Black-Scholes and information equilibrium

Continuing from here [1]

Let's say the price of a stock option depends on the stock price and time $Q = Q(p, t)$. If the price is our stochastic process from [1]

$$
\frac{dp}{p} = \mu dt + \sigma dx
$$

Ito's lemma tells us that

$$
dQ = \frac{\partial Q}{\partial t} dt + \frac{\partial Q}{\partial p} dp +\frac{1}{2} \frac{\partial^{2} Q}{\partial p^{2}} dp^{2}
$$

with

$$
dp^{2} = \sigma^{2} p^{2} dx^{2}
$$

Now let's look at the 'portfolio' $Q + \alpha p$, substitute $dp^{2}$) and use the 'hedge' $\alpha = - \partial Q/\partial p$ so that

$$
d(Q + \alpha p) = \left( \frac{\partial Q}{\partial t}  + \frac{1}{2}  \sigma^{2} p^{2} \frac{\partial^{2} Q}{\partial p^{2}} \right) dt
$$

That's where the stochastic bit proportional to $dx$ fell out. And now there's a model assumption. Since the stochastic bit fell out, we assume the return of the portfolio must be the risk free rate $r$. That means

$$
\frac{d}{dt} (Q + \alpha p) = r (Q + \alpha p) = r \left( Q - p \frac{\partial Q}{\partial p}  \right)
$$

And substituting, we get the Black-Scholes equation:

$$
\frac{\partial Q}{\partial t}  + r p \frac{\partial Q}{\partial p} + \frac{1}{2}  \sigma^{2} p^{2} \frac{\partial^{2} Q}{\partial p^{2}} = r Q
$$

That's just reiterating some stuff I found in vaious notes and appears to be a standard derivation.

So now some interesting (information equilibrium relevant) bits ...

From [1] we can show

$$
\sigma^{2} = \frac{1}{p^{2}} \left( \frac{\partial p}{\partial x} \right)^{2}
$$

So that

$$
\frac{\partial Q}{\partial t}  + r p \frac{\partial Q}{\partial p} + \frac{1}{2} \left( \frac{\partial p}{\partial x} \right)^{2} \frac{\partial^{2} Q}{\partial p^{2}} = r Q
$$

And using the second order chain rule (and rearranging) we obtain ...

$$
\frac{\partial Q}{\partial t}  + \left( r p  - \frac{1}{2} \frac{\partial^{2} p}{\partial x^{2}} \right) \frac{\partial Q}{\partial p} + \frac{1}{2} \frac{\partial^{2} Q}{\partial x^{2}} = r Q
$$

This equation is now independent of $\kappa$ and the supply and demand functions $S$ and $D$. Which is interesting -- the price of an option no longer depends on anything other than the stochastic process the price follows. But wait, there's more.

Let's say our option and our stock price follow diffusion processes in volatility space ($x$):

$$
\frac{\partial Q}{\partial t} = \frac{2}{\xi} \frac{\partial^{2} Q}{\partial x^{2}}
$$

$$
\frac{\partial p}{\partial t} = \frac{2}{\zeta} \frac{\partial^{2} p}{\partial x^{2}}
$$

I chose the diffusion coefficients to appear as the do because it allows a simplification. Substituting those equations into our IT-Black-Scholes equation we get ...

$$
(1 + \xi - \zeta) \frac{\partial Q}{\partial t}  + p \frac{\partial Q}{\partial p} - r Q = 0
$$

Which is a first order partial differential equation with solution:

$$
Q(p, t) = p f( t - \frac{1 + \xi - \zeta}{r} \log p)
$$

Where $f$ is an arbitrary function. Some observations ...

If $p = \exp \rho t$, then

$$
Q(p, t) = p f( a t)
$$

which is an arbitrary function of $t$ ($a$ is a combination of the other constants). If $\xi = \zeta$, i.e. the diffusion coefficients of the stock and option price are equal, we end up with the solution

$$
Q(p, t) = p f( t - \frac{1}{r} \log p)
$$

And if additionally $p = \exp \rho t$, then

$$
Q(p, t) = p f( t - \frac{\rho}{r} t)
$$

so if $\rho = r$

$$
Q(p, t) \propto p
$$

Price as a stochastic process

I mentioned the other day that I'd like to look into the Black Scholes equation in terms of information equilibrium. John Cochrane's Asset Pricing (2005) derives it using the asset pricing equation and a bit of stochastic calculus. I already looked at the basic asset pricing equation here, so I'm just going to set up the price in the information transfer framework as a stochastic process.

Let's start with the information equilibrium condition

$$
p \equiv \frac{dD}{dS} = \frac{1}{\kappa} \; \frac{D}{S}
$$

So that

$$
dp = \frac{1}{\kappa} \left( \frac{dD}{S} - \frac{D}{S^{2}} dS \right)
$$

$$
\frac{dp}{p} = \frac{dD}{D} - \frac{dS}{S}
$$

Using the information equilibrium condition again, we have:

$$
\frac{dp}{p} = \frac{dD}{D} - \kappa \frac{dD}{D}
$$

$$
\frac{dp}{p} = (1 - \kappa) \frac{dD}{D}
$$

Let's say $D$ has a deterministic part and a stochastic part ...

$$
\frac{dp}{p} = (1 - \kappa) \left( \frac{1}{D} \frac{\partial D}{\partial t} dt + \frac{1}{D} \frac{\partial D}{\partial x} dx \right)
$$

Now if we identify

$$
\mu \equiv (1 - \kappa) \frac{1}{D} \frac{\partial D}{\partial t}
$$
$$
\sigma \equiv (1 - \kappa) \frac{1}{D} \frac{\partial D}{\partial x}
$$

We end up with

$$
\frac{dp}{p} = \mu dt + \sigma dx
$$

Which is a geometric stochastic process that would become the price $p$ of the underlying stock in the Black-Scholes option pricing model.