Conditions under which $\int_{\mathbb{R}} x f_1(c - x) f_2(x) dx \geq 0$

Question

I have the integral $$\int_{\mathbb{R}} x f_1(c - x) f_2(x) dx$$ where $f_1$ and $f_2$ are both symmetric densities (symmetric about $x = 0$) and $c \geq 0$ is a constant. I would like to know if this integral is always nonnegative under these conditions. By breaking into the regions $x \geq 0$ and $x \leq 0$ and using the symmetry of $f_1$, we obtain $$\int_{\mathbb{R}} x f_1(c - x) f_2(x) dx = \int_{0}^{\infty} x (f_1(c - x) - f_1(c + x)) f_2(x) dx$$ From here, it seems like we also need the condition that the densities are decreasing in $|x|$ to guarantee that $f_1(c - x) \geq f_1(c + x)$ for all $x$. Thus, this might not be true for bimodal symmetric densities. I am wondering if I'm missing something obvious, though.

Answer 1

An example that does not hold is when the distributions have multiple peaks that fall on top of each other differently on the left and the right side.

For example

$$P_1(K = k) = \begin{cases} \frac{3}{8} & \quad \text{if} \quad k = -3 \\ \frac{1}{8} & \quad \text{if} \quad k = -2 \\ \frac{1}{8} & \quad \text{if} \quad k = 2 \\ \frac{3}{8} & \quad \text{if} \quad k = 3 \\ \end{cases}$$

$$P_2(K = k) = \begin{cases} \frac{1}{8} & \quad \text{if} \quad k = -3 \\ \frac{3}{8} & \quad \text{if} \quad k = -2 \\ \frac{3}{8} & \quad \text{if} \quad k = 2 \\ \frac{1}{8} & \quad \text{if} \quad k = 3 \\ \end{cases}$$

If you shift one of the distributions by one and multiply then you get that on the one side the high peaks get shifted on top of each other, but on the other side the low peaks.

With the above distributions for $c=1$ (where we can write $f(c-x) = f(x-c)$ and effectively you get a shift of $c$ to the right) you get

$$\sum k P_1(k-1)P_2(k) = \underbrace{-2 \cdot \frac{3}{8} \cdot \frac{3}{8} }_{k=-2} + \underbrace{3 \cdot \frac{1}{8} \cdot \frac{1}{8}}_{k=3} = - \frac{15}{64}$$

^{I use discrete distributions here and a sum instead of an integral. This is just to make the example simple and easy to compute. It will work the same for continuous distributions and you can turn the discrete example into a continuous example by turning each point mass into a tiny continuous uniform distribution around that point mass}

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Sufficient conditions when the integral is positive are:

• When $f_1(x)$ is non-increasing in $\vert x \vert$ as you noted yourselve
• When $f_1(x) = f_2(x)$, then the product $f_1(x−c)f_2(x)$ is symmetric around $x=c/2$, and the integral resembles the integral to compute the expectation of a density that is symmetric around $c/2$ (up to a positive normalisation constant) and will be positive.

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Another way to look at it:

We have

$$\int_{\mathbb{R}} x f_1(x-c) f_2(x) dx = 0$$

for $c=0$ and so we can rewrite:

$$\int_{\mathbb{R}} x f_1(x-c) f_2(x) dx = \int_{\mathbb{R}} x (f_1(x-c)-f_1(x)) f_2(x) dx $$

And in order for this integral to be negative you need to have the difference $f_1(x-c)-f_1(x)$ correlating/interacting in some way with $f_2(x)$.