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Let $0 < p < 100$. The $p$th percentile of a random variable $X$ is the value $x_p$ which separates the smallest $p\%$ of the values of $X$ from the largest $(100-p)\%$. In probabilistic terms, the $p$th percentile satisfies $$ \mathbb{P}(X \le x_p) = \frac{p}{100}. $$

Now suppose $X_1$ and $X_2$ are two random variables (not necessarily independent). The $99$th percentile of $X_1$ is $100$, and the $99$th percentile of $X_2$ is $200$. Consider the random variable $$ Y = X_1 + X_2. $$ My question is: where can the $99$th percentile of $Y$ lie, that is, the value $y$ such that $$ P(Y \leq y) = 0.99. $$

In the independent case, I know that the distribution of $Y$ is given by the convolution $$ F_Y(y) = \mathbb{P}(X_1 + X_2 \le y) = \int_{-\infty}^{\infty} F_{X_1}(y - x)\, f_{X_2}(x)\, dx, $$ but I am unsure how to use this to deduce the $99$th percentile of $Y$. I am also uncertain how to handle the general case when nothing is known about the dependence structure between $X_1$ and $X_2$. Any clarification, hints, or references would be greatly appreciated. Thank you sincerely for your time and support.

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    $\begingroup$ To get percentile from distribution, you need to solve for $F_Y(y) = 0.99$. But it's not possible to deduce quantile of sum from quantiles of terms, even in independent case - it can be arbitrary high or low. $\endgroup$ Commented Oct 28 at 10:20
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    $\begingroup$ A given percentile of the sum is not determined solely by the percentiles of the components, like for instance the expected value. You need the explicit sum density/distribution function $\endgroup$ Commented Oct 28 at 11:10

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Consider a bimodal distribution that looks like the following for $X_1$ (the small mode has been exaggerated for visual clarity), calibrated such that the 99th percentile is at $x = 100$.

A function with one large bump around x = 80 and a small bump around x = 110

Let $X_2$ be defined similarly.

Assuming $X_1$ and $X_2$ are independent, the distribution of $Y$ will have four modes. The largest leftmost one will be in a "bump" that contains roughly 98% of the integral, the smallest rightmost one will contain roughly 0.01% of the integral, and there will be two modes in-between that each account for about one percent of the integral (these might overlap into a single mode containing 2%). So the 99th percentile of $Y$ will be between the two smaller modes (or inside the single overlap mode).

If the two modes of $X_1$ are at $x_0$ and $x_1$, and the two modes of $X_2$ are at $x_2$ and $x_3$, the four modes of $Y$ will be at (or very close to) $x_0 + x_2$, $x_1 + x_2$, $x_0 + x_3$, and $x_1 + x_3$. By moving the two modes of the distributions of $X_1$ and $X_2$, we can get the two middle modes of $Y$, and therefore the 99th percentile, to appear wherever we want along the number line.

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  • $\begingroup$ I couldn't quite decide whether to use "bump" or "mode", and where to use which, or whether there is a better word for the thing I'm discussing here. $\endgroup$ Commented Oct 28 at 11:30

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