I have to solve the following problem:
The real random variables $X$ and $Y$ are independent and have a uniform distribution $U([0,1])$. Find
$$\mathbb{E}\left( \frac{3 X-Y+1}{\sqrt{X+Y+1}} | \quad e^{X+Y}-20 \log(X+Y) \right)$$
Answer: $\sqrt{X+Y+1}$.
My solution: In the condition there is the function $f(y)=e^y-20 \log(y)$. Its derivative is as follows: $f'(y)=e^y-\frac{20}{y}$. Let us note that for $y\in (0,2]$ this derivative is negative, i.e. $f$ is strictly decreasing in $(0,2]$ and $0 < X+Y < 2$ a.s. Hence follows that $$\sigma\left( e^{X+Y}-20 \log(X+Y) \right)=\sigma(X+Y),$$ so we have $$\mathbb{E}\left( \frac{3 X-Y+1}{\sqrt{X+Y+1}} | \quad e^{X+Y}-20 \log(X+Y) \right)=\mathbb{E}\left( \frac{3 X-Y+1}{\sqrt{X+Y+1}} | X+Y \right)$$ Hence $$\mathbb{E}\left( \frac{3 X-Y+1}{\sqrt{X+Y+1}} | X+Y \right)=\mathbb{E}\left( \frac{2 X-2Y+X+Y+1}{\sqrt{X+Y+1}} | X+Y \right)=$$ $$=\mathbb{E}\left( \frac{X+Y+1}{\sqrt{X+Y+1}} | X+Y \right)+2 \mathbb{E}\left( \frac{X-Y}{\sqrt{X+Y+1}} | X+Y \right)=$$ $$=\sqrt{X+Y+1}+\frac{2}{\sqrt{X+Y+1}}\mathbb{E}(X-Y|X+Y).$$
Now, I am not far from a complete solution, but I don't know how to prove that $\mathbb{E}(X-Y|X+Y)=0$. In the case of normality $N(0,1)$, it wouldn't be a problem because $$\mathrm{cov}(X+Y,X-Y)=E((X+Y)(X-Y))-E(X+Y)E(X-Y)=$$ $$E((X+Y)(X-Y))=E(X^2-Y^2)=1-1=0,$$ and for the normal distribution zero correlation implies indepedence hence we would have $\mathbb{E}(X-Y|X+Y)=\mathbb{E}(X-Y)=0$.
But how is it for the case of uniformity????
So basically the thing you want to prove is that:
$$\mathbb{E}(X-Y|X+Y)=0$$, where $X, Y$ are independent uniformly distributed random variables.
This is easy to see, since $\mathbb{E}(X|X+Y)=\mathbb{E}(Y|X+Y)$ due to the symmetry of the problem (i.e. just relabel $X$ and $Y$)! Now just use linearity of conditional expectation to finish the proof.
In particular, the only thing this proof needs is that $X$ and $Y$ are i.i.d. random variables in $\mathcal{L}^1$.