Is the sum of independent unimodal random variables still unimodal? Is the sum of independent unimodal random variables still unimodal? If yes, can you please give me some hint on why this holds? If no, can you show me some counter-example and suggest under what condition the sum remains unimodal? Thank you in advance.
 A: In addition to Henning's answer, here is a continuous distribution example of unimodal density such the sum is not unimodal:
$$
   f_X(x) = \frac{1}{182} \max\left( \frac{128}{x^2}, 42 - 5x \right) \mathbf{1}_{x \ge 1}
$$

The density $f_{X+Y}(z)$ is unsightly, so its explicit form is suppressed in the snapshot.
A: It's not true in general. Consider a discrete case where $P(X=0)=\frac12$ and $P(X=i)=\frac1{2n}$ for $1\le i\le n$, and let $Y$ have the same distribution.
Then $$P(X+Y=0)=\frac14$$
$$P(X+Y=1)=2\frac 12\frac1{2n}=\frac1{2n}$$
$$P(X+Y=n)=\frac1{2n}+\frac{n-1}{(2n)^2}$$
so the distribution of the sum is not unimodal (in the sense that the pdf has only one local maximum). This counterexample can be approximated by a smooth continuous distribution too.
A: As Henning Makholm points out, the result is not true in general.  I believe that if the independent random variables have identical unimodal distributions that are symmetrical about the mode, the sum will have unimodal distribution that is symmetric about the mode, but I don't have a proof worked out in detail.  The unimodality should follow from convolution and the Cauchy-Schwarz inequality.
