On the page 13 of this document[1],there is derivation of FOC of minimization of expected cost problem I am stuck with.
I believe that they are differentiated expected cost function
$\alpha \mathbb { E } \left( \left[ \tau ^ { i } - t + \sigma ^ { i } \tilde { \epsilon } ^ { i } \right] _ { + } \right) + ( \beta + \gamma ) \mathbb { E } \left( \left[ \tau ^ { i } - T + \sigma ^ { i } \tilde { \epsilon } ^ { i } \right] _ { + } \right) - \gamma \left( \tau ^ { i } - T \right)$
with respect to $\tau ^ { i }$
and let it equal to zero. I wonder how they earned $\mathbb { P }(*)$,which is cumulative distribution function, in
$\alpha \mathbb { P } \left( \tau ^ { i } - t + \sigma ^ { i } \tilde { \epsilon } ^ { i } > 0 \right) + ( \beta + \gamma ) \mathbb { P } \left( \tau ^ { i } - T + \sigma ^ { i } \tilde { \epsilon } ^ { i } > 0 \right) = \gamma$
. I think derivative of expected value does not yield distribution function. So how did they?
[1]http://mfglabs.com/publications/download/paris-princeton.pdf
Remember we are computing the expectation of the positive part of $\tau^i-t+\sigma^i\tilde\epsilon^i$, i.e., $$ E:=\int_{\tau^i-t+\sigma^iz>0}^\infty(\tau^i-t+\sigma^iz)\,\phi(z)\,dz,\tag1 $$ where $\phi$ denotes the standard normal density. We can differentiate (1) with respect to $\tau^i$ by differentiating under the integral sign: $$ \frac{d E}{d\tau^i}= \int_{\tau^i-t+\sigma^iz>0}^\infty\frac{\partial}{\partial \tau^i}\left[(\tau^i-t+\sigma^iz)\,\phi(z)\right]\,dz= \int_{\tau^i-t+\sigma^iz>0}^\infty\phi(z)\,dz=P(\tau^i-t+\sigma^i\tilde\epsilon^i>0).$$ If you are not convinced that this is permitted, you can also evaluate (1) directly, using $$ E(a+bZ)_+=\int_{a +b z>0}(a+bz)\phi(z)\,dz=aP(a+bZ>0)+b\phi(a/b), $$ then substitute $a:=\tau^i-t$, $b=\sigma^i$, $Z=\tilde\epsilon^i$, and then take the derivative.