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Hi, Could anyone explain to me how he wrote $(3.3)$ to $(3.4)$, in particular, why double integration? I am convinced with $(3.3)$ and another thing, I do not quite understand the role of $p(x,t)$.

Thank you very much for helping.


1 Answer 1


The function $p$ is the conditional density of $t_n$ given $x_n$ (with respect to Lebesgue measure on $[0, T]$). That is, equation (3.2) implies that $$ \Pr(t_n \in \mathcal{T} \mid x_n) = \int_{\mathcal{T}} p(x_n, u) \, du $$ for any Borel set $\mathcal{T} \subseteq [0, T]$.

It follows from some "usual" measure-theoretic arguments that $$ \tag{1} E[g(x_n, t_n) \mid x_n] = \int_0^T g(x_n, u) p(x_n, u) \, du $$ for any sufficiently nice (e.g., bounded and measurable) function $g : X \times [0, T] \to \mathbb{R}$.

Now we can verify equation (3.4): $$ \begin{aligned} E[h(x_{n+1})] &= E[h(S(x_n, t_n))] \\ &= E[E[h(S(x_n, t_n)) \mid x_n]] &&\text{double expectation theorem}\\ &= E\left[\int_0^T h(S(x_n, u)) p(x_n, u) \, du\right] &&\text{using (1) with $g(x, t)=h(S(x, t))$} \\&= \int_X \left[\int_0^T h(S(x, t)) p(x, t) \, dt\right] \, \mu_n(dx) &&\text{since $\mu_n$ is the distribution of $x_n$} \end{aligned} $$

  • $\begingroup$ Thank you very much for your help. $\endgroup$
    – Myshkin
    Commented Mar 20, 2019 at 11:45

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