Ito’s Partial Differential Equation

We will derive the Ito’s PDE for lognormal process and derive the solution.

Suppose given process A_t follows geometric brownian motion under \textbf{P} pricing measure,
$dA_t = \mu A_t dt + \sigma A_t dW_t$
The same process under \textbf{Q} pricing measure given by,
$dA_t = r A_t dt + \sigma A_t dW_t$

Now, let’s suppose we have process,

$dX_t = a(X_t, t) dt + b(X_t, t) dW_t$
Let’s suppose, we have Y_t = g(X_t, t)
By Ito’s process, we know that,
$dY_t = \frac{dg}{dt} + \frac{dg}{dX} X_t + \frac{1}{2} \frac{d^2g}{dX^2} (dX_t)^2$

Now, we from brownian motion properties, $dt^2 = 0, dW dt = 0, dW^2 = dt$ . Applying them,

$dY_t = \frac{dg}{dt} + \frac{dg}{dt}(a(X_t, t)dt + b(X_t, t) dW_t) + \frac{1}{2} \frac{d^2g}{dt^2} (b^2 (X_t, t)dt)$

Now, since we know $Y = ln A_t$ , therefore, $\frac{dg}{dx} = \frac{1}{X}, \frac{d^2g}{dX^2} = \frac{-1}{X^2}$ . We now apply,

$dY_t = \frac {1}{A} dA + \frac{1}{2} {-1}{A^2} (dA)^2$
$dY_t = \mu dt + \sigma dW_t - \frac{1}{2} \sigma^2 dt$
$dY_t = (\mu - \frac{1}{2} \sigma_2) dt + \sigma dW_t$