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Given a stochastic process $X: [0,T] \times \Omega \to \mathbb R$, I realized there are different meanings of $\int_0^T X(t) dt$.

  • $\int_0^T X(t, \omega) dt$, $\forall \omega \in \Omega$ or a.e., where the integral is Lebesgue integral wrt the Lebesgue measure on $[0,T]$. (This is the definition that seems most natural.)

  • $\int_0^T X(t, \omega) dt$, $\forall \omega \in \Omega$ or a.e., where the integral is Riemann integral over $[0,T]$. (This is the definition similar to Ito integral except the integrator $W(t)$ replaced with $t$.)

  • Given a partition $\mathcal P$ of $[0,T]$ into $0=t_0 < t_1 < t_2 < \dots < t_n=T$, define $\delta(\mathcal P) := \max_i (t_{i+1} - t_i)$. For each $i \leq n-1$, suppose that $t_i \leq \psi_i \leq t_{i+1}$. Define Riemann sum as $S_{\mathcal P} := \sum_i X(\psi_i) (t_{i+1} - t_i)$. If $S(\mathcal P)$ converges in $L^2$ norm as $\delta(\mathcal P)$ goes to $0$, then define $\int_0^T X(t) dt$ to be the limit. (This is the definition taken from Lamperti's Stochastic Processes.)

My questions are:

  • Is $\int_0^T X(t) dt$ always a random variable under each of the three definitions?

  • For a stochastic process $X$, does existence of one definition imply existence of another definition?

  • What if the first two definitions are relaxed to allow $\int_0^T X(t, \omega) dt$ exist a.e.?

  • In Ito isometry $E([\int_0^T X(t) dW(t)]^2) = E(\int_0^T X(t)^2 dt)$, how is $\int_0^T X(t)^2 dt$ on RHS defined?

Thanks and regards

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  • $\begingroup$ (This is the definition similar to Ito integral except the integrator W(t) replaced with t.)... Wow. This is wrong. $\endgroup$ – Did Feb 22 '13 at 14:41
  • $\begingroup$ @Did: Nice catch. I am sloppy here. I created the second definition using Riemann integral, because Ito integral is defined using randomized Riemann-Stieljes sum also. $\endgroup$ – Tim Feb 22 '13 at 14:47
  • $\begingroup$ Ito integral is defined using randomized Riemann-Stieljes sum... No it is not. $\endgroup$ – Did Feb 22 '13 at 14:50
  • $\begingroup$ @Did: Nice catch again. I am sloppy. I am using the definition from Shreve's Stochatic calculus in Finance. It define Ito integral first for a simple process using a Riemann-Stieljes sum with special points picked as the beginning of each interval in a partition. Then it generalizes the definition to a $L^2$ process. $\endgroup$ – Tim Feb 22 '13 at 14:55
  • $\begingroup$ If the Riemann integral of $X:[0,T]\to\Bbb R$ exists, it equals the Lebesgue integral of such function. $\endgroup$ – Ilya Feb 22 '13 at 14:59
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I am in the middle of taking a stochastic calculus course, so I can't guarantee that all my answers are 100% correct, but I can share some ideas that my instructor mentioned related to the first question:

$ \int_0^T X(t)dt $ is non-random since this is an integral with respect to T, and therefore we are not looking at the randomness of X(t) but rather we are looking at the entire path, and integrating over this given path.

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    $\begingroup$ It is random, since $$ \int_0^T X(t)\mathrm dt = \int_0^T X(t,\omega)\mathrm dt $$ to be precise. $\endgroup$ – Ilya Feb 22 '13 at 16:01

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