# Finding the stochastic differential equation satisfied by process Y

Consider the stochastic process $X$ that satisfies the SDE \begin{equation} X_t=1+\mu\int_0^t X_s \, ds+\sigma\int_0^t X_s \, dB_s, \end{equation} with $\mu$ and $\sigma$ constants. Consider another process $Y$ given by \begin{equation} Y_t=X_t^\beta, \end{equation} with $\beta\ge 2$. Find the stochastic differential equation for process $Y$.

I am looking at the following solution: \begin{equation} dY_t=\beta X_t^{\beta-1}dX_t+\beta(\beta-1)X_t^{\beta-2}(dX_t)^2\\[10pt] dY_t=\beta X_t^{\beta-1}(\mu X_t \, dt+\sigma X_t \, dB_t)+\beta(\beta-1) X_t^{\beta-2} \sigma^2 X_t^2 \, dt\\[10pt] dY_t=\beta\mu X_t^\beta \, dt+\sigma\beta X_t^\beta \, dB_t + \beta(\beta-1) \sigma^2 X_t^\beta \, dt\\[10pt] dY_t=(\beta\mu+\beta(\beta-1)\sigma^2)Y_t \, dt+\sigma\beta Y_t \, dB_t \end{equation}

Assuming this solution is right, my $\textbf{question}$, or the problem I'm having difficulty fully understanding is, why is the second spatial derivative not multiplied by $\frac{1}{2}$? Unless I'm totaly spacing, should that term not be \begin{equation} \left[\frac{\beta(\beta-1)}{2}\right]X_t^{\beta-2}(dX_t)^2 \end{equation} from an application of the Ito formula? Thank you.

You have a typo in your first equation, it should be $$dY_t=\beta X_t^{\beta-1}dX_t+\beta(\beta-1)X_t^{\beta-2}(dX_t)^2$$
If $Y_t =f(X_t)$, with $f(x)=x^{\beta}$, I do expect the $\frac{1}{2}$:
$$dY_t=df(X_t)=\frac{\partial f(X_t)}{\partial x}dX_t + \color{red} {\frac{1}{2}} \cdot\frac{\partial^2 f(X_t)}{\partial x^2}(dX_t)^2$$