Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

There is a theorem (Khinchin/Khintchine 1938) stating that a distribution is unimodal with mode at zero iff it is the distribution function of the product of two independent random variables one of which is uniform on (0,1). I am looking for a corresponding result for a unimodal distribution with mode at c > 0. This is not a homework. Thanks a lot in advance

share|cite|improve this question
Consider $X= U V$, where $U$ and $V$ are iid $U \sim \mathcal{U}_{(0,1)}$. Clearly $X$ is of the form you described. Yet $X>0$ with probability one, thus $X$ can not have mode at zero. Did you mean uniform on $[-1,1]$ instead? – Sasha Oct 16 '12 at 18:34
[Second edit] Sasha, thank you for looking into my question. The definition of the mode allows for a discontinuity at the mode. Khinchin's (sometimes spelled as Khintchine's] theorem is for U(0,1). To clarify, I am wondering whether there exists a theorem involving a decomposition of a unimodal distribution with a mode at c>0 into a product of two independent r.v. And/or, conversely, whether a product of two independent r.v., one of which is uniform on (a,b) with a>0 produces a unimodal distribution. Many thanks :) – TAK Oct 18 '12 at 15:14

There is a beautiful paper on the subject; both results are true, one can write X=WU, W\in R and U\in [0,1] and X=W_1U_1, W_1\in R^+ and U_1 \in [-1,1]; counter example does not hold, because Uniform is not unimodal in the sense of Khintchine's theorem.

M. C Jones (2002) On Khintchine's Theorem and its Place in Random Variate Generation, The American Statistician, 56:4, 304-307, DOI: 10.1198/000313002588

Yogen Chaubey

share|cite|improve this answer

I think a result like this is unlikely to hold for $c\gt0$. The reason that it holds for $0$ is that the contribution to the resulting distribution from each value $x$ of the unspecified distribution is spread over $[0,x]$, so whatever it contributes at some point it also contributes at all points closer to $0$ on the same side. You can't achieve the same effect with some other interval, since bounds other than $0$ change under multiplication.

A counterexample for the product of any distribution with a uniform distribution on $(a,b)$ with $a\gt0$ yielding a unimodal distribution is afforded by two peaks sufficiently far apart at $x_1$ and $x_2$ that their contributions at $[ax_1,bx_1]$ and $[ax_2,bx_2]$ don't overlap and hence form two separate modes. This effect doesn't occur for $a=0$ because in this case the contributions always overlap and add up at $0$.

share|cite|improve this answer
Joriki, many thanks, very intuitive. Do you think whether for a product with U(a,b) there will be at most two modes? Or would that depend on the other variable? And another question is what would be the decomposition for a distribution with a single mode at c>0? Many thanks – TAK Oct 19 '12 at 22:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.