# Expectation formula

Let $F(z)=P\{ Z \leq z\}$. Assume $F(c)=0$. It is well known that: $$E(Z)=\int_c^{\infty}(1-F(z))dz$$ and more generally: $$E(g(Z))=\int_c^{\infty}g(z)dF(z)$$ Is it also true for: $\tilde{F}(z)=P\{ Z < z\}$, i.e. $$E(g(Z))=\int_c^{\infty}g(z)d\tilde{F}(z)$$

How to prove this equality and more generally when these equalities holds? i.e. what are the regular conditions?

• Well, $\bar{F}$ differs from $F$ at $z$ if and only if $P(Z=z) >0$. So, start by looking at distributions which contain point masses. – Batman May 2 '16 at 11:00
• $F(z)-\tilde{F}(z)$ is non-zero only at mass points which are at most countable. So the integral should be the same. I am looking for a reference on this riemann stieltjes integral. – nir May 2 '16 at 11:05
• $Z$ is a continuous random variable? – BCLC May 2 '16 at 22:30