# Calculate expected value and variance of d, d = sqrt(x^2+y^2)

1. Jul 14, 2011

### tschoni

I'm bad at stochastics so really glad for any help

1. The problem statement, all variables and given/known data

I have two normally distributed NON INDEPENDENT stochastic variables X~N(muX,sigX^2) and Y~N(muY,sigY^2)
A third variable D is defined as D = sqrt(X^2 + Y^2).
Since Y and X are stochastic D will also be stochastic.

2. Relevant equations

But how to calculate expected value muD and variance sigD^2 properly?

3. The attempt at a solution

Calculate the variance sigD^2
D = sqrt(X^2 + Y^2) (1)
D^2 = X^2 + Y^2 (2)
E[D^2] = E[X^2 + Y^2] = E[X^2] + E[Y^2] (3)
sigD^2 = E[(D - muD)^2] = E[D^2] - muD^2 (4)
Using (3) in (4)
sigD^2 = E[X^2] + E[Y^2] - muD^2 (5)
sigD^2 = E[X]^2 + Var[X] + E[Y]^2 + Var[Y] - muD^2 (5)
sigD^2 = muX^2 + sigX^2 + muY^2 +sigY^2 - muD^2 (6)
So far so good.
Calculate muD
At this point I thought I'm done.
But what about muD? I dont have it!
At first tempting to assume
muD = sqrt(muX^2+muY^2)
But I dont think it's true, since X and Y are not independent. And even if it is true, how to show it.
If I start out
muD = E[D] = E[sqrt(X^2 + Y^2)]
I dont manage to come to a solution.

Really appreciate any help

2. Jul 15, 2011

### Ray Vickson

I think that finding Ed (d = sqrt(X^2 + Y^2) ) is possible only numerically. Your hoped-for result Ed = sqrt(muX^2 + muY^2) is false: it is an example of the so-called "fallacy of averages". In (x,y) space the function f(x,y) = sqrt(x^2 + y^2) is *convex*, so E f(X,Y) >= f(EX,EY), and for a spread out distribution like yours (which straddles the origin) the inequality will be strict. Google "Jensens inequality" for more on this.

RGV

Last edited: Jul 15, 2011
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook