$P(X+Y>t\mid aX+bY>u)\geq P(X+Y>t)$? Let $X,Y$ be independent continuous random variables with support on some neighbourhood of infinity. Let $a, b\geq 0$, and $t,u\in\mathbb{R}$ be constants. Then it holds that $$P(X+Y>t\mid aX+bY>u)\geq P(X+Y>t). $$
Do you know how to prove this? it sounds very intuitive and easy, but all that I tried failed.
 A: For simplicity we assume $a>0$ rather than only $a\geq0$.
By Bayes' formula, we may equivalently show that
\begin{align*}
\mathbb{P}(aX + bY > u \, | \, X + Y > t) \geq \mathbb{P}(aX + bY > u).
\end{align*}
Using the tower property in both the numerator and the denominator,
\begin{align*}
\mathbb{P}(aX + bY > u \, | \, X + Y > t)
&=
\dfrac{\mathbb{P}(aX + bY > u, X + Y > t)}{\mathbb{P}(X+Y>t)} \\
&=
\dfrac{\mathbb{E}[\mathbb{P}(aX + bY > u, X + Y > t \, | \, Y)]}{\mathbb{E}[\mathbb{P}(X + Y > t \, | \, Y)]}.
\end{align*}
Similarly,
\begin{align*}
\mathbb{P}(aX + bY > u) = \mathbb{E}[\mathbb{P}(aX + bY > u \, | \, Y)].
\end{align*}
Since $X$ and $Y$ are independent and $a>0$,
\begin{align*}
\mathbb{P}(aX + bY > u, X + Y > t \, | \, Y)
&=
S_X(Y_1 \vee Y_2), \\
\mathbb{P}(X + Y > t \, | \, Y) 
&= S_X(Y_2), \\
\mathbb{P}(aX + bY > u \, | \, Y)
&= S_X(Y_1),
\end{align*}
where $S_X(x) := 1 - \mathbb{P}(X \leq x)$ and $x_1 \vee x_2 = \max\{x_1,x_2\}$, while
\begin{align*}
Y_1 = \dfrac{u - bY}{a}, \hspace{5mm} Y_2 = t-Y.
\end{align*}
Note that $Y_1$ and $Y_2$ are functions of $Y$, and thus $S_X(Y_1 \vee Y_2)$, $S_X(Y_2)$, and $S_X(Y_1)$ are also random variables (more precisely, transformations of $Y$).
Collecting terms, we want to show that
\begin{align*}
\dfrac{\mathbb{E}[S_X(Y_1 \vee Y_2)]}{\mathbb{E}[S_X(Y_1)]\mathbb{E}[S_X(Y_2)]} \geq 1.
\end{align*}
Now obviously, $S_X(Y_1 \vee Y_2) \geq S_X(Y_1)S_X(Y_2)$. So it suffices to show that
\begin{align*}
\text{Cov}[S_X(Y_1),S_X(Y_2)] = \mathbb{E}[S_X(Y_1)S_X(Y_2)] - \mathbb{E}[S_X(Y_1)]\mathbb{E}[S_X(Y_2)] \geq 0.
\end{align*}
Note that $S_X(Y_1)$ is a non-decreasing function of $Y$ since $Y_1$ is a non-increasing as a function of $Y$ (recall $b\geq0$) and $S_X$ is a survival function and thus non-increasing. Similarly, $S_X(Y_2)$ is a non-decreasing function of $Y$. Consequently, see e.g. this question, we must have $\text{Cov}[S_X(Y_1),S_X(Y_2)] \geq 0$ as desired.
A: Is this a valid counterexample?

