# Find c in E[c|X-Y|] = 1

• Sep 12th 2012, 09:22 PM
Find c in E[c|X-Y|] = 1
Let X and Y by two independent identically distributed normal random variables with mean 1 and variance 1. Find c so that E[c|X-Y|] = 1.
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I went with the 'simple' route
E[c|X-Y|] = cE[X-Y] + cE[X+Y] = 0 + c(1+1) = 2c
And 2c = 1 so c = 1/2
But that was wrong...

I know there's a 1/sqrt(2pi) in the normal distribution but I can't see how that pi can get to the numerator in those integrals...
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• Sep 13th 2012, 04:50 AM
girdav
Re: Find c in E[c|X-Y|] = 1
The problem in what you did is in the first equality about the expectations. How would you justify that?

Hint: what is the distribution of \$\displaystyle X-Y\$?
• Sep 13th 2012, 11:01 AM
Re: Find c in E[c|X-Y|] = 1
Ah ok I made a mistake with the abs val...
E[c|X-Y|] = cE[X-Y] + cE[Y-X] = 0 + 0 = 0

The distribution of X-Y is another normal distribution with mean 0 and variance 2.

Normal Difference Distribution -- from Wolfram MathWorld
• Sep 14th 2012, 09:57 AM