# How to maximize profit of product with normal distribution

A quality characteristic of a product is normally distributed with mean µ and stdev σ = 1. Specs on the characteristic are 6 ≤ x ≤ 8. A unit that falls with spec results in profit C0. if x < 6 profit is –C1, if x > 8 profit is –C2. What is the value of µ that maximizes expected profit.

I guess the idea is to have product within spec as much of the time as possible, not really sure even where to begin on this. The value of –C1 and –C2 are not given either.

• Write down a mathematical expression for the expected profit in terms of $\mu$. – whuber Oct 2 '15 at 19:32
• Please add the [self-study] tag & read its wiki. – gung - Reinstate Monica Oct 2 '15 at 19:35
• The optimal $\mu$ will be expressed in terms of $C_0,C_1,C_2$, that's no issue (on the contrary, it is very educational because it lets you see what happens for different values of these magnitudes). Follow whuber's hint ("with probability XX I have this result, with probability YY the other", etc). It is not about maximizing probability mass in the interval $[6,8]$ -the economic consequences play also a part in determining the optimal $\mu$ – Alecos Papadopoulos Oct 4 '15 at 14:20

To close this one:
"Expected profit" is most probably understood as "the expected value of the random variable profit". There are three "states of the world", each with its profit and probability of occurring:

\begin{array}{| r | r | r | r|} \hline \text{State of the world} &\text{Profit} & \text {Prob} \\ \hline \hline X<6 & -C_1 & \Phi(6-\mu) \\ \hline 6\leq X \leq 8 & C_0 & \Phi(8-\mu)-\Phi(6-\mu) \\ \hline X > 8 & -C_2 & 1-\Phi(8-\mu) \\ \hline \end{array}

The middle column describes the support of the discrete random variable "profit" ($\equiv \pi$), and the last column is its probability mass function. $\Phi$ is the cumulative distribution function of the standard normal, and it determines here how probability mass is allocated to the three values in the support of $\pi$. So the expected value of $\pi$ is

$$E(\pi) = -C_1\cdot \Phi(6-\mu) + C_0 \cdot [\Phi(8-\mu)-\Phi(6-\mu)]-C_2\cdot [1-\Phi(8-\mu)]$$

Collecting terms,

$$E(\pi) = -\Phi(6-\mu)\cdot (C_0 + C_1) + \Phi(8-\mu) \cdot [C_0 +C_2] -C_2$$

To get the maximum with respect to $\mu$ we differentiate

$$\frac {\partial E(\pi)}{\partial \mu} = \phi(6-\mu)\cdot [C_0 + C_1] - \phi(8-\mu)\cdot [C_0 + C_2]$$

where $\phi$ is the probability density function of the standard normal. Setting this equal to zero we have

$$\frac {\partial E(\pi)}{\partial \mu} = 0 \implies \frac {\phi(6-\mu)}{\phi(8-\mu)} = \frac {C_0 + C_2}{C_0 + C_1}$$

Calculating the left-hand side and taking natural logarithms, we end up with

$$\mu* = 7 - \frac 12 \cdot \ln\left(\frac {C_0 + C_2}{C_0 + C_1}\right)$$

One should also verify that the second-order derivative is negative so that the sufficient condition for a maximum holds here (it does).

We can gain some intuition from the result, observing that if the losses are symmetric ($C_1 = C_2$), the logarithm term will be zero and the optimal mean will be $\mu^* = 7$, i.e. at the center of the "within specs" interval. But if they are not, the optimal mean is "pushed away from the comparatively worse deviation from specs": for example, if $C_2 > C_1$, meaning that the loss if we are "above specs" is worse than when we are "below specs", we will have $\ln\left(\frac {C_0 + C_2}{C_0 + C_1}\right) >0$ and so the optimal mean will be $\mu^* < 7$, i.e. closer to the lower bound of the "within specs" interval: the economic consequences play a role.

In fact they play a role so large, that we may even end up having an optimal mean outside the within specs interval.

For this to happen we want, for example,

$$\mu^* < 6 \implies \frac 12 \cdot \ln\left(\frac {C_0 + C_2}{C_0 + C_1}\right) >1 \implies \frac {C_0 + C_2}{C_0 + C_1} > e^2$$

$$\implies C_2 > C_0\cdot (e^2-1) + C_1\cdot e^2$$

This reflects that the loss if we are above specs is comparatively too big, so big in fact that we prefer to be "on average" below specs! Counter-intuitive it may be, but it is what the criterion we chose (maximization of expected profit), dictates that we should do.

• +1 Thank you for completing this thread with such a clear, well-explained, thorough answer. – whuber Dec 14 '15 at 19:54
• @whuber Thank you whuber, I do appreciate the kind words. – Alecos Papadopoulos Dec 14 '15 at 20:37