# Compute quantile of sum of distributions from particular quantiles

Let's assume $N$ independent random variables $X_1, ..., X_N$ for which the quantiles at some specific level $\alpha$ are known through estimation from data: $\alpha = P(X_1 < q_1)$, ..., $\alpha = P(X_N < q_N)$. Now let's define the random variable $Z$ as the sum $Z = \sum_{i=1}^N X_i$. Is there a way to compute the value of the quantile of the sum at level $\alpha$, that is, $q_z$ in $\alpha = P(Z < q_Z)$?

I think that in particular cases, such as if $X_i$ follows a Gaussian distribution $\forall i$ this is easy, but I'm not so sure for the case where the distribution of the $X_i$ is unknown. Any ideas?

• are these $q_i$ estimated from data or theoretically known? Commented Jan 26, 2015 at 15:35
• This is not possible without making specific assumptions about the distributions of the $X_i$. Do you have a family of distributions in mind?
– whuber
Commented Jan 26, 2015 at 17:53
• @chuse the $q_i$ are estimated from data, as the distribution of the $X_i$ is not known but samples are available. I have updated the question with this fact. Commented Jan 27, 2015 at 8:51
• @whuber I have no prior knowledge about the family of distributions the $X_i$ might be following, though data samples are available. Would assuming a family of distributions (aside from Gaussian) make this easier? Commented Jan 27, 2015 at 8:54

$q_Z$ could be anything.

To understand this situation, let us make a preliminary simplification. By working with $Y_i = X_i - q_i$ we obtain a more uniform characterization

$$\alpha = \Pr(X_i \le q_i) = \Pr(Y_i \le 0).$$

That is, each $Y_i$ has the same probability of being negative. Because

$$W = \sum_i Y_i = \sum_i X_i - \sum_i q_i = Z - \sum_i q_i,$$

the defining equation for $q_Z$ is equivalent to

$$\alpha = \Pr(Z \le q_Z) = \Pr(Z - \sum_i q_i \le q_Z - \sum_i q_i) = \Pr(W \le q_W)$$

with $q_Z = q_W + \sum_i q_i$.

What are the possible values of $q_W$? Consider the case where the $Y_i$ all have the same distribution with all probability on two values, one of them negative ($y_{-}$) and the other one positive ($y_{+}$). The possible values of the sum $W$ are limited to $ky_{-} + (n-k)y_{+}$ for $k=0, 1, \ldots, n$. Each of these occurs with probability

$${\Pr}_W(ky_{-} + (n-k)y_{+}) = \binom{n}{k}\alpha^k(1-\alpha)^{n-k}.$$

The extremes can be found by

1. Choosing $y_{-}$ and $y_{+}$ so that $y_{-} + (n-1)y_{+} \lt 0$; $y_{-}=-n$ and $y_{+}=1$ will accomplish this. This guarantees that $W$ will be negative except when all the $Y_i$ are positive. This chance equals $1 - (1-\alpha)^n$. It exceeds $\alpha$ when $n\gt 1$, implying the $\alpha$ quantile of $W$ must be strictly negative.

2. Choosing $y_{-}$ and $y_{+}$ so that $(n-1) y_{-} + y_{+} \gt 0$; $y_{-}=-1$ and $y_{+}=n$ will accomplish this. This guarantees that $W$ will be negative only when all the $Y_i$ are negative. This chance equals $\alpha^n$. It is less than $\alpha$ when $n\gt 1$, implying the $\alpha$ quantile of $W$ must be strictly positive.

This shows that the $\alpha$ quantile of $W$ could be either negative or positive, but is not zero. What could its size be? It has to equal some integral linear combination of $y_{-}$ and $y_{+}$. Making both these values integers assures all the possible values of $W$ are integral. Upon scaling $y_{\pm}$ by an arbitrary positive number $s$, we can guarantee that all integral linear combinations of $y_{-}$ and $y_{+}$ are integral multiples of $s$. Since $q_W \ne 0$, it must be at least $s$ in size. Consequently, the possible values of $q_W$ (and whence of $q_Z$) are unlimited, no matter what $n\gt 1$ may equal.

The only way to derive any information about $q_Z$ would be to make specific and strong constraints on the distributions of the $X_i$, in order to prevent and limit the kind of unbalanced distributions used to derive this negative result.

• Thanks a lot @whuber, for the explaining and the illustrative example. Even though the answer is negative, I can't say this was unexpected. Then I will try to find out which family of distributions suits my data and see if with that I can work out the quantiles of the sum. Commented Jan 28, 2015 at 7:49
• What if the variables were 100% correlated instead? Under Gaussian law, the quantile of the sum would then be equal to the sum of quantiles - is this true in general or all laws, or for some family of laws (alpha-stable?), or is Gaussian an exception? Thank you Commented Oct 5, 2020 at 12:16
• @Confounded When all the variables are correlated, they are almost surely the same variable. The question becomes one of how to combine estimates of quantiles based on estimates from several samples. The interesting case concerns when those samples are independent. A great deal can be said about that even in very general cases (such as when no distributional assumptions are made) starting by generalizing methods to find confidence intervals for the median.
– whuber
Commented Oct 5, 2020 at 13:46

(version 3)

Since you say you have sample data, you could use the following numerical method:

a) fit pdfs to the data for each $$X_i$$ variable...maybe using kernel densities

b) take the DFT (discrete Fourier transform) of each kernel density

c) multiply the DFTs together

d) take the inverse DFT

That would give you an estimate of the pdf of $$Z$$.

It's a standard technique for finding the distribution of the sum of independent random variables, covered by many authors, and not too hard to derive. It's used in the insurance industry for combining distributions of possible insurance claims. For the sake of providing a citation I googled it, and found this one: doi.org/10.1016/S0167-4730(96)00032-X (although I have only read the abstract).

• Because, in this thread, I have demonstrated the solution could be anything, how do you reconcile your recommendation with that? Further, how do you propose the OP would take the DFT of a variable when they have specifically stated the distribution is unknown and the only information they have is a specific quantile?
– whuber
Commented Oct 9, 2022 at 13:10
• @whuber: I'm iterating my suggestion to make it clearer. Hopefully it makes more sense now, but let me know if not. Commented Oct 10, 2022 at 16:17