I am really struggling to come up with the correct SDE for the stochastic process:

$Y(t) = a[Z(t)]^2$

where $Z(t)$ is a Brownian Motion. According to my Prof, the SDE is:

$dY(t) = adt + 2aZ(t)dZt $

Can anyone explain how he got to that solution? Thanks in advance, any help appreciated


1 Answer 1


If you apply Ito to $Y_t=aZ_t^2$ it simple to arrive at $dY_t$:

$$ f(t,z):=az^2 \\ dY_t = f_t(t,Z_t)dt+f_z(t,Z_t)dZ_t+\frac{1}{2}f_{zz}(t,Z_t)*[dZ_t]^2 \\ dY_t = 0*dt+ 2aZ_t dZ_t+\frac{1}{2}2a[dZ_t]^2 \\ dY_t = 2aZ_t dZ_t+adt \\ $$ Which is your desired result. Recall that $dZ_t^2=dt$.

  • $\begingroup$ Thanks a lot! Just to make sure, lets suppose it said that $Z(t)$ followed a geometric Brownian Motion instead of only a brownian Motion. Would it then be allowed to change $dZ_t$ with: $\begin{equation} dZ_t=\mu Z_t dt + \sigma Z_t dW_t \end{equation}$ Thanks a lot again $\endgroup$ May 11, 2019 at 16:09
  • 1
    $\begingroup$ Yes, that is correct $\endgroup$
    – Sanjay
    May 11, 2019 at 17:59

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