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Define $X_t=\int^t_0 W_s^2ds$, what will be the distribution of $X_t$? My approach is as follow: Let $f(s)=W_s^2s$, by Ito's lemma we have $X_t=W_t^2t-2\int^t_0W_ssdW_s-\frac{t^2}{2}$.

  1. Discretize the integral $\int^t_0W_ssdW_s$, gives $lim_{n\rightarrow\infty}\sum_{i=1}^{n}((W_{t_i}-W_{t_{i-1}})^2\times(t_{i}-t_{i-1}))$, which is an infinite sum of iid gamma dist. What is the distribution ?

  2. Overall, what is the distribution of $X_t$ ?

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  • $\begingroup$ It does not have a known distribution. Why do you need this? For example do you need the moments? $\endgroup$ – Math-fun Apr 8 '15 at 12:56
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Out of Borodin and Salminen Handbook of Brownian Motion - Facts and Formulae:

$$ E_x\left[\exp\left(-\frac{\gamma^2}{2} \int_0^t W_s^2 ds \right) \right] = \frac{1}{\sqrt{\cosh(t\gamma)}}\exp\left( -\frac{x^2\gamma \sinh(t\gamma)}{2 \cosh(t\gamma)} \right) $$

You can obtain it by using Girsanov's theorem. Really nifty formula when $x=0$. I also recommend Yor and Pitman on all matters dealing with Bessel processes.

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As I commented already, this does not have a known distribution.

To see this let's derive the characteristic function of $X_t$:

\begin{align} \phi(u)&=E\exp{\Big(iuX(t)\Big)}\\ &=E\exp{\Big(iu\int_0^tW(s)^2ds\Big)}\\ &=E\exp{\Big(iu\int_0^tJ_c(s)^2ds+c\int_0^tJ_c(s)dJ_c(s)+\frac12c^2\int_0^tJ_c(s)ds \Big)}\\ \end{align} where I have used Girsanov's theorem with $dJ_c(s)=-cJ_c(s)ds+dW(s)$. Now choose $c=\sqrt{-2iu}$ to obtain \begin{align} \phi(u)&=E\exp{\Big(c\int_0^tJ_c(s)dJ_c(s)\Big)}\\ &=E\exp{\Big(\frac c2(J^2_c(t)-t)\Big)}\\ &=\exp{\Big(-\frac{tc}{2}\Big)}E\exp{\Big(\frac c2J^2_c(t)\Big)}\\ &=\frac{\exp{\Big(-\frac{tc}{2}\Big)}}{\sqrt{1-c\frac{1-e^{-2ct}}{2c}}}\\ &=\frac{\exp{\Big(-\frac{tc}{2}\Big)}}{\sqrt{1+\tanh(ct)}}\\ \end{align}

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  • $\begingroup$ Since he didn't ask for the pdf or cdf of this r.v, why isnt the Fourrier or the Laplace transform enough? $\endgroup$ – user3371583 Apr 8 '15 at 13:51
  • $\begingroup$ the only thing I could think of for $X_t$ was the char. function. As the char. function does not correspond to a known distribution, we could possibly think that this is not a known distribution. But then if we really need the pdf, we could (at leas theoretically) invert the transform to have the pdf. so having this transform could be helpful. $\endgroup$ – Math-fun Apr 8 '15 at 13:53
  • $\begingroup$ But then the OP tries to match his expansions with some dist, hence a pdf is desired. $\endgroup$ – Math-fun Apr 8 '15 at 13:55

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