Let $\{X_t: t\geq 0\}$ be a stochastic process on the probability space $(\Omega,\mathcal{A},\mathbb{P})$ with values in $\mathbb{R}$ and $T:(\Omega,\mathcal{A},\mathbb{P})\rightarrow\mathbb{R}^{+}_0$ a random variable such that T is stochastically independent from the process $\{X_t: t\geq 0\}$. Furthermore there should hold that $\frac{\mathrm{d}\mathbb{P}_{X_t}}{\mathrm{d}\mu}=f_t(x)$ and $\frac{\mathrm{d}\mathbb{P}_{T}}{\mathrm{d}\nu}=g(x)$.

I want to show that $\mathbb{E}(X_T)=\int_{\mathbb{R}}\int_{\mathbb{R}} xf_t(x)\mathrm{d}\mu(x)\,g(t)\mathrm{d}\nu(t)$. To do this I used a transformation to write

$$\mathbb{E}(X_T)=\int_\Omega X_{T(\omega)}(\omega)\mathbb{P}(\omega)=\int_{\mathbb{R}\times\Omega}X_t(\omega)\mathrm{d}\mathbb{P}_{(T,\mathrm{id})}(t,\omega)$$

I now want to use the independence of $T$ and $X_t$ and then Fubini's theorem but unfortunately there doesn't hold $\mathbb{P}_{(T,\mathrm{id})}=\mathbb{P}_T\otimes \mathbb{P}$ in which case I could then conclude that

\begin{align} \int_{\mathbb{R}\times\Omega}X_t(\omega)\mathrm{d}\mathbb{P}_{(T,\mathrm{id})}(t,\omega)&=\int_{\mathbb{R}\times\Omega}X_t(\omega)\mathrm{d}(\mathbb{P}_T\otimes \mathbb{P})(t,\omega)=\int_{\mathbb{R}}\int_\Omega X_t(\omega)\mathrm{d}\mathbb{P}(\omega)\mathrm{d}\mathbb{P}_T(t)\\ &=\int_{\mathbb{R}}\int_\Omega x\mathrm{d}\mathbb{P}_{X_t}(x)\,g(t)\,\mathrm{d}\nu (t)=\int_{\mathbb{R}}\int_{\mathbb{R}} xf_t(x)\mathrm{d}\mu(x)\,g(t)\mathrm{d}\nu(t) \end{align} How can I show that in this case I can write $\mathbb{P}_{(T,\mathrm{id})}=\mathbb{P}_T\otimes \mathbb{P}$?

  • $\begingroup$ Do you assume that the process has right-continuous sample paths? $\endgroup$ – saz Dec 13 '15 at 14:09
  • $\begingroup$ in my case $X_t$ is a Poisson process but I wanted to show a more general statement. $\endgroup$ – lbf_1994 Dec 13 '15 at 14:11
  • $\begingroup$ I see. If I'm not mistaken, you need some additional assumption to ensure that $X_T$ is measurable. A sufficient condition is that the process has right-continuous sample paths. (If $(X_t)_{t \geq 0}$ has right-continuous sample paths, it is not difficult to see that the formula for $\mathbb{E}(|X_T|)$ holds; the idea is to approximate $T$ by discrete random variables.) Note that we also need some additional assumption to ensure that the expectation exists.) $\endgroup$ – saz Dec 13 '15 at 14:15
  • $\begingroup$ In my case $X_t$ has right continuous paths because its a Poisson process. So this condition is fulfilled. If I get you right I now have to approximate T by step functions and then use convergence theorems for the lebesgue integral, don't I? $\endgroup$ – lbf_1994 Dec 13 '15 at 14:20
  • 1
    $\begingroup$ Yeah, exactly. (I know that the Poisson process has right-cts. sample paths + moments of arbitrary order. Just keep in mind that you need additional assumptions if you want to prove it for a larger class of processes.) $\endgroup$ – saz Dec 13 '15 at 14:22

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