Let $\left( {{X_t}:t \in \left[ 0 \right.\left. {, + \infty } \right\rangle } \right)$ be a continuous time Markov chain on a probability space $\left( {\Omega ,\mathcal{F},\mathbb{P}} \right)$ with a finite state space $S$, defined by jump chain/holding times definition. Suppose $A,B \in \sigma \left( {{X_0}} \right)$ and $f\left( t \right) = \mathbb{E}\left[ {\mathbb{P}\left( {A|{X_t}} \right)|B} \right]$ is a decreasing function. Suppose I want to calculate $f'\left( t \right)$. We can assume that $f'\left( t \right)$ is always finite.

There are two ways to go about this, and the other one is probably wrong, but I want to know why.

1) Since $S$ is finite,

$\mathbb{E}\left[ {\mathbb{P}\left( {A|{X_t}} \right)|B} \right] = \sum\limits_{x \in S} {\mathbb{P}\left( {A|{X_t} = x} \right)\mathbb{P}\left( {{X_t} = x|B} \right)} \Rightarrow f'\left( t \right) = \sum\limits_{x \in S} {\frac{d}{{dt}}\left( {\mathbb{P}\left( {A|{X_t} = x} \right)\mathbb{P}\left( {{X_t} = x|B} \right)} \right)} $.

2) Since ${f'}$ is finite, $\frac{d}{{dt}}\mathbb{E}\left[ {\mathbb{P}\left( {A|{X_t}} \right)|B} \right]$ is finite so $\frac{d}{{dt}}\mathbb{P}\left( {A|{X_t}} \right)$ is finite $\mathbb{P}\left( { \cdot |B} \right)$-almost surely. Lebesgue dominated convergence theorem then implies $f'\left( t \right) = \mathbb{E}\left[ {\frac{d}{{dt}}\mathbb{P}\left( {A|{X_t}} \right)|B} \right] = \sum\limits_{x \in S} {\frac{d}{{dt}}\left( {\mathbb{P}\left( {A|{X_t} = x} \right)} \right)\mathbb{P}\left( {{X_t} = x|B} \right)} $.

Suppose 1) and 2) are both true. That would imply $\sum\limits_{x \in S} {\mathbb{P}\left( {A|{X_t} = x} \right)\frac{d}{{dt}}\left( {\mathbb{P}\left( {{X_t} = x|B} \right)} \right)} = 0$, which I have strong reasons to believe is wrong.

Suppose that my line of reasoning is wrong when I conclude that differentiation and expectation commute. Does 2) hold when they do commute? Under what (usual) applicable assumptions do they commute?

EDIT: The following holds for a discrete random variable $X$: $\mathbb{E}\left[ {g\left( X \right)} \right] = \sum\limits_x {g\left( x \right)\mathbb{P}\left( {g\left( X \right) = g\left( x \right)} \right)} = \sum\limits_x {g\left( x \right)\mathbb{P}\left( {X = x} \right)} $, where the first equality follows from definition, and the second is the law of unconscious statistician.

Let $g\left( x \right) = \frac{d}{{dt}}\mathbb{P}\left( {A|{X^{\left( t \right)}} = x} \right)$, then $\mathbb{E}\left[ {\frac{d}{{dt}}\mathbb{P}\left( {A|{X^{\left( t \right)}}} \right)|B} \right] = \mathbb{E}\left[ {g\left( {{X^{\left( t \right)}}} \right)|B} \right] = \sum\limits_x {g\left( x \right)\mathbb{P}\left( {{X^{\left( t \right)}} = x|B} \right)} = \sum\limits_x {\left( {\frac{d}{{dt}}\mathbb{P}\left( {A|{X^{\left( t \right)}} = x} \right)} \right)\mathbb{P}\left( {{X^{\left( t \right)}} = x|B} \right)} $.

  • 1
    $\begingroup$ Your 2. is wrong. How would you justify the second = sign when computing $f'(t)$? $\endgroup$ – Did Nov 7 '14 at 9:17
  • $\begingroup$ $\sum\limits_{y = g\left( x \right)} {y\mathbb{P}\left( {g\left( X \right) = y} \right)} = \sum\limits_x {g\left( x \right)\mathbb{P}\left( {X = x} \right)} $ $\endgroup$ – Alen Nov 7 '14 at 9:23
  • $\begingroup$ ?? Please be much more specific (and I am sorry but the formula in your comment is not true). $\endgroup$ – Did Nov 7 '14 at 9:24
  • 2
    $\begingroup$ To begin with, the LHS depends on $x$ while the RHS does not. Anyway this identity, even once corrected, does not obviously provide your 2., does it? $\endgroup$ – Did Nov 7 '14 at 15:55
  • 1
    $\begingroup$ You might be using $$\sum_{y=g(x)}A(y)$$ with in mind something different from what it actually means. (Point already explained.) Hence you should explain what you mean by this notation. (As already said.) Let me add that I have no incentive to force you to see the light if your main concern is to stonewall the approach (2.) against criticisms, although (2.) is at present squarely wrong. $\endgroup$ – Did Nov 7 '14 at 16:51

The equality $$\mathbb{E}\left[ {\frac{d} {{dt}}\mathbb{P}\left( {A|{X_t}} \right)|B} \right] = \sum\limits_{x \in S} {\mathbb{P}\left( {{X_t} = x|B} \right)\frac{d} {{dt}}\mathbb{P}\left( {A|{X_t} = x} \right)} $$ is wrong since $$\mathbb{P}\left( {A|{X_t}} \right) = \sum\limits_{x \in S} {\mathbb{P}\left( {A|{X_t} = x} \right){1_{\left\{ {{X_t} = x} \right\}}}} $$ so $$\frac{d} {{dt}}\mathbb{P}\left( {A|{X_t}} \right) = \sum\limits_{x \in S} {\frac{d} {{dt}}\left( {\mathbb{P}\left( {A|{X_t} = x} \right){1_{\left\{ {{X_t} = x} \right\}}}} \right)} $$

The random variable ${{1_{\left\{ {{X_t} = x} \right\}}}}$ also depends on $t$, which I've accidentally ignored.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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