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.

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The key concept (and term) here is strong unimodality. For continuous distributions, see I. A. Ibragimov (1956), On the composition of unimodal distributions, Theory Prob. Appl., vol. 1, pp. 255-260, and for the discrete counterpart, J. Keilson and H. Gerber (1971), Some results for discrete unimodality, JASA, vol. 66, no. 334, pp. 386-389. – cardinal Oct 11 '11 at 3:09

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.

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Nice work. This counterexample also (almost) demonstrates that something like "smooth with exactly two inflection points" would not be a sufficient condition either. – Henning Makholm Oct 7 '11 at 20:23

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.

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Seems this does not hold for all cases. Is there any hint on under what condition it holds? Examples also abound for it to hold, e.g., both X and Y are constants. – sinoTrinity Oct 7 '11 at 17:55

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.

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