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I recently started reading ''Lectures On Stochastic Programming'' by Alexander Shapiro, Darinka Dentcheva & Andrzej Ruszczyński. On the introduction they adress the News Vendor Problem:

Suppose that a company has to decide about order quantity $x$ of a certain product to satisfy a demand $d$. The cost of ordering is $c>0$ per unit. If the demand $d$ is larger than $x$, then the company makes an additional order for the unit price $b\geq 0$. The cost of this is equal to $b(d-x)$ if $d>x$ and 0 otherwise. On the other hand, if $d<x$, then a holding cost of $h(x-d) \geq 0$ is incurred.

The Objective Function is:

$$F(x,d) = max \big\{(c-b)x + bd, (c+h)x -hd\big\}$$

Say demand $D$ is a random variable, then the expectation of the objective function is:

$$\mathbf{E}[F(x,D)] = b\mathbf{E}[D] + (c-b)x+(b+h)\cdot \int_0^xH(z)dz$$

with $H(x):=Pr(D\leq x)$, cdf of D.

How do you get to $\mathbf{E}[F(x,D)]$?.

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  • $\begingroup$ What do you mean "How do you get to $\mathsf E(F(x,D))$?" It says what to do right there. $\endgroup$ – Graham Kemp Mar 7 '17 at 0:05
  • $\begingroup$ Whats the proof that $E[F(x,D)]$ is what is stated there, sorry for my english $\endgroup$ – Tomás Arturo Herrera Castro Mar 12 '17 at 17:58
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We begin by observing the second term in the max of $F$ can be rewritten as

$$ (c - b)x + (h+ b)x - hD$$

so

$$ F(x,D) = (c - b)x + bD + (h+b) (x - D)_+$$

where $(x - D)_+ := \max\{0, x - D\} $. Therefore

$$\mathbb E[F(x,D)] = b \mathbb E[D] + (c - b) x + (h+b)\mathbb E[(x-D)_+] $$

so it suffices to show

$$ \mathbb E[(x-D)_+] = \int_0^x H(z) dz $$

Note that

$$ \mathbb E[(x-D)_+] = \mathbb E[\textbf {1} _{x \ge D} (x - D)] = \mathbb E[\textbf 1_{x \ge D} x] - \mathbb E[\textbf 1_{x \ge D} D] = H(x)x - \mathbb E[\textbf 1_{x \ge D} D]$$

where $\textbf 1_{x \ge D}$ denotes the indicator random variable that is 1 if and only if $D$ is no more than $x$, and 0 otherwise.

Finally, for positive $z$ we define $F(z):= P(\textbf 1_{x \ge D} D \le z)$. Together with the fact that the complementary CDF of a nonnegative random variable integrates to expectation, we are able to obtain

$$\mathbb E[\textbf 1_{x \ge D} D] = \int_0^\infty 1 - F(z) dz$$

and

$$ F(z) = \begin{cases} 1 & z \ge x \\ H(z) + 1 - H(x) & z < x \end{cases} $$

Note that $\textbf 1_{x \ge D}D \le x \le z$ so the first case in the above piecewise function holds. Verifying the second one is left as an exercise ;) (it's only slightly more elaborate).

Substituting this definition for $F(z)$ back into our integral tells us for $z\ge x$, $-F(z)$ cancels the 1 to give 0, so we are left with

$$ \int_0^x 1 - (H(z) + 1 - H(x)) dz = \int_0^x H(x) - H(z) dz$$

By evaluating part of the above integral, we are able to at last conclude that

$$ \mathbb E[(x-D)_+] = H(x)x - \Big((H(x)x - \int_0^x H(z) dz\Big) = \int_0^x H(z) dz$$

which is the claim we said it sufficed to show, completing the proof.

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