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If $W_0=0$, then $W_t\sim N(0,t)$, so $W_1^2\sim\chi_1^2$. But Ito's lemma for $f(x)=x^2$ states $\mathrm{d}\left(W_t^2\right)=\mathrm{d}t+2W_t\mathrm{d}W_t$, so $W_t^2=t+2\int_0^t{W_s\mathrm{d}W_s}$ and hence $\int_0^1{W_t\mathrm{d}W_t}=\frac{\chi^2-1}{2}$, which seems incorrect. Where am I making a mistake?

Junyong Kim
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  • $\int_0^1 W_t dW_t = \frac{W_1^2-1}{2}$. The term $W_1^2$ is indeed chi-squared distributed. Why do you think it is something else? Maybe you think it shouldn't have mean zero? – Ian Apr 24 '20 at 00:39
  • I assumed that there is no closed-form solution because could not find anywhere. I was also calculating $\int_0^1{W_t^2\mathrm{d}W_t}$ so would be able to find a similar closed-form solution in this way using $\mathrm{d}\left(W_t^3\right)=3W_t\mathrm{d}t+3W_t^2\mathrm{d}W_t$ if this approach is correct. – Junyong Kim Apr 24 '20 at 00:51
  • With $\int_0^1 W_t^2 dW_t$ it's not so closed form because it can't be separated from $\int_0^1 W_t dt$. – Ian Apr 24 '20 at 00:56
  • My mistake—I was thinking not $\int_0^1{W_t^2\mathrm{d}W_t}$ but $\int_0^1{W_t^2\mathrm{d}t}$. My apologies. – Junyong Kim Apr 24 '20 at 02:45
  • This is not about $\int_0^1{W_t^2}\mathrm{d}t$, but isn't $\int_0^1{W_t^2\mathrm{d}W_t}$ closed? From $\mathrm{d}\left(W_t^3\right)=3W_t\mathrm{d}t+3W_t^2\mathrm{d}W_t$, $W_1^3=3\int_0^1{W_t\mathrm{d}t}+3\int_0^1{W_t^2\mathrm{d}W_t}$, where $3\int_0^1{W_t\mathrm{d}t}=3\mathrm{Normal}\left(0,\frac{1}{3}\right)=W_{3}$. Therefore, $\int_0^1{W_t^2\mathrm{d}t}=\frac{W_1^3-W_{3}}{3}$? – Junyong Kim Apr 28 '20 at 05:11
  • You can get the distribution of $\int_0^1 W_t dt$ but that doesn't make it equal to $W_3$. – Ian Apr 28 '20 at 11:23
  • What if just $3\int_0^1{W_t\mathrm{d}t}=\mathrm{Normal}(0,3)$ then? Won't it be $\int_0^1{W_t^2\mathrm{d}t}=\frac{\left[\mathrm{Normal}(0,1)\right]^3+\mathrm{Normal}(0,3)}{3}$? – Junyong Kim Apr 28 '20 at 19:10
  • You can't just add dependent random variables by knowing their marginal distributions. It is true that $d(W_t^3)=3W_t dt + 3 W_t^2 dW_t$, and so it is also true that $3\int_0^1 W_t^2 dW_t=W_1^3-3\int_0^1 W_t dt$. It is also true that the distribution of $3 \int_0^1 W_t dt$ is $N(0,3)$. But the two are dependent, so it means nothing to add them up without taking the joint distribution into account. – Ian Apr 28 '20 at 19:28
  • Crystal clear. Thanks! – Junyong Kim Apr 28 '20 at 20:15

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