I am reading through an introduction to h-transform (available on https://linbaba.wordpress.com/2010/06/02/doob-h-transforms/), and came upon the following equality: $$P\left(X_{t+s}=y;X_T\in A|X_t=x\right)=P\left(X_{t+s}=y|X_t=x\right)P\left(X_T\in A|X_{t+s}=y\right)$$ $X_t$ is a Markov process and $A$ is a subset of the state space. From other books (Brownian motion and martingale in analysis) where the same step appears, it is said to be an application of Markov property. However, Markov property is commonly found under this form : $$ E\left[f\left(X_{t+s}\right)\right|\mathcal{F}_s]=E^{X_s}\left[f\left(X_{t}\right)\right]$$ I could rewrite $P\left(X_{t+s}=y;X_T\in A|X_t=x\right)$ as $E\left[1\{X_{t+s}=y;X_T\in A\}|\mathcal{F}_t\right]=E^{X_t}\left[1\{X_{t+s}=y;X_T\in A\}\right]$, but that already feels weird and does not seem to bring me any closer to the proposed result.

Any suggestions appreciated, thanks :)


Following your approach you can write :

$E[1\{X_{t+s}=y,X_T\in A \}|\mathcal{F}_t]$
$=(1)E^{X_t=x}[1\{X_{t+s}=y\}.1\{X_T\in A\}]$
$=(2)E^{X_t=x}[E^{X_{t+s}=y}[1\{X_{t+s}=y\}.1\{X_T\in A\}]]$
$=(3)E^{X_t=x}[1\{X_{t+s}=y\}.E^{X_{t+s}=y}[1\{X_T\in A\}]]$ $=(4)E^{X_t=x}[1\{X_{t+s}=y\}.P(X_T\in A|X_{t+s}=y)]$ $=(5)E^{X_t=x}[1\{X_{t+s}=y\}].P(X_T\in A|X_{t+s}=y)$ $=(6)P(X_{t+s}=y|X_t=x).P(X_T\in A|X_{t+s}=y)$

With the following justifications :
(1) : we just write the intersection as the product of two indicator functions
(2) : we are conditioning on $X_{t+s}=y$ (using Markov Property)
(3) : we pull out the $X_{t+s}$-measurable function of the conditional expectation
(4) : We rewrite the conditional expectation as a conditional probability.
(5) : The conditional probability has the property for fixed $A$ to be a function of $y$ so we can pull it out of the expectation (there exists a good version of the conditional probability s.t. for fixed $A$, $P(X_T\in A|X_{t+s}=y)$ can be written a.s. as measurable a function $g$ of $y$).
(6) : Rewrite conditional expectation into conditional probability form.

Note that for (6) which is your desired result it is only a formal expression as $P(X_{t+s}=y|X_t=x)$ is null. As a matter of fact, you cannot conditionate on an event of null probability. To do this more rigorously you need at some step to use the good version of conditional probability and rewrite the whole thing using an event of the form "$X_{t+s}\in [y,y+dy)$" until you can rewrite (6) into something like :

$=(6)'P(X_{t+s}=dy|X_t=x).P(X_T\in A|X_{t+s}=y)$

Best regards

  • $\begingroup$ Thank you I'm studying your note right now, really helpful. In between step (1) and (2) you rewrite $E[1\{X_{t+s}=y\}1\{X_T\in A\}]$ as $E[1\{X_{t+s}=y\}1\{X_T\in A\}|F_{t+s}]$ using that both variables are $F_{t+s}$ measurable, is that right ? Also, regarding the step 5, you consider the conditional probability as a $F_{t+s}$ measurable function, but it would have to be $F_t$ measurable to be pulled out of the expectation conditioned on $X_{t}$, which isn't the case automatically. Isn't that correct ? Thank you. $\endgroup$ – zebullon May 12 '15 at 5:47
  • $\begingroup$ @ zebullon : Hi in step (1-2) I use the fact that $E[Y]=E[E[Y|\mathcal{G}]]$ for any good random variable and any sigma algebra, and the Markov property : $E[1\{X_{t+s}=y\}1\{X_T\in A\}|F_{t+s}]=E[1\{X_{t+s}=y\}1\{X_T\in A\}|X_{t+s}=y]$ (or with your notation $=E^{X_{t+s}=y}[1\{X_{t+s}=y\}1\{X_T\in A\}]$). Best regards $\endgroup$ – TheBridge May 12 '15 at 8:15

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