Joint Distributions CS1 04 Page 54

[Andrew Brown]

The solution of E[XY] is double integral of 4/5(3x^3y +x^2*y^2)dydx

Is the XY the product of the two marginal density functions fY and fX?

how is XY arrived at?

 

[CapitalActuary]

I don't have the reading I'm afraid, thought I'd try and point you in the right direction anyway.

XY will be the product of two random variables X and Y, and the density function for XY won't just be the product of the two densities (apart from in special cases).

How I'd derive E[XY] depends on what X and Y are, and whether they are independent or not. But you can use the law of total expectation to write E[XY] = E[Y * E[X|Y]]. This might help (it's a double integral over the density of Y, and the density of X given Y). If X and Y are independent then E[XY] = E[X] * E[Y].

Or if you have the joint density function f_XY(x, y) you can just integrate x*y*f_XY(x, y) (which is E[XY] by definition).

 

[Andrew Brown]

I don't have the reading I'm afraid, thought I'd try and point you in the right direction anyway.

XY will be the product of two random variables X and Y, and the density function for XY won't just be the product of the two densities (apart from in special cases).

How I'd derive E[XY] depends on what X and Y are, and whether they are independent or not. But you can use the law of total expectation to write E[XY] = E[Y * E[X|Y]]. This might help (it's a double integral over the density of Y, and the density of X given Y). If X and Y are independent then E[XY] = E[X] * E[Y].

Or if you have the joint density function f_XY(x, y) you can just integrate x*y*f_XY(x, y) (which is E[XY] by definition).

thanks A lot for your reply. You provided clear answer . This is the trend I observed from the notes and from the flash cards issued but the ActEd . I just needed a solid basis for the answer. Once again . I am grateful for your answer.