Suppose we have $X_1,X_2,...X_n\overset{\text{iid}}{\sim}f(x-\mu)$ where, $f$ is symmetric about $0$. So, $\mu=$ median of $X$ and (if it exists) mean of $X$.
Since, $f(x)$ is symmetric around $0$, $f(x-\mu)$ is symmetric around $\mu$, its median. But, why must this be equal to the mean of the population, if it exists?
This is probably a really simple doubt but I am not totally convinced for some reason? Maybe a small explanation of it would help?
On
Since $f(x)$ is symmetric, the probability of $x$ is equal to the probability of $-x$, so these terms cancel out when calculating the mean. This is true of all $x$, so the mean is $0$. The same argument can be applied to $f(x - \mu)$, where the probability of $\mu + x$ is the same as $\mu - x$, and the $x$ terms all cancel in the same way to leave a mean of $\mu$.
I am assuming $f$ is a valid density function. If the mean exists, we can write:
$$\mathbb{E}[X] = \int_{-\infty}^\infty xf(x-\mu)dx = \int_{-\infty}^{\infty} (u+\mu)f(u)du = \int_{-\infty}^\infty uf(u)du + \mu\int_{-\infty}^\infty f(u)du$$
where we used the change of variable $u = x-\mu$. The first integral is equal to $\mathbb{E}[Y]$ where $Y\sim f$. This is equal to $0$ since $f$ is symmetric around $0$ (you could see it either by splitting the integral or by noticing that $Y\stackrel{d}{=}-Y$). The second integral is equal to $1$ since $f$ is a density function. This shows that $\mathbb{E}[X] = \mu$.