Let $x,y \geq 0$ be independent random variables with the property that $E[x] = E[y]$. Can I infer that
$$P(x \geq E[x] \mid x + y \geq E[x+y]) = 1/2~?$$
My heuristic reasoning is that, since $x$ and $y$ have equal expectations, the weight should be split "fairly" between them. But that, of course, falls far short of a solution.
How can I analyze this rigorously?
On
No, you can't infer that. For simplicity, assume $E[x]=E[y]=0$. Then your equation reads
$$ P(x\ge0\mid x+y\ge0)=\frac12\;. $$
Consider $x=\pm2$ with probability $\frac12$ each, and $y=\pm1$ with probability $\frac12$ each. Then
$$ P(x\ge0\mid x+y\ge0)=1\;. $$
On
No. Take the case of $X$ being a Bernoulli with parameter $1/3$, and $Y$ being constant equal to $1/3$.
We have $$\mathbb{P}\{ X \geq \mathbb{E}[X] \} = \mathbb{E}[X]= \mathbb{E}[Y] = 1/3\,,$$
but $$ \mathbb{P}\{ X \geq \mathbb{E}[X] \mid X+Y \geq \mathbb{E}[X+Y] \} = \mathbb{P}\{ X \geq 1/3 \mid X \geq 1/3 \} =1\,. $$
An even simpler example would be to take a deterministic $X$, i.e., take for example $X=c$ (a.s), you could take $c$ any number you want. Then, $\mathbb{E}[X] = c$ and $\mathbb{P}(X \geq \mathbb{E}[X]) = 1$ which implies that $\mathbb{P}(X\geq \mathbb{E}[X] | X+Y \geq \mathbb{E}[X+Y]) =1$ because $X\geq \mathbb{E}[X]$ always holds (in the deterministic case).