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What is the conditional distribution of $$X_t = \int_0^t W_s \mathrm{d}s$$with respect to $W_t = x$?

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Note that \begin{align*} X_t = tW_t -\int_0^t sdW_s = \int_0^t (t-s)dW_s, \end{align*} and \begin{align*} W_t = \int_0^t dW_s. \end{align*} Then, for any real numbers $a$ and $b$, \begin{align*} aX_t + b W_t = \int_0^t (at-as+b)dW_s, \end{align*} is normal. That is, $W_t$ and $X_t$ are jointly normal. Moreover. note that \begin{align*} E\bigg(W_t\bigg(X_t - \frac{Cov(X_t, W_t)}{Var(W_t)}W_t\bigg)\bigg) = 0. \end{align*} That is, $\frac{Cov(X_t, W_t)}{Var(W_t)}W_t= \frac{1}{2}tW_t$ and $X_t-\frac{1}{2}tW_t$ are independent. Given that \begin{align*} E(X_t^2) &= \int_0^t(t-s)^2 ds = \frac{1}{3}t^3, \end{align*} then \begin{align*} Var\big(X_t - \frac{1}{2}tW_t\big) = \frac{1}{12}t^3. \end{align*} Therefore, \begin{align*} P(X_t \le y \mid W_t) &= P\bigg(X_t - \frac{1}{2}tW_t + \frac{1}{2}tW_t \le y \mid W_t \bigg)\\ &= P\big(X_t - \frac{1}{2}tW_t \le y - \frac{1}{2}tW_t \mid W_t \big)\\ &=\int_{-\infty}^{y-\frac{1}{2}tW_t} \frac{1}{\sqrt{\frac{1}{6}\pi t^3}}e^{-\frac{z^2}{\frac{1}{6} t^3}} dz. \end{align*} That is, the conditional distribution of $X_t$ given $W_t=x$ is normal with the density function \begin{align*} \frac{1}{\sqrt{\frac{1}{6}\pi t^3}}e^{-\frac{(y-\frac{1}{2}tx )^2}{\frac{1}{6} t^3}}. \end{align*}

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  • $\begingroup$ Thanks so much for your Answer... $\endgroup$
    – Alicia
    Apr 27 '20 at 0:21
  • $\begingroup$ You are welcome. $\endgroup$
    – Gordon
    Apr 27 '20 at 11:17

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