Let me first state what I mean by a random measure with orthogonal increments.

Definition: A random measure with orthogonal increments $Z$ is a collection $\left(Z(B): B \in \mathcal{B}_{(-\pi,\pi]}\right)$ of zero-mean, complex random variables $Z(B)$ indexed by the Borel sets $\mathcal{B}_{(-\pi,\pi]}$ defined on some probability space $(\Omega,\mathcal{U},P)$ such that for some finite measure $\mu$ on $(-\pi,\pi]$ $$\mathrm{cov}(Z(B_1),Z(B_2))=\mu(B_1\cap B_2) \quad \forall{B_1,B_2}\in \mathcal{B}_{(-\pi,\pi]}$$

Honestly, I don't think I understand this definition. If $Z$ is a measure then it must be true that $Z(\emptyset) = 0$. How does this follow from the definition above exactly? Do I take $B_1 = B_2 = \emptyset$ and argue that given $\mathrm{cov}(Z(B_1),Z(B_2))=\mu(B_1\cap B_2) = 0$ and $E[Z(B)] = 0$ for any $B \in \mathcal{B}_{(-\pi,\pi]}$, $Z(\emptyset) = 0$ $\ P$-almost surely?

Next I would like to show that $Z(\cup_j B_j) = \sum_jZ(B_j)$ in mean square whenever $B_1,B_2,\ldots$ is a sequence of pairwise disjoint Borel sets. I am not sure what the mathematical formulation of this result is. My guess is $$E[\lvert Z(\bigcup_j B_j) - \sum_jZ(B_j)\rvert^2] = 0$$ as this would imply $Z(\cup_j B_j) = \sum_jZ(B_j)$ $\ P$-almost surely, which I think is the most you can expect from a random measure. (If it were an ordinary measure, we would insist on algebraic equality.)

Assuming that my guess is right I proceed. From the definition I have (after a few steps) $$\mathrm{var}\left(Z\left(\bigcup_j B_j\right)\right) = \mathrm{var}\left(\sum_jZ(B_j)\right) $$

I don't know how to continue from this point on. I would really appreciate some help. Thanks.

  • $\begingroup$ I removed my answer since then I guess you are interpreting the definition correctly. So this is a rather unusual "measure", $Z(B)$ is a random variable valued random variable defined on a probability space whose outcomes are Borel measurable sets? What is the text? $\endgroup$
    – muaddib
    Jul 19, 2015 at 18:42
  • $\begingroup$ @muaddib Thanks for trying to help. The point of this unusual measure is to define a stochastic integral which is different than the Ito integral. This new stochastic integral will later be used to prove the spectral representation theorem. $\endgroup$
    – Calculon
    Jul 19, 2015 at 19:48

1 Answer 1


You essentially have proven the result. I will restate and embellish upon your calculations to show that $Z$ satisfies the definition of a Vector Measure.

Let $\mathcal{F}$ be the Borel field for the interval $(-\pi, \pi]$. Let $X$ be the set of complex valued $L^2$, mean-zero, random variables on the probability space $(\Omega, \mathcal{U}, P)$. We then have the function $$Z : \mathcal{F} \to X$$

$Z(\emptyset) = 0$

By definition $Z(\emptyset)$ is the random variable in $X$ s.t. $$\mathrm{cov}(Z(\emptyset), Z(\emptyset)) = \mu(\emptyset \cap \emptyset) = 0$$ The only function in $X$ that satisfies this (i.e. has $L^2$ norm of $0$) is the zero function.

