User:PAR/Work3

Pricing Algorithm

Sales of the object is a Poisson process. The probability of a sale occurring in the time interval t to t+dt is

$dP=\lambda (p)dt\,$

where p is the price of the object and $\lambda (p)$ is the average rate of sales at some fixed price p via the demand curve. A simple linear demand curve will be assumed:

$\lambda (p)=2\lambda _{0}\left(1-{\frac {p}{2p_{0}}}\right)\,\,\,\,(0\leq p<2p_{0})$

$\lambda (p)=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(p\geq 2p_{0})$

where $\lambda _{0}$ is a constant equal to the rate of sales at optimum price $p_{0}$. The optimum price $p_{0}$ is the price at which the rate of income $p\lambda$ is maximum. For prices above $2p_{0}$ the sales rate will be zero.

The pricing algorithm will be to have a linearly decreasing price, decreasing to zero at time $\tau$ or until a sale is made, at which point the price jumps to $1+\alpha$ times the sale price, and again begins a linear decline. That is, if $p_{n}=p_{s}(1+\alpha )$ where $p_{s}$ is the sale price, then the price p as a function of time t after that sale is

$p(t)=p_{n}\left(1-{\frac {t}{\tau }}\right)\,\,\,\,\,\,$ for $t\leq \tau$

$p(t)=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ for $t\geq \tau$

$\lambda (t)$ divides into two cases. When $p_{n}$ is greater than $2p_{0}$, then $\lambda (t)$ remains zero until $p_{n}=2p_{0}$, at which point it begins to rise linearly. It does so until $t=\tau$, at which point it remains at $2\lambda _{0}$

$\lambda (t)=0\,$ for $0\leq t\leq \tau (1-2p_{0}/p_{n})$

$\lambda (t)=2\lambda _{0}\left(1-{\frac {p_{n}}{2p_{0}}}\left(1-{\frac {t}{\tau }}\right)\right)$ for $\tau (1-2p_{0}/p_{n})\leq t\leq \tau$

$\lambda (t)=2\lambda _{0}\,$ for $t\geq \tau$

When p_n is less than $2p_{0}$, $\lambda (t)$ rises linearly until $t=\tau$, at which point it remains at $2\lambda _{0}$.

$\lambda (t)=2\lambda _{0}\left(1-{\frac {p_{n}}{2p_{0}}}\left(1-{\frac {t}{\tau }}\right)\right)$ for $0\leq t\leq \tau$

$\lambda (t)=2\lambda _{0}\,$ for $t\geq \tau$

Approximate Equilibrium

Depending on the initial price, the price function will take a certain amount of time to equilibrate. (This does not mean it is constant, of course, only that its average behavior gives no clue as to the amount of time elapsed since time zero.)

An approximate equilibrium condition is that the average time between sales $(T=1/{\overline {\lambda }}(t))$ is such that the price after a sale decays to the price before the sale.

$p_{s}=p_{s}(1+\alpha )(1-T/\tau )\,$

${\frac {1}{T}}=?$

These are two equations in two unknowns ($p_{s}$ and $t$). Solving:

$T=\tau {\frac {\alpha }{1+\alpha }}\,$

${\frac {p_{s}}{p_{o}}}=?$

Note the problems when $2\lambda _{0}T<1$

Exact equilibrium

The probability that the price is p at time t+dt is the probability that the price was $pe^{t/\tau }$ at time t and a sale was not made, plus the probability that the price was $p/(1+\alpha )$ at time t and that a sale was made. Normalizing to unity $p_{0}$ and $\lambda _{0}$

$P(p,t+dt)=(1-\lambda (pe^{dt/\tau })dt)P(pe^{dt/\tau },t)+\lambda (p/(1+\alpha ))P(p/(1+\alpha ),t)dt\,$

$=(1-\lambda (p)dt)P(p(1+dt/\tau ),t)+\lambda (p/(1+\alpha ))P(p/(1+\alpha ),t)dt\,$

$=(1-\lambda (p)dt)P(p+pdt/\tau ,t)+\lambda (p/(1+\alpha ))P(p/(1+\alpha ),t)dt\,$

or

$P(p,t)+{\frac {\partial P}{\partial t}}dt=\left(1-\lambda (p)dt\right)\left(P(p,t)+{\frac {\partial P}{\partial p}}{\frac {p}{\tau }}dt\right)+\lambda (p/(1+\alpha ))P(p/(1+\alpha ),t)dt\,$

or

${\frac {\partial P}{\partial t}}=\lambda (p)P(p,t)+{\frac {p}{\tau }}{\frac {\partial P}{\partial p}}+\lambda \left({\frac {p}{1+\alpha }}\right)P\left({\frac {p}{1+\alpha }},t\right)\,$

The sale price

The sale price is a random variable, but its not a Poisson process. The sale price probability is dependent on the previous sale price.

Given that the last sale price was $p_{i}$ at time $t_{i}$ the probability that the next sale will occur between time t and t+dt is

$P(t)dt=e^{-{\overline {\lambda }}t}\lambda (t)dt$

$\lambda (t)=2\lambda _{0}\left(1-{\frac {p_{i}(1+\alpha )e^{-t/\tau }}{2p_{0}}}\right)\,\,\,\,\,\,\,\mathrm {for} \,\,p_{i}(1+\alpha )e^{-t/\tau }\leq 2p_{0}\mathrm {\,\,zero\,\,otherwise}$

$\lambda (t)$ will be zero when

$p_{i}(1+\alpha )e^{-t/\tau }\geq 2p_{0}$

or, equivalently,

$t\leq \tau \ln \left({\frac {p_{i}(1+\alpha )}{2p_{0}}}\right)$

as long as that t>0. For $p_{i}(1+\alpha )e^{-t/\tau }\leq 2p_{0}$ then, we have:

${\overline {\lambda }}={\frac {1}{t}}\int _{0}^{t}\lambda (t')dt'=2\lambda _{0}\left(1+{\frac {\tau }{t}}\,{\frac {p_{i}(1+\alpha )\left(e^{-t/\tau }-1\right)}{2p_{0}}}\right)$

and for $p_{i}(1+\alpha )e^{-t/\tau }>2p_{0}$

${\overline {\lambda }}={\frac {1}{t}}\int _{t_{x}}^{t}\lambda (t')dt'=2\lambda _{0}\left(1-{\frac {t_{x}}{t}}+{\frac {\tau }{t}}\,{\frac {p_{i}(1+\alpha )\left(e^{-t/\tau }-e^{-t_{x}/\tau }\right)}{2p_{0}}}\right)$

and that sale price will be

$p_{i+1}(t)=p_{i}(1+\alpha )e^{-t/\tau }\,$

The expected value of $p_{i+1}$ is

$\langle p_{i+1}\rangle =\int _{0}^{\infty }P(t)p_{i+1}(t)dt=\int _{0}^{\infty }p_{i}(1+\alpha )e^{-t/\tau }e^{-{\overline {\lambda }}t}\lambda (t)dt$

Given a sale at [p,0], and given that there is a sale at time t, what is the probability distribution for that sale price? The sale is not necessarily the first sale after the original.

References

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• Dynamic pricing with real-time demand learning K.Y. Lin