How can I understand the mean of convoluted Gaussian function is $\mu_1 + \mu_2$.

Question

I struggle to understand the mean of convoluted two Gaussian functions is $\mu_1 + \mu_2$ instead of $(\mu_1 + \mu_2)/2$. Could someone provide visual explaination?

Answer 1

If $X$ and $Y$ are two independent continuous random variables with means $E[X]=\mu_X$ and $E[Y] = \mu_Y$, then the density of their sum $X+Y$ is the convolution of the densities of $X$ and $Y$. In the special case when $X$ and $Y$ are Gaussian random variables, $X+Y$ is also a Gaussian random variable. But, regardless of whether $X$ and $Y$ are Gaussian or not, independent or not, or continuous or not, it is true that $$E[X+Y] = E[X] + E[Y].$$ This result is referred to as the linearity of expectation, and explains why the mean of $X+Y$ is $\mu_X + \mu_Y$, and not the arithmetic average $\frac{\mu_X+\mu_Y}{2}$ that you are hoping to see.

Visually, the densities of independent Gaussian $X$ and $Y$ are bell-shaped curves centered at $\mu_X$ and $\mu_Y$ respectively, and the value of the convolution of these densities at $z$ (a.k.a the numerical value of the density of $X+Y$ at $z$) is obtained by

• "flipping" the density of $Y$ about the origin the result of which operation looks for all the world just like the density of $Y$ centered at $-\mu_Y$ instead of at $\mu_Y$.
• sliding the flipped density to the right by $z$
• multiplying the density of $X$ with the flipped and slid density of $Y$ and integrating the product from $-\infty$ to $\infty$.

If your brain has not already boggled at all these shenanigans and can still visualize a tad more, consider that the maximum value of the integral will occur when the two peaks align, that is, when the flipped density of $Y$ has been slid so that its peak at $-\mu_Y$ has shifted due to the slide so that it coincides with the peak of the $X$ density which is at $\mu_X$, i.e. a total slide of $\mu_X+\mu_Y$. So, the peak of the convolution is at $\mu_X+\mu_Y$ and so the mean of $X+Y$ is $\mu_X+\mu_Y$.