Additivity: Suppose $B_1, B_2$ are disjoint. Then $$Z(B_1 \cup B_2) = Z(B_1) + Z(B_2)$$

From the definition of $Z$ we know that $$\textrm{cov}(Z(B_1 \cup B_2), Z(B_i)) = \mu(B_i)$$ Therefore, by linearity of covariance, and additivity of $\mu$, and definition of $Z$: \begin{eqnarray*} \textrm{cov}(Z(B_1 \cup B_2), Z(B_1) + Z(B_2)) &=& \mu(B_1 \cup B_2) \\ &=& \sqrt{\mu(B_1 \cup B_2)} \sqrt{\mu(B_1) + \mu(B_2)}\\ &=& \sqrt{\textrm{var}(Z(B_1\cup B_2))}\sqrt{\textrm{var}(Z(B_1)) + 2\textrm{cov}(Z(B_1), Z(B_2)) + \textrm{var}(Z(B_2))} \\ &=& \sqrt{\textrm{var}(Z(B_1\cup B_2))}\sqrt{\textrm{var}(Z(B_1) + Z(B_2))} \\ \end{eqnarray*} This can only happen if $Z(B_1 \cup B_2)$ and $Z(B_1) + Z(B_2)$ are colinear. Additionally, since they have the same norm, they must be equal.

Countable Additivity: Suppose $B_i \in \mathcal{F}$, disjoint, then $$Z(\cup_{i=1}^\infty B_i) = \sum_{i=1}^\infty Z(B_i)$$
where the right hand side converges wrt to the norm of $X$.

The approach we can take here is to show $S_n = \sum_{i=1}^n Z(B_i)$ is cauchy wrt the $L^2$ norm.
\begin{eqnarray*} \int_\Omega \overline{(S_n - S_m)}(S_n - S_m) dP &=& \mathrm{Cov}(\sum_{i={m+1}}^nZ(B_i), \sum_{i={m+1}}^nZ(B_i)) \\ &=& \sum_{i={m+1}}^n \mathrm{Var}(Z(B_i)) \\ &=& \sum_{i={m+1}}^n \mu(B_i) \\ \end{eqnarray*} Here we used $Z(B_i)$ are uncorrelated since the $B_i$ are disjoint. Now since $\sum_{i=1}^\infty \mu(B_i) = \mu(\cup_{i = 1}^\infty B_i) < \infty$ (since $\mu$ is a finite measure) we know that the partial sums $\sum_{i=m}^n \mu(B_i)$ become arbitrarily small so we are done showing they are Cauchy.

Let $S$ be the limit of the Cauchy sequence $S_n = \sum_{i=1}^n Z(B_i) = Z(\cup_{i=1}^n B_i)$. It remains to show that $S = Z(\cup_{i=1}^\infty B_i)$. \begin{eqnarray*} ||Z(\cup_{i=1}^\infty B_i) - S||_2 &=& ||(Z(\cup_{i=1}^\infty B_i) - S_n) + (S_n - S)||_2 \\ &=& ||(Z(\cup_{i=n+1}^\infty B_i) + (S_n - S)||_2 \\ &\leq& ||(Z(\cup_{i=n+1}^\infty B_i)||_2 + ||S_n - S||_2 \\ &=& \mu(\cup_{i=n+1}^\infty B_i) + ||S_n - S||_2 \\ \end{eqnarray*} As above, $\mu(\cup_{i=n+1}^\infty B_i)$ becomes arbitrarily small as $n$ becomes large. Furthermore, $||S_n - S||_2$ does as well since $S_n$ converges to $S$.

  • $\begingroup$ I don't see why $\sum_i \mu(B_i) \leq 1$. $\endgroup$
    – Calculon
    Jul 20, 2015 at 12:39
  • $\begingroup$ $\mu$ is not a probability measure but it is a finite measure. So the sum is finite at least. I see it now thanks. $\endgroup$
    – Calculon
    Jul 20, 2015 at 12:42
  • $\begingroup$ Another question: $(S_n)$ is Cauchy and hence converges to something in $L^2(P)$. But how do you conclude that it converges to $Z(\cup_i B_i)$? $\endgroup$
    – Calculon
    Jul 20, 2015 at 12:43
  • $\begingroup$ @Calculon - I added in that part of the argument. $\endgroup$
    – muaddib
    Jul 20, 2015 at 13:03
  • $\begingroup$ Why did you define $f_n$ that way? I thought we wanted to show $S_n$ converges to the random variable $Z(\cup_iB_i)$. $\endgroup$
    – Calculon
    Jul 20, 2015 at 13:25

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