# How can we say that $B_n$ is a Markov process (or something)?

From Probability with Martingales:

I chose $$\mathscr F_n = \sigma(B_1, B_2, ..., B_n)$$.

My argument assumes that $$E[M_n | \mathscr F_{n-1}] = E[M_n | B_{n-1}]$$. I was able to show that $$M_{n-1} = E[M_n | B_{n-1}]$$.

How does one show that $$E[M_n | \mathscr F_{n-1}] = E[M_n | B_{n-1}]$$ or $$E[M_n | \mathscr F_{n-1}] = M_{n-1}$$?

Any chance we have something like $$B_n \le B_{n+1}$$?

If not, might it help if I define $$B_n = X_1 + ... + X_n$$ where $$X_i$$'s are iid Bernoulli conditioned on some probability p (or $$\Theta$$?) and then use a similar argument to the one here?

• Can you prove that $(B_n)$ is a Markov chain ? In this case, the equality you want to prove is easy to derive. Commented Dec 27, 2015 at 10:28
• @StéphaneLaurent You mean prove that $P(B_n = k_n | B_1 = k_1, ..., B_{n-1} = k_{n-1}) = P(B_n = k_n | B_{n-1} = k_{n-1})$ for $k_i \in \{0, 1, ..., n\}$ like here ?
– BCLC
Commented Dec 27, 2015 at 10:40

Intuition: this problem illustrates a fairly common strategy for constructing a martingale. First, a collection of random variables $$(B_n)_{n\geq 1}$$ is generated sequentially, with $$B_n$$ defined in terms of $$B_{n-1}$$. The method of construction ensures that, though they are not independent, the dependence structure is relatively simple in that they form a Markov chain. However, typically the conditional expectation requirement for a martingale is not satisfied yet, so we seek functions $$f_n$$ such that the random variables $$M_n=f_n(B_n)$$ do satisfy the martingale condition. In this case $$f_n(x)=\frac{x+1}{n+2}$$.
Showing that we have a Markov chain: To show that $$(B_n)_{n\geq 1}$$ is a markov chain, we want to show that $$P(B_n = k_n | B_1 = k_1, ..., B_{n-1} = k_{n-1}) = P(B_n = k_n | B_{n-1} = k_{n-1})$$ where the numbers $$k_i$$ are non-decreasing with $$k_i\leq i$$.
If we condition on $$B_1 = k_1, ..., B_{n-1} = k_{n-1}$$ then we know that there are $$k_{n-1}+1$$ black balls in the urn (out of a total of $$n+1$$ balls) just after time $$n-1$$. So the conditional probability that $$B_n=k_n$$ is either $$1-\frac{k_{n-1}+1}{n+1}$$ (if $$k_n=k_{n-1}$$ i.e. white ball chosen) or $$\frac{k_{n-1}+1}{n+1}$$ (if $$k_n=k_{n-1}+1$$ i.e. black ball chosen). Now, you might ask, what about our knowledge of $$B_1$$,$$B_2$$,...,$$B_{n-2}$$? Well, yes we know them, but their values don't affect the conditional probabilties we just computed. This is just a consequence of the way the sequence $$(B_n)_{n\geq 1}$$ is constructed.
If we condition only on $$B_{n-1} = k_{n-1}$$ then... we obtain exactly the same probabilities! The probabilities are the same because $$k_1,...,k_{n-2}$$ are irrelevant once we know $$k_{n-1}$$. So $$(B_n)_{n\geq 1}$$ is indeed a Markov chain.