# Stochastic calculus rules $d(B_t^2) = 2B_t\,dB_t + dt$ - why?

Let $B_t$ = Brownian motion at time $t$

I know that $(dB_t)^2 = dt$ and $d(f(x)) = f'(x)\,dx$ for some differentiable function.

Now, I have that $$M_t = B_t^2 - t$$

$$dM_t = d(B_t^2) - d(t)$$

$$dM_t = d(B_t^2) - 1\,dt$$

$$dM_t = 2B_t\,dB_t + dt - dt$$

I am struggling to see how the last line comes abot.

Why does $d(B_t^2) = 2B_t\,dB_t + dt$ ?

Edit: wait, this is just chain rule with Ito's formula for $f(B_t)$!!

In case anyone else has the same question;

Ito's formula for $f(B_t)$ is as follows;

We have $(dB_t)^2 = dt$

Then for some differentiable function $f(x)$

$$df(x) = f'(x)\,dx + \frac{1}{2}f''(x)(dx)^2$$ With the above being the result of a 2nd order taylor expansion.

Letting $x = B_t$

$$df(B_t) = f'(B_t)\,dB_t + \frac{1}{2}f''(B_t)\,dt$$

So, in our case, we use $f(x) = x^2$

$$d(x^2) = 2x\,dx + \left(\frac{2 \, dx}{2}\right)^2$$

Substitute $B_t^2 = x^2$

$$d(B_t^2) = 2B_t\,dB_t + (dB_t)^2 = 2B_t\,dB_t + dt$$

• Where does the $dt$ term come from? Commented Jun 3, 2015 at 17:24
• I have edited the original question with an answer. I feel cheap answering my own question, so feel free to post it as an answer and i will accept it. Commented Jun 3, 2015 at 17:34
• For future reference, you can also post an answer and click the box underneath it: "community" and it will attributed to a general user. Cheers. Commented Jun 3, 2015 at 17:34
• Cool, didn't know you could do that.Thanks Commented Jun 3, 2015 at 17:36

Ito's formula for $f(B_t)$ is as follows;

We have $(dB_t)^2 = dt$

Then for some differentiable function $f(x)$

$$df(x) = f'(x)dx + \frac{1}{2}f''(x)(dx)^2$$ With the above being the result of a 2nd order taylor expansion.

Letting $x = B_t$

$$df(B_t) = f'(B_t)dB_t + \frac{1}{2}f''(B_t)dt$$

So, in our case, we use $f(x) = x^2$

$$d(x^2) = 2xdx + (\frac{2 dx}{2})^2$$

Substitute $B_t^2 = x^2$

$$d(B_t^2) = 2B_tdB_t + (dB_t)^2 = 2B_tdB_t + dt$$