Proof of Chapman Kolmogorov Equation

[joy9696]

Hi,

I am unable to follow the proof on Chapman Kolmogorov equation (Ch3 - Markov Chains). Here they have derive the proof as an analogy from the combination of Markov property and total probability in the conditional form:

Wanted to understand the total probalility in the conditional form, especially the derivation of the equation:

P[B/C] = ∑ P[B/C, Ak] P[Ak/C]

The derivation has not been explained in CT3 while discussing total probability. Or have I missed something. Please help.

 

[jensen]

It does not look like something they'd ask you in exam. Can you please provide the full equation? I cant understand the notation.

 

[joy9696]

Hi,

Thank you for your response. This is with reference to the proof of the Chapman Kolmogorov Equation. This appears at the start of the chapter on Markov Chains.

The equation being proved is the Chapman Kolmogorov Eqn:

Pij (m,n) = ∑ k in S Pik (m,l) Pkj (l,n)

Proof: This is based on the Markov property (3.1) and on the law of total probability in its conditional form.

(The Markov Propery being referred to is:

P[Xn = j/X0 = i,X1=i1, ....,Xm-1=im-1,Xm=i] = P[Xn=j/Xm= i]
for all integers n>m, and states i0,i1,i2,....im-1, im, i,j in S )


If A1, A2,….Ak,….. form a complete setof disjoint events, ie:

Union(k from 1 to infinity) Ak = omega, Ak intersection Aj = empty set , k not equal to j.


Then for any two events B, C:
P[B/C] = ∑k=1 to infinity P[B/C, Ak] P[Ak/ C]

I would like to understand how the above equation/ property was derived.

Regards,

Joy

 

[MarkC]

(First, some notation. Each of the following summations is over all k. I'll use the & symbol to denote an intersection of sets.)

It sounds like you're happy with the "CT3 version" of the Law of Total Probability. This states that, for any event X (and sets A_k as you've already defined),

P(X) = ΣP(X|A_k)P(A_k)
... = ΣP(X&A_k).​

Putting X = B&C gives

P(B&C) = ΣP(B&C&A_k). <-- (1)​

By definition,

P(B|C) = P(B&C) ÷ P(C).​

Using (1),

P(B|C) = [ ΣP(B&C&A_k) ] ÷ P(C)
... = Σ[ P(B&C&A_k) ÷ P(C) ]
... = Σ[ [P(B&C&A_k)÷P(C&A_k)] * [P(C&A_k)÷P(C)] ]
... = Σ[ P(B|C&A_k) * P(A_k|C) ],​

which is the result we were looking for. I'll admit that I've missed out some of the technicalities (e.g. P(C) non-zero) but otherwise I think this works.

I would've thought the proof of this result was unlikely to come up. The author of the Core Reading seems to regard it as an "obvious" (hmm) extension of the CT3 formula, with a "|C" tacked on the end of everything. So it may be safe to just state it.

But I guess you never know what the Examiners want to throw at you! I hope this helps.