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The following question is based on the slides 3-5 in http://www.maths.lancs.ac.uk/~belton/www/notes/23iv14.pdf. The material is consistent with the book "Levy Processes and Stochastic Calculus", section 3.1.

Let $S$ be a topological space and $X_t: S\to S$ a stochastic process on $S$ where $t\in [0,\infty)$. Moreover, $X_t$ satisfies the Markov property $$\mathbb{E}[f(X_{s+t})| (X_r)_{0\leq r\leq s}] = \mathbb{E}[f(X_{s+t})|X_s]$$ for $s,t\geq 0$ and $f: S\to S$ measurable and bounded. In addition, $X_t$ is time homogeneous, i.e. $$\mathbb{E}[f(X_{s+t})|X_s=x] = \mathbb{E}[f(X_t)|X_0=x]$$ for $s$, $t$ and $f$ as above.

Then the claim is that $(T_t f)(x)=\mathbb{E}[f(X_t)|X_0=x]$ defines a semigroup of operators, i.e. $T_t T_s f = T_{t+s} f$.

As said, the calculation for this can be found in the slides I've linked above. I don't understand the line $$\mathbb{E}[\mathbb{E}[f(X_{s+t})|X_s]|X_0=x] = \mathbb{E}[(T_t f)(X_s)|X_0 = x].$$ Any help is appreciated!

Sebastian Bechtel
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1 Answers1

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For the mentioned line, time homogenity has been used. Namely, $E[f(X_{s+t})|X_s](\omega)=E[f(X_{s+t})|X_s=X_s(\omega)]=E[f(X_{t})|X_0=X_s(\omega)]=(T_tf)(X_s(\omega))$ for all $\omega\in \Omega$. Hence, $E[f(X_{s+t})|X_s]=(T_tf)(X_s)$

Manan
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  • Thanks! My mistake was that I thought $EX|Y=E[X|Y=\omega]$ but this seems to be only true in the discrete case, whereas in the continuous case one uses the conditional expectation w.r.t the initial $\sigma$-algebra of $Y$ and then has $E[X|Y=\omega]=E[X|\sigma(Y)]\circ Y$. – Sebastian Bechtel Jun 05 '18 at 07:54
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    I think that $E[X|Y=y]$ is calculated by using the regular conditional distribution, whereas $E[X|Y]$ is a conditional expectation, which is a random variable. Maybe you can see Stefan Hansen's answer in https://math.stackexchange.com/questions/496608/formal-definition-of-conditional-probability – Manan Jun 05 '18 at 12:13