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!
Too long for a comment.
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 $$ \alpha\beta+\frac{1}{2}\sigma^2\beta (\beta -1)\,. $$