Course 1 - Conditional Variance of y Question

[Al]

Hi,

A few of us that are taking course 1 are stumped on a question in the Averbach study manual - Practice Test 1, question 5.

Looking at the solution, we cannot derive how they came up with that simple of a formula to solve the problem. Can anyone help explain how they came up with this?

The problem is:

Let X and Y have a bivariate normal distribution with mean of x and y = 0 and variance x = 1 and variance y = 2, and correlation coefficient = 1/2. What is the conditional variance of Y, given X = x?

The solution is:

The conditional variance of Y given X = x, i.e., Var[YlX=x) = var(1-p^2) = 2[1-(1/2)^2] = 3/2.

[msqa]

var(Y | X = x) = σ2^2 (1 − ρ^2).

= 2*( 1-0.5^2)

= 3/2

[Sladewski]

I don't have the Averbach study manual and I am clueless about where to begin with this problem.

Could you send me some details?

I'd really appreciate it.

Steve Sladewski

[Al]

I'd be happy to send details, but I'm not exactly sure what you're looking for - all of the information I have related to this problem is what I originally wrote, which is why we were so confused as to how they got their answer....Let me know what you're looking for and I'll do what I can to get it to you!

[msqa]

Hi Al,

the conditional distribution of Y given X=x is normal with this variance. it can be found out by integrating joint distribution over the range if x and finding the conditional distribution by dividing joint by marginal. by making certain manipulations (to make the conditional distribution look like a p.d.f. of normal distribution) you can observe that it has variance that is provided in the solution. to find proof of this u can see some standard book in probability and statistics say by hogg & craig. i hope it helps!

msqa