Distribution with mean $\mu$ and variance $\sigma^2$

Question

Suppose there is a p.d.f. $f(x)$ with mean $\mu$ and variance $\sigma^2$. If we want to compute $$g(a)=\int_{-\infty}^{+\infty} [x-a]^+ f(x) dx ,$$ is it possible for to write $g(a)$ as a function of $\mu$ and $\sigma$? Currently I can obtain $$ g(a)=\int_{a}^{+\infty} (x-a) f(x) dx = \int_{a}^{+\infty}xf(x)dx - a \int_{a}^{+\infty}f(x)dx \\ \qquad\qquad\qquad\qquad\quad=\mu-\int_{-\infty}^{a}xf(x)dx-aF(a) $$

Answer 1

No. Take $X$ to be $N(0,1)$ and $Y=\sqrt {12} (U-\frac 1 2)$ where $U$ is uniformly distributed on $(0,1)$. Both have mean 0 and variance 1. The values of $g(0)$ for these two random variables are different. Hence $g$ cannot be expressed as a function of $\mu$ and $\sigma$.

Answer 2

Not only can this not be expressed as a function of $\mu$ and $\sigma$, but for any value of $a$ greater than or equal to $\mu$ you can't even find a positive lower bound for $g(a)$ if $a\ge\mu$. Just take a distribution on 3 points, one of which is $\mu$ and the other of which are symmetric about $\mu$, with equal probabilities assigned to the extreme points. Let those extreme points fly off to infinity while holding $\sigma$ (and obviously $\mu$) constant. g(a) will tend toward 0 for all a, since the probability of being at the positive extreme point has to decrease quadratically to keep the variance constant, and the integrand in the definition of $g$ is merely linear.

A fun exercise would be to find an upper bound for $g(a)$ given $\mu$ and $\sigma$.