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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

3 Answers 3

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

In addition to Henning's answer, here is a continuous distribution example of unimodal density such the sum is not unimodal: $$ f_X(x) = \frac{1}{182} \max\left( \frac{128}{x^2}, 42 - 5x \right) \mathbf{1}_{x \ge 1} $$

enter image description here

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

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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