characterize the distribution of $\int_0^T f(t)Z_tdt$. In particular, verify that it is a Gaussian distribution and compute its moments.


It was shown in another question that \begin{align} \int^T_0 f(t) Z_t dt &= Z_T \int^T_0 f(u) du - \int^T_0 \Bigl( \int^s_0 f(u) du \Bigr) dZ_s\\ &= Z_T \cdot F(T) - \int^T_0 F(s) \cdot dZ_s \end{align} where I've defined the integral $F(T) = \int^T_0 f(t) dt$, and note this is just another deterministic function of time.

Both of the two terms in the solution are normally distributed with mean 0, so their difference must also be normally distributed with mean 0.

In order to calculate the variance of the combined quantity, we remember that: \begin{align} {\mathbb V}[A - B] &= {\mathbb E}[(A-B)^2] - {\mathbb E}[(A-B)]^2 \\ &= {\mathbb E}[A^2] + {\mathbb E}[B^2] - 2 \cdot {\mathbb E}[AB] \end{align} The first two terms are the variances of the two terms above, which are $F(T)^2 \cdot T$ and $\int_0^T F(t)^2 dt$ (as shown in this answer), which just leaves the remaining ${\mathbb E}[AB]$ term to calculate, for which we use the Ito Isometry: \begin{align} {\mathbb E}[AB] &= {\mathbb E}\Bigl[ F(T) \cdot Z_T \cdot \int_0^T F(t) dZ_t \Bigr] \\ &= {\mathbb E}\Bigl[ F(T) \cdot \int_0^T dZ_t \cdot \int_0^T F(t) dZ_t \Bigr] \\ &= F(T) \cdot \int_0^T F(t) dt \end{align}

so the variance is $F(T)^2 \cdot T + \int_0^T F(t)^2 dt - 2 \cdot F(T) \cdot \int_0^T F(t) dt$, and \begin{align} \int^T_0 f(t) Z_t dt \sim {\mathbb N}\Bigl( 0, F(T)^2 \cdot T + \int_0^T F(t)^2 dt - 2 \cdot F(T) \cdot \int_0^T F(t) dt \Bigr) \end{align}

Since this is gaussian in $T$, it is completely characterised by its mean and variance, and higher moments can be calculated in the usual way (odd moments are all 0, even moments as given in the table in the link)

  • $\begingroup$ I suppose we're assuming that $f(t)$ is deterministic: correct? $\endgroup$ – Jan Stuller Oct 13 '20 at 8:05
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    $\begingroup$ Yes... in the post above the notation used was $f(t)$, I've assumed that this is NOT $f(t, Z_t)$ $\endgroup$ – StackG Oct 13 '20 at 8:07
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    $\begingroup$ Cool, that makes sense. Nice answer btw. $\endgroup$ – Jan Stuller Oct 13 '20 at 8:08
  • $\begingroup$ Many thanks!!!! $\endgroup$ – StackG Oct 13 '20 at 8:15

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