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I have got one question about Ito decomposition. Suppose $W_t$ is a Brownian Motion:

$X_t = W_t^2 + \int_0^t(W_t^3-1)du$

How to get $dX_t$? I am quited comfused by the integral. Should we calculate the integral first since it is just a normal integral. Or we take differential on both sides like: $dX_t=dW_t^2+(W_t^3-1)du$ and use Ito's formula to calculate $dW_t^2$? If it is the latter case, why can't we do the integral first? If it is the former case, why can we do the differential first? Thanks so much!

Brain Zhang
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2 Answers2

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By the very nature of paths of unbounded variation (such as those of Brownian motion), an expression such as $dW_t$ must be interpreted in integral form. In other words, when someone writes an expression of the form $$dX_t = b dt + \sigma dW_t, \quad t \geqslant 0,$$ what they really mean is $$ X_t - X_0 = \int_0^t b ds + \int_0^t \sigma dW_s, \quad t\geqslant 0.$$ The differential notation is shorthand for the integral form and is defined to be the process $X_t$ which satisfies the integrated form in a suitable sense.

To answer your question, you can do either. That is, you can compute the integral $\int_0^t (W_s^3 - 1) ds$ directly, and then apply Itô's formula when applying the differential (as you must do with the $W_t^2$ term). However, if all you want is to write the process in differential form, there is no need to do this. Indeed,

\begin{align*} d(X_t) &= d(W_t^2) + (W_t^3 - 1)dt \\ &= 2W_t dW_t + dt + (W_t^3 - 1)dt \\ &= 2W_t dW_t + W_t^3 dt. \end{align*} Again, in the above, the first equality only makes sense because $$W_t^2 - W_0^2 = \int_0^t 2W_s dW_s + \int_0^t ds.$$

snar
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  • hi, then my question is how to compute the integral $\int _0^t(W_s^3-1)ds$? – Brain Zhang Feb 05 '15 at 04:46
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    @EEEEVannnnnn Define "compute the integral" for a random quantity. – Did Feb 09 '15 at 08:37
  • @Did Would you mind giving a hint as to how to compute the variance of the mean zero Gaussian process $\int_0^t W_s^3 ds$ in a "simple" manner? (i.e., without going through the definition of the Riemann-Stieltjes integral) – snar Feb 09 '15 at 17:37
  • (1) This is not a Gaussian process. (2) Indeed each such integral is centered since $E(W_s^3)=0$ for every $s$. (3) To compute the second moment, try $$E\left(\left(\int_0^tW_s^3ds\right)^2\right)=2\int_0^t\int_0^sE(W_u^3W_s^3)duds,$$ which reduces the problem to the computation of $E(W_u^3W_s^3)$ for every $u\lt s$ (which you might be able to complete by yourself). – Did Feb 09 '15 at 23:50
  • Right, $W_s \rightarrow \int_0^t W_s^3 ds$ is not linear, my mistake. Thank you for (2) and (3). Do you then have any suggestions about how to identify the distribution of $\int_0^t W_s^3 ds$, or perhaps tell-tale signs of things to try when dealing with integrals or a Brownian Motion? Again, thanks for your time. – snar Feb 10 '15 at 00:22
  • This would make for a nice m.se question. (Unrelated: Please use @.) – Did Feb 11 '15 at 00:41
  • @Did Ah, thanks. Perhaps, but not so sure I did a great job: http://math.stackexchange.com/questions/1142932/identifying-distributions-of-integrated-stochastic-processes . – snar Feb 12 '15 at 02:48
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X(t) = F(X,t) = X^2 + (X^3-1)t

Now you can use taylor expansion in terms of t and x and 1/2x^2 thats it.

raj
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