Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $Z$ be a random variable with finite mean and consider the function $F: R \rightarrow R$ defined by $$ F(x) = E[\max(x+Z,0)]$$ How to prove that $F(x)$ is a continuous function of $x$?

This is not homework. A textbook I am reading states that $F(x)$ is continuous without providing any explanation, and I can't seem to fill the gap. I would be happy with a hint.

share|cite|improve this question
up vote 2 down vote accepted

If $\lvert x-y\rvert\leq\delta$, then $\lvert (x+Z)-(y+Z)\rvert\leq\delta$, and also $$\lvert \mathrm{max}(x+Z,0)-\mathrm{max}(y+Z,0)\rvert\leq\delta,$$ right? (Notice that $\mathrm{max}(\bullet+Z,0)$ is 1-Lipschitz.)

Hence we also have $\lvert F(x)-F(y)\rvert\leq\delta$.

Remark: In order for $\mathrm{max}(x+Z,0)$ to be well-defined, we should assume that $Z$ is bounded, I guess. (According to the comments this is not neccessary.)

share|cite|improve this answer
That was pretty simple. – robinson Nov 30 '10 at 9:05
Shouldn't it suffice to assume $Z$ has a finite mean? This implies $|Z|$ has a finite mean, which implies that $\int_0^{\infty} P(|Z|>t) dt$ converges, which implies that $\int_0^{\infty} P(Z>t) dt$ converges, which implies that $\int_0^{\infty} P(Z>t-x) dt$ converges for any x. This is equivalent to the finiteness of $E[\max(x+Z,0)]$. – robinson Nov 30 '10 at 9:07
@robinson: It suffices to assume $Z$ has finite mean. Indeed, by conditioning on $Z$, we have $E[\max(x+Z,0)] = \int_{( - x,\infty )} {(x + s)F_Z (ds)}$, where $F_Z$ is the distribution of $Z$. – Shai Covo Nov 30 '10 at 9:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.