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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.

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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.)

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That was pretty simple. –  robinson Nov 30 '10 at 9:05
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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
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@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
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