If $X$ and $Y$ are independent, is it true that $\mathbb P (X+Y>x | X+Y>0) \geq \mathbb P (Y>x | Y>0)$ for every $x>0$?

Question

$Y$ is a random variable with symmetric distribution around 0 and both random variables are continuous with existing density

It seems true, because the random variable $X$ should just bring further volatility.

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It basically breaks down to the following: Is it true that

$$\mathbb P (Y \geq 0)\mathbb P (X +Y \geq x) \geq \mathbb P (Y \geq x)\mathbb P (X +Y \geq 0)$$

where $Y$ is symmetric but I do now know how to proceed from here; Oh and for everyone downvoting, you could at least give me a reason to understand why my question is that bad...

Answer 1

The claim is false in general.

Here is a counterexample. Let $X$ be uniform over the interval $[-5,-4]$ and $Y$ be uniform over the interval $[-10,10]$. Then $P(X+Y>7|X+Y>0)=0$, but $P(Y>7|Y>0) = 0.3 > 0$.

Answer 2

No, this is not true. Consider $Y$ such that $\mathbb{P}(|Y| \le x) = 0$. Then $\mathbb{P}(Y > x | Y > 0) = 1$, but we can easily find an $X$ with $\mathbb{P}(X+Y > x | X+Y > 0) < 1$.