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We denote with $K_{s,t}$ a stochastic kernel from measurable space $(S,\mathcal{S})$ to itself. Further we assume that $P[X_t\in A|\mathcal{F}_s]=P[X_t\in A|X_s] = K_{s,t}(X_s,A)$, where $X=(X_t)$ is a stochastic process with values in $S$. I know the following result:

Let $X_i:(\Omega,\mathcal{F})\to (S_i,\mathcal{S}_i)$ with $i=1,2$ r.v. and $F:S_1\times S_2\to [0,\infty]$ is $\mathcal{S}_1\times \mathcal{S}_2$ measurable. Look at the vector $(X_1,X_2)$ as $S_1\times S_2$ valued r.v. We assume that the distribution of $(X_1,X_2)$ on $(S_1\times S_2,\mathcal{S}_1\times \mathcal{S}_2)$ is of the form $P_1\times K$ where $P_1$ is a probability measure on $(S_1,\mathcal{S}_1)$ (the distribution of $X_1$ under $P$) and $K$ a stochastic kernel from $(S_1,\mathcal{S}_1)$ to $(S_2, \mathcal{S}_2)$. Then we have $$E[F(X_1,X_2)|X_1](\omega)=\int_{S_2}F(X_1(\omega),x_2)K(X_1(\omega),dx_2)$$

Now why is the following equation true:

$$E[K_{t,u}(X_t,A)|\mathcal{F}_s]=\int_S K_{t,u}(y,A)K_{s,t}(X_s,dy)$$

for $s\le t\le u$. It seems that we take $F=K$, however this makes no sense for me. It would be appreciated if someone could explain this equation.


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1 Answer 1

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You know that, for every bounded measurable function $\varphi$, $$ E(\varphi(X_s,X_t)\mid\mathcal F_s)=E(\varphi(X_s,X_t)\mid X_s)=\int_S\varphi(X_s,y)K_ {s,t}(X_s,\mathrm dy). $$ Apply this to $\varphi:(x,y)\mapsto K_{t,u}(y,A)$.

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So in fact, your $\varphi$ does not depend on $x$? This was also my thought, but I was not sure about it. Is this for a fixed $A$? What I mean is, $\varphi$ depends on $y$ and $A$. –  math Jun 22 '12 at 7:18
Yes, one fixes $t$, $u$ and $A$ and then $\varphi(x,y)$ does not depend on $x$. –  Did Jun 22 '12 at 7:37

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