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.
- 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) $.
- 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?