
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 < 1x
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(1x) using the limits of 0 to 1x
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 1x I get E(Y) = (1x)^3  3(1x)^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 $\displaystyle f(x) = 3x(1x)$ 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:
$\displaystyle E[Y] = \displaystyle \int_0^{1x} y\;f(y)\;dy$

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(x1) which also integrate to one.

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]

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

You need to work on your integration.
Work properly on figuring out the value of E[Y]

Yes, ino, i have nt done it for a while, thanks btw.
Finally, i've got it 3(1x)^2 / 2

okay so you have"
$\displaystyle E[X]=\dfrac{3}{2}(1x)^2$
$\displaystyle Var(Y) = E[X^2](E[X])^2$
find $\displaystyle E[X^2]$ now, and then the variance

Yep,
E(Y^2) = (1x)^3
Therefore Var(Y) = (1x)^3 [ 1  9(1x)/4 ]
