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 $\displaystyle X-Y$?

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.

I found this page but it's not helping..

Normal Difference Distribution -- from Wolfram MathWorld

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

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... :/