What is the the expected value of the process $Y = X^{3}$, where X satises the SDE $$ dXt = −X_tdt + σX_tdB_t $$ $(σ > 0)$ and $X_0 = 1$
I have two different answers: 1) I know that $X_t$ is a GBM with the following form $dX_t = \mu X_tdt + σX_tdB_t$ and i know it's $\mu$ and $\sigma$, so $$ X_t = e^{-(1+\frac{\sigma^2}{2})t + \sigma B_t} $$ Because $Y = X^{3}$ I can find (Is is correct to do?): $$ Y_t = e^{-3(1+\frac{\sigma^2}{2})t + 3\sigma B_t} $$ With this result which is the variance and the mean of $Y_t$ 2)Another way that opens in front of me is to apply the Ito's Lemma for $Y = X^{3}$, and I find: $$ \frac{dY_t}{Y_t}=3(\sigma - 1)dt + 3\sigma dB_t $$ So (if my computations are correct) I find a new $\mu$ and a new $\sigma$ compared to the standard form of a GBM, and are respectively $3(\sigma - 1)$ and $3\sigma$. So i must find the form of the process $Y_t$ in the same way of $X_t$ and I find: $$ Y_t = e^{((3(\sigma - 1)) - \frac{9\sigma^2}{2}) + 3\sigma B_t} $$ If this computation is correct now i need to find the expected value of $Y_t$ that is: $$ E[Y_t] = e^{((3(\sigma - 1)) - \frac{9\sigma^2}{2})t} $$
Please can you tell me which of this solutions is the right one? My professor helped me too much to increase the confusion on my head, i hope someone there could clear my mind a little. Thanks in advance.
Your first solution is correct; you didn't apply Itô's formula correctly.
Recall that for
$$X_t = \sigma(t) \, dB_t + b(t) \, dt$$
we have
$$f(X_t)-f(X_0) = \int_0^t f'(X_s) \, dX_s + \frac{1}{2} \int_0^t f''(X_s) \sigma^2(s) \, ds.$$
Applying this with $\sigma(t) = \sigma X_t$, $b(t) = - X_t$ and $f(x) = x^3$ gives
$$\begin{align*} Y_t - Y_0 &= 3 \int_0^t X_s^2 \, dX_s + 3 \int_0^t X_s (\sigma X_s)^2 \, ds \\ &= 3\sigma \int_0^t X_s^3 \, dB_s + \int_0^t (-3 X_s^3 + 3\sigma^2 X_s^3) \, ds. \end{align*}$$
i.e.
$$dY_t = 3(\sigma^2-1) Y_t \, dt + 3 \sigma Y_t \, dB_t.$$
Consequently, $(Y_t)_{t \geq 0}$ is also a geometric Brownian motion and we obtain
$$\begin{align*} Y_t &= \exp \left( \left( 3 (\sigma^2-1) - \frac{1}{2} (3\sigma)^2 \right) + 3 \sigma B_t \right) \\ &= \exp \left( -3 \left(1+ \frac{\sigma^2}{2} \right) + 3 \sigma B_t \right) \end{align*}$$