# Solution of a GBM

Looking to check my solution to the below :

Suppose that $$X$$ satisfies the SDE

$$dX_t = αX_tdt+σX_tdW_t$$

Now define $$Y$$ by $$Y_t=X_t^{\beta}$$ ⁠, where β is a real number.

Then $$Y$$ is also a GBM process.

Compute $$dY$$ and find out which SDE Y satisfies.

My attempt :

Since $$Y_t = X_t^{\beta}$$, compute $$dY_t$$ using Ito's Lemma

$$dY_t = \frac{dY_t}{dt}dt + \frac{dY_t}{dX_t}dX_t + \frac{d^{2}X_t}{dt^{2}}dX_t^{2}$$

$$dY_t = 0.dt + \beta X_t^{\beta -1}dX_t + \beta (\beta -1)X_t^{\beta - 2}dX_t^{2}$$

and given $$dX_t = αX_tdt+σX_tdW_t$$, this simplifies to :

$$dY_t = Y_t ( \beta \alpha + \beta^{2}\sigma^{2} - \beta\sigma^{2})dt + \sigma\beta Y_t dW_t$$

Which to me looks like a GBM(?) - since the drift and volatility term both contain $$Y_t$$.

Is this the correct way to find out the SDE of $$Y_t$$?

Thank you!

• There is a minor mistake in the drift of $Y_t\,.$ Hint looking at it from another angle: $$X_t=X_0e^{\alpha t+\sigma W_t-\sigma^2t/2}\,.$$ Now raise this to the power of $\beta$ and bring it into the form of a GBM. What is its drift? Commented Mar 14, 2023 at 5:39
• You forgot a factor 1/2 in front of the second-order derivative. Commented Mar 14, 2023 at 6:49
• Working backwards, it looks like the solution should be something like: $$Y_t=X_0^{\beta}e^{\alpha \beta t+\sigma \beta W_t - \beta \sigma^2t/2}\,.$$ And hence I have an additional $\beta^2 \sigma^2$ term in the drift. However in the second order derivative, an additional term will exist because of the $\beta(\beta-1)$, so I am uncertain as to how to remove this? Commented Mar 14, 2023 at 6:58
• See answer. Now please finish that exercise here and please don't say "something like" when writing about mathematics. Commented Mar 14, 2023 at 13:01

The equation $$dY_t = 0.dt + \beta X_t^{\beta -1}dX_t + \color{red}{\frac{1}{2}}\beta (\beta -1)X_t^{\beta - 2}dX_t^{2}$$ had the term $$\frac{1}{2}$$ missing. If you plug in $$dX_t=\alpha X_t\,dt+\sigma\,X_t\,dW_t$$ you get \begin{align} dY_t=\alpha\beta X_t^\beta\,dt+\sigma\,\beta\,X_t^\beta\,dW_t+\frac{1}{2}\sigma^2\beta(\beta-1)X_t^\beta\,dt \end{align} which shows that $$Y_t=X_t^\beta$$ is a GBM with volatility $$\sigma\,\beta$$ and drift $$\alpha\beta+\frac{1}{2}\sigma^2\beta (\beta -1)\,.$$