Consider a function $$g=E[\max(a+X,d+Y)]$$ where $a,d\in R$ and $X$ and $Y$ are independent and identically distributed standardized random variables with mean $\mu$, variance $\sigma^2$, continuous normal density $f$ and cumulative distribution function $F$.

Then, the quantity: $$\Phi(b)≡\frac{∂g}{∂a}=\int \int_ {-\infty}^b f(X)f(Y)dYdX=\int F(b)f(X)dX$$ where $b=a-d+ X$.

How can one interpret $\Phi$ , if, for instance $g$ indicates the time at which a meeting between two people starts.

  • $\begingroup$ Be aware that $\partial g/\partial a$ is a function of $(a,d)$, certainly not of $b$. You might want to explain. $\endgroup$ – Did Jan 25 '13 at 13:48
  • $\begingroup$ @Did That is true. I used the notation to conserve space. That is $$\Phi= \int f(X) \left(\int _{-\infty}^{a-d+X}f(Y)dY\right)dX$$ and $F(a-d+X)$ denotes $$ \int _{-\infty}^{a-d+X}f(Y)dY$$ I hope my calculations is correct. $\endgroup$ – Daniel Lårs Jan 25 '13 at 14:13
  • $\begingroup$ Then what is the meaning of $\Phi(b)=\partial g/\partial a$? $\endgroup$ – Did Jan 25 '13 at 14:23

Since $\max(u,v)=v+(u-v)^+$, $$ g(a,d)=\mathbb E(d+Y+(X-Y+a-d)^+)=d+\mu+\mathbb E((Z+c)^+), $$ where $c=a-d$ and $Z$ is normal centered with variance $2\sigma^2$. Note that $c\mapsto(z+c)^+$ is differentiable at each given $c$ with derived function $c\mapsto\mathbf 1_{z+c\gt0}$, for Lebesgue almost every $z$, and that $\mathbb P_Z$ is absolutely continuous with respect to the Lebesgue measure. Hence, $$ \frac{\partial g}{\partial a}=\mathbb P(Z+c\gt0)=\frac12+\frac12\mathrm{erf}\left(\frac{a-d}{2\sigma}\right), $$ where erf is the error function.

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