If $ S_t $ follows a log-normal Brownian motion, what SDE does the square of $ S_t $ follow?

Question

If $ S_t $ follows a log-normal Brownian motion, what SDE does the square of $ S_t $ follow?

I have found two possibles ways of solving it. But, they diverge with respect to the drift.

1. First solution: We recall the stochastic equation: $ dS_t = \mu S_t dt + \sigma S_t \, dW_t $ and its solutions $ S_t = S_0 \exp((\mu-\sigma^2/2)t + \sigma W_t) $. Applying directly to $ S_t^2 $, we got: $ S_t^2 = S_0^2 \exp((2\mu-\sigma^2)t + 2\sigma W_t) $. The drift component is $ (2\mu-\sigma^2) $.
2. Second solution: We say that $ f(S_t)=S_t^2 $. Applying Itô's formula, we get: $ dS_t^2 = 2S_td \, S_t + \sigma^2 S_t^2 \, dt \rightarrow dS_t^2 = (2\mu + \sigma^2)S_t^2 \, dt + 2\sigma S_t^2 \, dW_t $. The drift component is $ (2\mu + \sigma^2) $

Which is the correct solution? Where am I making a mistake?

Thanks!

Answer 1

There is no contradiction between your two approaches. If we define

$$X_t := S_0^2 \exp \left( (2 \mu - \sigma^2)t +2 \sigma W_t \right)$$

(which is the process which you got in your first solution) then a straight-forward application of Itô's formula shows that

$$dX_t = 2 \sigma X_t \, d W_t + \left( 2\mu + \sigma^2 \right) X_t \, dt,$$

and this is exactly the SDE which you got in your second approach.

To get from the second solution to the first one you may note that

$$d(S_t^2) = (2\mu+\sigma^2) S_t^2 \, dt + 2 \sigma S_t^2 \, dW_t$$

shows that $X_t := S_t^2$ solves the SDE

$$dX_t = (2\mu+\sigma^2) X_t \, dt + 2 \sigma X_t \, dW_t,$$

and as you pointed out in your first approach the solution to this SDE is given by

$$X_t = X_0 \exp \left((2\mu+\sigma^2-(2\sigma)^2/2)t + (2\sigma) W_t \right)$$

which is exactly the process which you got in your first solution.