct by integrating it from 0 to 1. The pdf should equal to 1.

In this case:

$\displaystyle \displaystyle \int_0^1 f(x)\;dx = 1$

and $\displaystyle f(x) = 3x(1-x)$ does not satisfy the above stated property of the pdf.

Therefore, your expected value of X also happens to be incorrect.

To find the expected value of Y, you would integrate:

$\displaystyle E[Y] = \displaystyle \int_0^{1-x} y\;f(y)\;dy$

3. Thanks, but I thought to find the marginal p.d.f of X, you have to integrate with respect to y.

edit: Sorry my mistake, my integration was incorrect, I got f(x) = -6x(x-1) which also integrate to one.

4. Thats right.

Lastly your calculation of E[Y] is completely wrong.

you have f(y)= 3. Use this and info in post #2 to find E[Y]

5. Sorry I meant Var(Y) = (1-x)^3 - 3(1-x)^2
for E(Y) i got 3(1-x)^2 / 2

Thank you.

6. You need to work on your integration.

Work properly on figuring out the value of E[Y]

7. Yes, ino, i have nt done it for a while, thanks btw.

Finally, i've got it 3(1-x)^2 / 2

8. okay so you have"

$\displaystyle E[X]=\dfrac{3}{2}(1-x)^2$

$\displaystyle Var(Y) = E[X^2]-(E[X])^2$

find $\displaystyle E[X^2]$ now, and then the variance

9. Yep,

E(Y^2) = (1-x)^3

Therefore Var(Y) = (1-x)^3 [ 1 - 9(1-x)/4 ]

10. that looks good.