Home > Mathematics, Pricing, Statistical Economics > Some Thoughts On Pricing, Part II

## Some Thoughts On Pricing, Part II

Let's say instead of a uniform distribution, as described in "Some Thoughts On Pricing, Part I" and its simulation, comparable products' (say apple juice) prices hover around 10.00 pesos, with standard deviation 3.00 pesos, and the distribution is normal (by omniscience).  If there is just one other competing company (F1) against me I could have ascertained the mean and variance (or standard deviation) by sampling its pricing across 1000 shops, and obtained them by maximum likelihood.  If there are more than one companies (several tens) in the market competing against me perhaps I could have derived the distribution parameters by maximum likelihood across a shelf of a single shop (by ergodicity), always assuming prices across companies are realizations of the same distribution.

An advantage of the Gaussian distribution is that the domain is the entire real line, and so probabilities are assigned for every conceivable price (even negative prices, meaning I as a supplier choose to pay you the consumer to take the product!).  Expected profit is likewise therefore defined everywhere of practical interest.

Following as before, let's say again that the cost of producing a single unit of product is 1 for me (*).  The probability of my selling by  being the lowest price at a particular shop is one minus the cumulative density of a Gaussian:

$1 - \frac{1}{2}\big(1+erf(\frac{x - \mu}{\sigma \sqrt{2}})\big)$,

or in this case:

$1 - \frac{1}{2}\big(1+erf(\frac{x - 10}{3 \sqrt{2}})\big)$,

and the probability of failing at a sale is the cumulative density:

$\frac{1}{2}\big(1+erf(\frac{x - 10}{3 \sqrt{2}})\big)$

As before, for one competing company, my expected profit is:

$E_p(x) = \Big(1 - \frac{1}{2}\big(1+erf(\frac{x - 10}{3 \sqrt{2}})\big)\Big) \cdot (x-1) + \frac{1}{2}\big(1+erf(\frac{x - 10}{3 \sqrt{2}})\big) \cdot (-1)$

and if there are n competing companies,

$E_p(x) =\Big(1 - \frac{1}{2}\big(1+erf(\frac{x - 10}{3 \sqrt{2}})\big)\Big)^n (x-1) +$

$+ \Big(1 - \Big(1 - \frac{1}{2}\big(1+erf(\frac{x - 10}{3 \sqrt{2}})\big)\Big)^n\Big)(-1)$

Here I have graphed price to expected profit for 1...5, 10, 20, and 40 competing companies.  I also used a computer to obtain approximations of zeroes and extrema.

One of the remarkable things about this family of functions is that maximum expected profit can be achieved for any competition in the market, no matter if there are few companies competing against me or thousands, although if there are several hundreds or thousands at some point my maximum expected profit is negative (but never less than -1, of course, since I cannot lose more than what I cannot sell).  Here's a graph of this phenomenon:

Another remarkable thing is that maxima seem to lie on a line defined for $E_p(x) > -1$, although I am not entirely sure of the behavior at near $E_p(x) = -1$ (maxima may in fact be asymptotic to such).  Break-even prices become lower and lower the more competition there is, until finally there is a point at which, even though I may sell some product of apple juice, I can never make up for my expense (graph above).  Lastly, here again maximal expected profit is achieved at lower pricing the more competition there is in the market.

At any rate, the lesson would seem to be: the more competition there is, the more the piece of the pie is divided up, the lesser the maximal profit for me.

In my next post, I will describe what happens to expected profit when the Gaussian distribution of prices across competing firms is more certain: by this I mean the standard deviation or variance is smaller, which translates to there being a general agreement or consensus across suppliers about what the price of a product should be.

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1. July 16th, 2009 at 13:54 | #1

I think one of the ideas here is that a particular market will admit me as a competitive agent (*) as long as my expected profit is positive, and everyone is pricing all over the place, Gaussianly.

*