Consider two unimodal probability density functions $ f(x)$ and $ g(x)$ on $\mathbb{R}$, both symmetric around their modes $\mu_f$ and $\mu_g$ which are also their means and medians.

Given the density function obtained as their normalized product $$ h(x) = \frac{f(x) g(x) }{\int dy f(y) g(y)},$$ prove that its mean $\mu_h$ lies between $\mu_f$ and $\mu_g$.

Edit: I began by considering what would happen with two gaussians. According to my calculation, the result is a gaussian with mean

$$ \mu = \frac{\mu_f \sigma_f^2 + \mu_g \sigma_g^2 }{ \sigma_f^2 + \sigma_g^2} $$

and with variance

$$ \sigma^2 = \frac{\sigma_f^2 \sigma_g^2}{\sigma_f^2 + \sigma_g^2} $$

From the above formula for $\mu$ it's easy to see that indeed it lies between $\mu_f$ and $\mu_g$.

But how to show that it's true with all unimodal distributions?

  • $\begingroup$ Post your efforts and attempts to solve the problem, that way, users can respond based on your level of understanding and not just give you the answer. $\endgroup$ Commented Apr 20, 2022 at 4:33

1 Answer 1


There are counter examples: \begin{align} f(x)&=\Big(-(x+25)^2+1\Big)^++\Big(-(x+17)^2+5\Big)^++\Big(-(x+9)^2+1\Big)^+\,,\\ g(x)&=\Big(-(x+25)^2+1\Big)^++\Big(-x^2+5\Big)^++\Big(-(x-25)^2+1\Big)^+\,. \end{align} With normalizations such that all integrals $\int_{-\infty}^{+\infty}f(x)\,dx$, $\int_{-\infty}^{+\infty}g(x)\,dx$, $\int_{-\infty}^{+\infty}g(x)f(x)\,dx$ are one the expected values are $\mu_f\approx -17.0,\mu_g=0\,,\mu_h\approx\color{red}{-25.0}$. This is also intuitively clear when one looks at the plots:

enter image description here

Note that $f(x)$ and $g(x)$ overlap only around the left most hump at $-25$.


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