Effect of conditioning on quantiles Cross posted on Cross Validated https://stats.stackexchange.com/questions/565661/effect-of-conditioning-on-quantiles
Suppose that we have three continuous, independent, non-negative random variables $X,Y,Z$. Fix $q\in(0,1)$ and suppose that $$q=\mathbb{P}(X+Y\leqslant \tau)=\mathbb{P}(X+Y+Z\leqslant \pi).$$
For $\mu>0$ such that $\mathbb{P}(X\leqslant \mu)>0$, is it true that $$\mathbb{P}(X+Y\leqslant \tau|X\leqslant \mu) \geqslant \mathbb{P}(X+Y+Z\leqslant \pi|X\leqslant \mu)?$$
In other words, what effect does replacing the unconditional sums $X+Y, X+Y+Z$ with $(X+Y)|(X\leqslant \mu)$, $(X+Y+Z)|(X\leqslant \mu)$ have on the quantiles?
 A: Here is a handwaving argument for the inequality, illustrated with an example using identical uniform distributions but I thing it could be extended to the general case:
To have $q=\mathbb{P}(X+Y\leqslant \tau)=\mathbb{P}(X+Y+Z\leqslant \pi)$ you need $\tau$ to be a tighter constraint than $\pi$ as shown in the pink and cyan here: remember that a diagonal line from the vertices across a square cuts off half the area while a similar diagonal plane from the vertices across a cube cuts off a third of the volume.

So when you condition on $X \leqslant \mu$, the removed part has a lower ratio of original probability corresponding to $X+Y\leqslant\tau$  in the 2D case than corresponding to $X+Y+Z\leqslant \pi$  in 3D when compared to their complements of $X+Y\gt \tau$ of $X+Y+Z\gt \pi$.  This has the direct consequence that in the retained part (illustrated in green) there is a higher ratio in the 2D case than the 3D case; both ratios increase but the in the 2D case it increases more.  This means $\mathbb{P}(X+Y\leqslant \tau|X\leqslant \mu) \geqslant \mathbb{P}(X+Y+Z\leqslant \pi|X\leqslant \mu)$

