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How do you prove Standard Brownian Motion (SBM) is a Markov Process using the independence lemma?

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asked May 29 in BUS 4028F - Financial Economics by mike_symmonds (280 points)

Attached is my attempt to prove that SBM is a Markov process using the independence lemma. My question is whether my definition of g(z) is appropriate/suitable. In the explanation of the independence lemma in the slides [slide 19/40 of the Binomial Model (2) slides], g(z) is defined as an unconditional expectation. However, if this is not suitable, I am not sure how to prove that SBM is a Markov process by the definition on slide 22/40 of the Binomial Model (2) slides. If this is the case, how should one go about proving the result? Thank you!

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2 Answers

+1 vote
answered May 30 by ChrisS (380 points)

Hi Mike

So you seem to be on the right track. Just a few notes:

The independence lemma, given all the criteria gives us the result \(g(Z) = \mathbb{E}^\mathbb{P}[f(X,Y)|\mathcal{G}]\). Thus, by the independence lemma, we can conclude that \(\mathbb{E}^\mathbb{P}[W_t|\mathcal{F}^x_s] = g(W_s)\). Since you've shown that \(g(Z) = W_s\), we can then conclude that \(\mathbb{E}^\mathbb{P}[W_t|\mathcal{F}^x_s] = W_s\). Now we can repeat the above process, conditioning instead on \(W_s\), and show that we get the same result. Thus, we can conclude that a Brownian Motion depends only on the most recent history, and not the history as a whole (which is the Markov property).

+1 vote
answered May 30 by ABackwell (560 points)

Chris's note is in the correct direction, but I've had a few questions about this issue, so I think I'll give a full solution. Firstly, note what we're trying to show: $$\mathbb{E}[f(W_t) |\mathcal{F}^W_s]=\mathbb{E}[f(W_t) |W_s],$$ for all \(0\leq s \leq t\) and for all functions \(f(\cdot)\). This is the definition of the Markov property.

Let's use the independence lemma to express the left-hand side. One should rewrite \(f\) so that it takes two arguments, and then apply the independence lemma:$$\mathbb{E}[f(W_t) |\mathcal{F}^W_s]=\mathbb{E}[f(W_s,W_t-W_s) |\mathcal{F}^W_s]=g(W_s),$$

where \(g(x)=\mathbb{E}[f(x,W_t-W_t)]\). In the first step, I am essentially defining a new function that takes two arguments, adds them together, and then applies the old function (strictly speaking, this new function should have a new name, but it is conventional to retain the old symbol \(f\) because the function ends up doing the same thing - you could give a new name if you prefer: \( h(x,y)=f(x+y) \) ). The correct definition of this new function justifies the first step. The second step is justified by the independence lemma - it is important that you understand that it is applicable and that this is the correct application.

Now let's consider the right-hand side. Again we can write $$\mathbb{E}[f(W_t) |W_s]=\mathbb{E}[f(W_s,W_t-W_s) |W_s]=g(W_s),$$

where \(g\) is defined as above. Once these two steps are justified, the result is proved. The first step relies on the same definition of the modified version of \(f\). The second step uses the independence lemma. The crucial insight is that using the independence lemma in this slight different circumstance still gives the same form for unconditional expectation (which we have been calling \(g\)).

If you cannot follow this argument, I think it will be worthwhile to try to figure out exactly where you are struggling and to address those points.

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