# Mean of a normalized product of densities

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?

• 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. Commented Apr 20, 2022 at 4:33

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