conditional expectation - problem with solution at the finish

Question

I'm not sure how to finish the excercise:

Let $\Omega=[0,1]^2$, $P=dxdy$ - Lebesgue measure on $\Omega$. Let $X(x,y)=x$ and $Y(x,y)=y$. Compute $E(f(x,y)|G)$, when $f(x,y)=x^2-y^2$ and $G=\sigma(X+Y)$.

So:

$E(f(x,y)|\sigma(X+Y))=E(X^2-Y^2|X+Y)=E((X+Y)(X-Y)|X+Y)$

now, since $X+Y$ is $\sigma(X+Y)$-measurable we get

$(X+Y)E(X-Y|X+Y)=(X+Y)E(X-Y+2Y-2Y)=(X+Y)E(X+Y|X+Y)-2E(Y|X+Y)=(X+Y)^2-2E(Y|X+Y)$

And now I'm not sure. Is $Y$ $\sigma(X+Y)$-measurable or not?

Answer 1

I am suggesting here a different approach:

1. First, it is easy to see that the couple (X,Y) has a uniform distribution on the square defined by the point $A = (0,0), B=(1,0), C=(1,1)$ and $D=(0,1)$. In particular $X$ and $Y$ are independent uniform variables on the interval $[0,1]$

2. The transformation $(x,y) \mapsto (x+y,x-y)$ is the composition of

   a. The axial symmetry of axis (AC)
   b. The rotation of center A and angle 45° clockwise.
   c. The homothety of center A and ratio $\sqrt{2}$
   

   And maps the square (ABCD) to the square (AEFC) which we will denote by the domain $\mathcal{D}$. Note that each of the three transformation, maps a uniform distribution to uniform distribution.

We deduce that the couple $(U,V) := (X+Y, X-Y)$ is having a uniform distribution on the domain $\mathcal{D}$ of density $f(u,v) = \frac{1}{2} 1_{ (u,v)\in\mathcal{D}}$

Intuitively studying $E[X-Y|X+Y] = E[V|U]$ consists in looking at a vertical band centered on a fixed $u$, of infinitesimal width $du$ and averaging the values of the ordinates of the points inside that band (in green in the figure). By the symmetry of the figure, we expect the average to be 0.

More formally, note that the variable $U = X+Y$ has the density $f(u) = u 1_{u\in[0,1]} + (2-u) 1_{u \in[1,2]}$ (because the density of the sum of two independent random variables is the convolution product of their densities)

Let us now compute the quantity $E[X-Y|X+Y] = E[V|U]= g(U)$ where

$$ \begin{eqnarray} g(u) &=& \int_{-1}^1 v \frac{f(u,v)}{f(u)} dv \\ & =& \frac{1}{2} \int_{-1}^1 v \frac{f(u,v)}{u} 1_{0\leq u \leq 1} dv + \frac{1}{2} \int_{-1}^1 v \frac{f(u,v)}{2-u} 1_{1\leq u \leq 2} dv \\ &=&\frac{1}{2} \int_{-1}^1 v \frac{1_{-u \leq v \leq u}}{u} 1_{0\leq u \leq 1} dv + \frac{1}{2} \int_{-1}^1 v \frac{1_{u-2 \leq v \leq 2-u}}{2-u} 1_{1\leq u \leq 2} dv\\ &=& 0\\ \end{eqnarray} $$ As the functions we are integrating in every integral are odd functions.

Thus , $E[X^2 - Y^2|X+Y] = E[(X+Y)(X-Y)|X+Y] = (X+Y) E[X-Y|X+Y] = 0$

Answer 2

This one is short but in order to make it rigorous you will need a formula from the theory of regular conditional distribution.

By symmetry of the measure and the subspace $X+Y=z$ with respect to switching the two coordinates $x$ and $y$, you have $$ E[X^2|X+Y] = E[Y^2|Y+X] $$ And therefore the answer is zero.

To make this rigorous, you can consider the regular conditional distribution of $X^2$ given $X+Y=z$, which will be the same as the one for $Y^2$. The regular conditional distribution is parametrized by $z$ and is given by $$ P(X^2\in[a,b]|Z=z) = \lim_\Delta \lim_\delta P(X^2\in[a-\Delta,b+\Delta]|Z\in[z-\delta,z+\delta])$$