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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2The 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 03:09
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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.
Sasha
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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. – hmakholm left over Monica Oct 07 '11 at 20:23
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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.
hmakholm left over Monica
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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 07 '11 at 17:55
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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.
Dilip Sarwate
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2Here's the proof. Let $f(x)$ and $g(x)$ be the two pdfs, both of which are symmetric about zero. Fix $0<a\le b$. You can show that symmetry and unimodality implies $$\forall x\in \mathbb R,\qquad \big[f(x)-f(x-a-b)\big]\cdot \big [g(a-x)-g(b-x)]\ge 0$$Namely, both factors are positive when $x < (a+b)/2$, and both are negative when $x>(a+b)/2$. If you then integrate that expression over all $x\in \mathbb R$, you get $h(a)+h(-a)-h(b)-h(-b)\ge 0$, where $h(y)=\int f(x) g(y-x)$ is the pdf of the sum. Therefore, $h(a)\ge h(b)$, proving unimodality for the sum. $\square$ – Mike Earnest Sep 05 '21 at 20:11