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Original Question:

Derive an SDE for $B^2(t)$, where $B(t)$ is standard Brownian Motion.

Attempt at an answer:

Apply Ito's calculus over $f(t,b):= B^2(t)$.

$$df(t,b) = \frac{\partial f(t,b)}{\partial t}dt + \frac{\partial f(t,b)}{\partial b}dB^2(t) + \frac12 \frac{\partial^2 f(t,b)}{\partial b^2}d\langle\, B^2\rangle_t$$

$$ = 2B(t)dB^2(t) + d\langle\, B^2\rangle_t.$$

MY Question:

What does $d\langle\, B^2\rangle_t$ equal, why, and how do I arrive at it?

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1 Answer 1

up vote 2 down vote accepted

Sorry but almost none of the identities in your post is correct... To straighten things out, let us first recall Itô's formula: if $X_t=F(t,B_t)$ for some suitable function $F:(t,x)\mapsto F(t,x)$ and a standard Brownian motion $(B_t)$, then $$ \mathrm dX_t=G(t,B_t)\mathrm dt+H(t,B_t)\mathrm dB_t, $$ where $G(t,x)=\frac{\partial F}{\partial t}(t,x)+\frac12\frac{\partial^2F}{\partial^2x}(t,x)$ and $H(t,x)=\frac{\partial F}{\partial x}(t,x)$.

In the present case $F(t,x)=x^2$ hence $\frac{\partial F}{\partial t}=0$, $\frac{\partial^2F}{\partial^2x}=2$ and $\frac{\partial F}{\partial x}=2x$. Thus, if $X_t=B_t^2$ then $$ \mathrm dX_t=2B_t\mathrm dB_t+\mathrm dt. $$

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