I'm trying to find an analytical expression for the following. Suppose $X$ is a geometric Brownian motion, such that: $dX_{t} = \mu X_{t} dt + \sigma X_{t} dW_{t}$. Suppose furthermore, that the process $Y$ is defined by $Y_{t} = X_{t}^{n}$. I have found the dynamics of $Y$ which becomes:

$dY_{t} = X^{n}_{t}\left(n\mu + \frac{1}{2}n(n-1)\sigma\right)dt + nX^{n}_{t}\sigma dW_{t}$.

In integral form we have,

$Y_{t'} = Y_{t} + \left(n\mu + \frac{1}{2}n(n-1)\sigma\right)\int^{t'}_{t}X^{n}_{s}ds + n\sigma\int^{t'}_{t}X^{n}_{s} dW_{s}$,

assuming that $t' > t$. My problem is how to evaluate the two integrals. For the first one I thought about doing the following:

$\int^{t'}_{t}X^{n}_{s}ds = \left[X_{s}s\right]^{t'}_{t} = X_{t'}t' - X_{t}t = (X_{t'}-X_{t})(t'-t)$.

The stochastic integral is causing me some trouble. I would appreciate any kind of help to evaluate both the integrals.

  • 3
    $\begingroup$ Either solve use the solution for $X_t$ and simply set $Y_t=X_t^n$ or realise that $\text{d}Y$ looks like a GBM with a modified drift and volatility. Using the usual steps (log transformation) you can then get a solution for $Y_t$. $\endgroup$
    – Kevin
    Jul 20, 2022 at 12:10
  • $\begingroup$ @Kevin Thank you a lot. I managed to do it with realising that $dY$ is a GBM. I wondered how one would go about it the "long" way without transformation. But I think this suffices. $\endgroup$ Jul 20, 2022 at 23:22
  • 1
    $\begingroup$ You can simply set $y = ln(X^n)$ and follow its Ito calculus. $\endgroup$ Jul 21, 2022 at 12:57
  • $\begingroup$ @Kermittfrog Thank you. I figured it out. $\endgroup$ Jul 23, 2022 at 19:03


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