1
$\begingroup$

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!

$\endgroup$
4
  • 2
    $\begingroup$ 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? $\endgroup$
    – Kurt G.
    Commented Mar 14, 2023 at 5:39
  • 1
    $\begingroup$ You forgot a factor 1/2 in front of the second-order derivative. $\endgroup$
    – Abezhiko
    Commented Mar 14, 2023 at 6:49
  • $\begingroup$ 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? $\endgroup$
    – Sushiix
    Commented Mar 14, 2023 at 6:58
  • $\begingroup$ See answer. Now please finish that exercise here and please don't say "something like" when writing about mathematics. $\endgroup$
    – Kurt G.
    Commented Mar 14, 2023 at 13:01

1 Answer 1

1
$\begingroup$

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 $$ \alpha\beta+\frac{1}{2}\sigma^2\beta (\beta -1)\,. $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .