# Is the pth percentile of a mixture distribution a weighted average of the pth percentiles of it's components?

Suppose Y is a mixture distribution defined as follows:

$Y=w_1X_1+w_2X_2+w_3X_3;&space;w_1+w_2+w_3=1$

If $X_1,&space;X_2,&space;X_3$ are beta distributed random variables (making Y a mixture of three beta distributions); and $Y^{[0.95]},&space;X_{1}^{[0.95]}$ are the 95th percentiles of the densities of Y and X1 respectively (similarly for X2 and X3), is it necessarily true that:

$Y^{[0.95]}&space;=&space;w_1X_{1}^{[0.95]}+w_2X_{2}^{[0.95]}+w_3X_{3}^{[0.95]}$

And does this hold for any rv's X1, X2, X3 that are defined over a positive range? Intuitively it makes sense that it does, but there may be something I haven't considered.

Thanks.

• Consideration of cases where there are just two variables and one of them is certainly less than the other should quickly and clearly indicate why such a formula cannot possibly be correct.
– whuber
Commented Aug 8, 2017 at 17:05
• Yes, you are of course correct (sorry, it's 3AM here) Commented Aug 8, 2017 at 17:21

In general, NO.

Consider the simpler case of two different (independent) beta variables -- say a beta(1,2) and a beta(2,1). In that particular case a 50-50 mixture is standard uniform, which has 0.95 quantile equal to 0.95.

The weighted average of the two quantiles of the components is smaller than that (it's about 0.8755):

> qbeta(.95,1,2)
[1] 0.7763932
> qbeta(.95,2,1)
[1] 0.9746794


(This was the first example I tried, but knowing that quantiles are not linear in general, I expected almost any random example should fail unless it was a special case.)

[whuber's example in comments is an especially simple and compelling one -- but I wanted to show that it was still clearly the case for beta random variables - which overlap completely, since they're all on the unit interval]