# Analytical expression for SDE

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.

• 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 $Y_t$. Jul 20, 2022 at 12:10
• @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. Jul 20, 2022 at 23:22
• You can simply set $y = ln(X^n)$ and follow its Ito calculus. Jul 21, 2022 at 12:57
• @Kermittfrog Thank you. I figured it out. Jul 23, 2022 at 19:03