2

Assume the joint distribution for X, Y is also normal. I have no clue how to approach this problem.

Follow up question: Without knowing the joint distribution of X, Y can you still calculate it?

butterbetter
  • 351
  • Note that $E[X|X+Y=1]=E[Y|X+Y=1]$, clearly. Call the common value $E$. Adding gives $2E=E[X+Y|X+Y=1]=1$ whence $E=\frac 12$. – lulu Dec 09 '15 at 02:19
  • 1
    You haven't told us the joint distribution. If, for example, $X$ and $Y$ were independent, we'd know what we need. – Michael Hardy Dec 09 '15 at 02:20
  • @lulu : That's true even of $X$ and $Y$ are not independent, if $(X,Y)$ and $(Y,X)$ both have the same distribution, but as I said we don't have the joint distribution, and we're not even told they're jointly normal, so the word "clearly" is exaggerated. – Michael Hardy Dec 09 '15 at 02:22
  • @MichaelHardy Agreed. I expect that the OP intended to say that $X,Y$ were independent, but you are correct that this ought to be spelled out. – lulu Dec 09 '15 at 02:24

2 Answers2

2

$\begin{align}\mathsf E(X\mid X+Y=1) & = \dfrac{\int_\Bbb R x f_{X,Y}(x,1-x)\operatorname d x}{\int_\Bbb R f_{X,Y}(x,1-x)\operatorname d x} \\[1ex] & = \dfrac{\int_\Bbb R xe^{-x^2/2}e^{-(1-x)^2/2}\operatorname d x }{\int_\Bbb R e^{-x^2/2}e^{-(1-x)^2/2}\operatorname d x} \\[1ex] & =\dfrac{\Big[-\tfrac 1 2 e^{-(x^2+(1-x)^2)/2} -\sqrt\pi\operatorname {erf}(\tfrac 1 2-x)\big/2\sqrt[4]e\Big]_{x=-\infty}^{x=+\infty}}{\Big[-\sqrt\pi\operatorname {erf}(\tfrac 1 2-x)\big/2\sqrt[4]e\Big]_{x=-\infty}^{x=+\infty}} \\[1ex] & = \dfrac{1}{2}\end{align}$

Which can more easily be obtained by symmetry.

Graham Kemp
  • 129,094
1

If we assume that

  • $X,Y$ are independent; or that
  • $X,Y$ are jointly normal and $(X,Y)$ has the same distribution as $(Y,X)$ (a weaker assumption consistent with nonzero correlation); or that
  • $(X,Y)$ has the same distribution as $(Y,X)$ (which can be true with each of them normally distributed, even if they're not jointly normal);

then $$ \operatorname{E}(X\mid X+Y=1) = \operatorname{E}(Y\mid X+Y=1) $$ and $$ \operatorname{E}(X\mid X+Y=1) + \operatorname{E}(Y\mid X+Y=1) = \operatorname{E}(X+Y\mid X+Y=1) = 1, $$ so $$ \operatorname{E}(X\mid X+Y=1) = \frac 1 2. $$

Michael Hardy
  • 1
  • Someone commented using integration, he deleted it somehow but I wanted to ask if this is correct: $E(X|X+Y=1)$ will only depends on the value of Y, so $E(X|X+Y=1) = \int (1-y)dy$ – butterbetter Dec 09 '15 at 02:31
  • @Dreamer I was mistaken, I did not normalize correctly. – Ian Dec 09 '15 at 02:33
  • No. See my answer for how integration would work. Symmetry is the much easier approach. – Graham Kemp Dec 09 '15 at 02:34
  • Thanks! Without knowing X, Y are independent or (X,Y) has the same distribution as (Y, X) can you still calculate it? – butterbetter Dec 09 '15 at 02:44