# Properties of expectation

#### kgan

##### New Member
I am working on a pricing methodology under certain conditions. Suppose we have a population of size N and this population has a distribution of reservation prices $$f(p)$$. Now the problem of the firm becomes to maximize

$$R = p N \int_p^{\infty} f(p)$$

Or to maximize

$$log(R) = log(p) + log( N ) + log[1-F(p)]$$

Now, I take the first order derivative with respect to p, to find the optimal value of p.

$$\frac{\partial log(R)}{\partial p} = \frac{1}{p} -\frac{f(p)}{1-F(p)} =0$$

This implies that the optimal value of p satisfies the equation

$$1 - F(p) = pf(p)$$

Now, under the assumption that $$f(p)$$ is distributed as uniform or exponential, the optimal value of $$p$$ is the mean of the distribution. This also makes sense from intuitively. However, how do I show this for any general (perhaps non-negative) f(p)?

Additionally, integrating both sides of the equation, we get $$E(p)$$ on both sides of the equation. I dont think that is very helpful but thought I would point that out.

#### chiro

##### New Member
Hey kgan.

Since you have F(p) = 1 - pf(p),

This means that the value 1 - pf(p) must be between 0 and 1 (or 0 <= 1 - pf(p) <= 1). So we have
-1 <= -pf(p) <= 0 or
0 <= pf(p) <= 1

Now f(p) > 0 so we must restrict p >= 0.

But if you assume the mean property in general, then it means that the mean can never be negative and this can clearly happen. So you get one contradiction for the answer always being equal to the mean.