0
$\begingroup$

enter image description here

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.

$\endgroup$

1 Answer 1

1
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ Thank you very much for your help. $\endgroup$
    – Myshkin
    Commented Mar 20, 2019 at 11:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.