Here is a confusion regarding stochastic integrals. Let $Y_t=\int_0^tW_sds$ where $W_t$ is a Brownian Motion. Now $dY_t=W_tdt$. So from this expression one can conclude that $dY_t \cdot dY_t=d[Y_t,Y_t]$ where $[\ ]$ is the quadratic variation process and it must be zero as $dt \cdot dt =0$. But people apply Fubini-stochastic integral exchange trick and see $Y_t=\int_0^t(t-u)dW_u$. So by applying Ito isometry one can see that $E(Y_t^2)=\int_0^t(t-u)^2du \neq 0$. Can someone tell me where I'm going wrong? Why this discrepancy?

  • $\begingroup$ $\mathsf E(Y^2_t)$ is not its quadratic variation. Also, I am not sure that you did an integration by parts in a correct way. There shall be a term $tW_t$ $\endgroup$ – Ilya Jan 31 '13 at 11:23
  • $\begingroup$ Got something from an answer below? $\endgroup$ – Did Feb 9 '13 at 10:38

Every step is correct, including the identities $Y_t=\displaystyle\int_0^tW_s\mathrm ds=\int_0^t(t-s)\mathrm dW_s$ and $[Y]_t=0$, except the underlying assertion that the process $Y^2-[Y]$ should be a martingale and consequently that $\mathbb E(Y_t^2)$ and $\mathbb E([Y]_t)$ should coincide for every $t$.

Recall that the quadratic variation process $[Y]$ is defined as $[Y]_t=Y_t^2-2\displaystyle\int_0^tY_s\mathrm dY_s$. In particular, $Y^2-[Y]=2\displaystyle\int_0^\cdot Y\mathrm dY$ is a martingale when $Y$ is, but not otherwise.

Here, $Y$ is not a martingale since, for every $s\lt t$, $\mathbb E(Y_t\mid\mathcal F_s)=Y_s+(t-s)W_s\ne Y_s$.

Edit: A general method (which is not a trick) to compute $\mathbb E(Y_t^2)$ is indeed to use Itô's formula for $Y^2$. One gets $$ \mathrm d(Y_t^2)=2Y_t\mathrm dY_t+\mathrm d[Y]_t=2Y_tW_t\mathrm dt. $$ Fubini on $\Omega\times[0,t]$ with the measure $\mathbb P\otimes\mathrm{Leb}$ yields $$ \mathbb E(Y_t^2)=\mathbb E\left(\int_0^t2Y_sW_s\mathrm ds\right)=\int_0^t2\mathbb E(Y_sW_s)\mathrm ds. $$ Fubini again yields $$ 2\mathbb E(Y_sW_s)=2\int_0^s\mathbb E(W_uW_s)\mathrm du=\int_0^s2u\mathrm du=s^2, $$ hence $$ \mathbb E(Y_t^2)=\int_0^ts^2\mathrm ds=\frac{t^3}3. $$

  • $\begingroup$ Then how can one use Ito formula to compute $E(Y_t^2)$ though another method could be given by approximating $W_t$ by some representations in dyadic intervals and then computing the lebesgue integral $\int_0^tW_sds$ and finally taking its square and limit.But I do not want to use this method and instead want to use some trick.Is there any..?? $\endgroup$ – user39646 Jan 31 '13 at 16:52
  • $\begingroup$ See Edit. $ $ $ $ $\endgroup$ – Did Jan 31 '13 at 17:44
  • $\begingroup$ Thank you very much Did...But my last doubt is what about applying Ito isometry if the integrand involves "t".For example the same $t^3/3$ result can be got if i apply ito isometry to $\int_0^t(t-s)dW_s$.please clarify ...!! $\endgroup$ – user39646 Feb 1 '13 at 13:30
  • $\begingroup$ Then what about it? I do not feel responsible for unsubstantiated assertions in other answers. Of course the same result is obtained by Ito isometry even if the integrand involves $t$! $\endgroup$ – Did Feb 1 '13 at 15:22
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    $\begingroup$ @Monolite Because $Y$ is, and because $y\mapsto y^2$ is regular ($C^2$). $\endgroup$ – Did Jan 30 '18 at 6:37

Your second representation is correct, but you cannot treat it as a stochastic integral (so will not be a martingale) as the integrand depends on the end point of the integral, $t$. Even if it were a martingale, you cannot apply the Itou isometry to it for the same reason. Basically, the problem is that $Y_t$ are integrands against completely different processes in the second representation.

  • $\begingroup$ The result of your partial copying of my answer is an odd mixture of true (but irrelevant) assertions and of false results. For example, one can very well treat it as a stochastic integral even if the integrand depends on the end point of the integral, one can very well apply the Itou (sic) isometry and both representations yield exactly identical processes. $\endgroup$ – Did Jan 31 '13 at 13:47
  • $\begingroup$ Apologies, I didn't mean to copy your answer (I posted without refreshing for a while then noticed you posted and admittedly edited it to note it won't be a martingale). Yes, sorry. I should have just said "we cannot treat it as a stochastic integral process (only a stochastic integral up to a single fixed time), so it will not be a martingale". I generally use Itou because it conforms with the most logical and easy to type transliteration system for Japanese. Wikipedia uses Itō, and a variety of others have been used, so I don't think Ito is the "correct" one. $\endgroup$ – user52119 Feb 1 '13 at 2:18

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