Thread: 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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Answer is sqrt(pi)/2

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 ?

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.

I found this page but it's not helping..
Normal Difference Distribution -- from Wolfram MathWorld

From that site in my last post:
1/[sqrt(2pi(sigx^2 + sigy^2)] * exp(-[u-(muX-muY)]^2/[2(sigx^2 + sigy^2)]

In my question, since sigx^2 = sigy^2 = 1 and muX = muY = 0 this simplifies to:
1/[sqrt4pi] * exp(-[u]^2/[4])

Going backwards like this, I can sort of see that taking the double integral from the original Px-y with these values for mu and sig^2 could have led to:
4/[sqrt(4pi)] * exp(-[u]^2/[4]) where exp(-[u]^2/[4]) = 1
so E[X] = 4c/[sqrt(4pi)] = 2c/[sqrt(pi)]
Setting E[X] = 1 gives the book's answer.

But...I don't see myself getting a question like this right next time and I also didn't even take into account the absolute value. Honestly I have no idea... :/