Again, I am not sure how the following works; Could someone please give me an almost stupidly detailed explanation of why/what is happening in the part below. First, the question itself;

Q. $B_t$ is a standard Brownian motion. Use Ito's formula to show the following identity holds,



A. By Ito's formula, one obtains $dB_s^4=4B_2^3dB_2+6B_s^2ds$ and which by integrating both sides ...etc

The rest is fine. I just don't know how "Ito's formula" gives "$dB_s^4=4B_2^3dB_2+6B_s^2ds$". I thught Ito's formula was the super tedious stochastic equation.

Briefly, it says for a Ito process $X_t$ which is defined $dX_t=\mu dt+\sigma dW_t$ with $X_0=x_0$, we have for some new Ito Process $Y_t$,

$$dY_t=[\frac{\partial f}{\partial t}+\mu \frac{\partial f}{\partial t}+\frac{\sigma^2}{2}\frac{\partial^2 f}{\partial t^2}]dt+[\sigma \frac{\partial f}{\partial x}]dW_t$$

I thought this differential equation was the "Ito's Formula". But how does this lead to the answer above??? I guess it's something very elementary but as a newbie, no, I don't see what is going on at all. Can someone elaborate on what the answer above has skipped and abbreviated? The bit "one obtains" as to how?

Thank you very much in advance for your help


First you have to decide on the notation, first you denoted Brownian motion by $(B_t)$ and then by $(W_t)$. I will use the notation $(B_t)$ for a (standard) Brownian motion.

Ito's formula applies to Brownian motion and if you want to see that it is indeed an Ito process note that $$ B_t = \int_0^t dB_s$$ and to adjust it to the form that you have written note that $$dB_t = 0 \cdot dt + 1 \cdot dB_t.$$ I would say that a it is worth to look at Ito's formula in the following form, for an Ito process $X_t$ if $f$ is twice continuously differentiable on $[0, \infty) \times \mathbb{R}$ then $$ df(t, X_t) = \frac{\partial f}{\partial t}(t, X_t)dt +\frac{\partial f}{\partial x}(t, X_t)dX_t +\frac{1}{2}\frac{\partial^2 f}{\partial x^2}(t,X_t) (dX_t)^2.$$ Now let $f(x, t)= x^4$ then $\frac{\partial f(x,t)}{\partial t} = 0$,$\frac{\partial f(x,t)}{\partial x} = 4x^3$, $\frac{\partial^2 f(x,t)}{\partial x^2} = 12x^2$, therefore by Ito's formula and having in mind Ito's table (that is, $(dt)^2 = dtdB_t = dB_tdt=0 \mbox{ and } (dB_t)^2 = dt$) $$dB_t^4 = 0 dt + 4B_t^3 dB_t + \frac{1}{2}\cdot 12 B_t^2 (dB_t)^2 =4B_t^3 dB_t + 6 B_t^2dt.$$

  • $\begingroup$ Hi, sorry about the notation; the question itself was from a paper and the $W_t$ was from my notes, i'll be careful. Thank you for answering too, I'll read through and let you know if I have anything still unclear. Otherwise, I'll accept your answer. thanks so much! $\endgroup$ – John Trail Dec 5 '15 at 22:13
  • $\begingroup$ Actually, yes, can I ask something about the Ito process? Is an Ito process normally distributed with $N(0,1)$ just as the Brownian motion? I ask this since in the part where you have introduced a new form of the formula, the $X_t$ you have is an Ito process, but the formula seems to be obtained by allowing $\mu=0,\sigma=1$ in the Ito formula. So does this mean $X_t$ is (standard) normally distributed? Or for convenience, in this particular context, are you saying $X_t=W_t$? Thanks $\endgroup$ – John Trail Dec 5 '15 at 22:24
  • $\begingroup$ It is not true what you have written above, Brownian motion $(B_t)$ has the property that $B_t - B_s \sim \mathcal{N}(0,t-s)$, and in particular $B_t \sim \mathcal{N}(0, t)$. The Ito process which you have mentioned is only a specific example and if $dX_t = \mu dt + \sigma dB_t$ then $X_t = X_0 + \mu t + B_t$, assuming that $X_0 = 0$, we have that $X_t = \mu t + B_t$. Thus $X_t$ is a Brownian motion with drift parameter $\mu$ and scale parameter $\sigma$ and $X_t \sim \mathcal{N}(t\mu, t\sigma^2)$. Does this answer your question? $\endgroup$ – m_gnacik Dec 6 '15 at 1:36
  • $\begingroup$ You might want to read math.uah.edu/stat/brown there are four sections that you should check and you will know all about Brownian motion :). Are you happy with my answer to accept it? $\endgroup$ – m_gnacik Dec 7 '15 at 0:07
  • $\begingroup$ Ah I think I got it now, thank you so much for the help! Sorry I took some time to get back in accepting it, indeed yes, I am very happy with it :) Appreciate it a lot! $\endgroup$ – John Trail Dec 7 '15 at 1:39

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