Stochastic Calculus, Fall 2002 (http://www.math.nyu.edu/faculty/goodman/teaching/StochCalc/)

Assignment 2.

Given September 12, due September 19. Last revised, September 15.

Objective: Conditioning and Markov chains.

1. Suppose that F and G are two algebras of sets and that G adds information to F in the sense that any F measureable event is also G measurable. Since F and G are collections of events, this may be written F ⊂ G. Suppose that the probability space Ω is finite and that X(ω) is a variable defined on Ω (that is, a function of the random variable ω). The conditional expectations (in the modern sense) of X with respect to F and G are Y = E[X | F] and Z = E[X | G]. In each case below, state whether the statement is true or false and explain your answer with a proof or a counterexample.

a. Z ∈ F.

b. Y ∈ G.

c. Z = E[Y | G].

d. Y = E[Z | F].

2. For any event A ⊆ Ω we can define the “indicator function” (also called the “characteristic function”, particulatly by people who learned probability late in life), 1_{A}(ω) = 1 if ω ∈ A and 1_{A}(ω) = 0 if ω ∉ A. People who call this the “characteristic function” (usually people who learned probability late in life) write χ_{A}(ω) for 1_{A}(ω).

a. Show that E[1_{A}] = P (A).

b. For any event, B, show that the classical E[1_{A} | B] is the same as the Bayes’ rule conditional probability of A.

The “modern” definition of conditional probability, conditioning on an algebra of sets rather than a single set, is P (A | F) = E[1_{A} | F]. The relationship between the classical and modern definition of conditional probability is more or less the same as the relation between classical and modern expected value.

3. Let S be a finite state space and Ω be the set of paths of length T from S. Let P (X) be the probability of a path X ∈ Ω. For any t in the range 1 < t < T, let F_{t} be the algebra of events in Ω generated by the values of X_{s} for 1 ≤ s ≤ t. Let G_{t} be the smaller algebra generated only by X_{t}. Finally, let H_{t} be the “complementary” algebra (based on complementary information) generated by the values X_{t}, …, X_{T}. An event in H_{t} is a statement about the path from time t on without saying anything about the beginning values X_{1}, …, X_{t−1}. Show that the Markov property is equivalent to either of the following, a.

E[1_{A} | F_{t}] = E[1_{A} | G_{t}] for any A ∈ H_{t+1}.

b.

E[F (X) | F_{t}] = E[F (X) | G_{t}] for any F ∈ H_{t+1}.

Notes:(i) Part a is a special case of part b (why?). (ii) Part b implies that E[F (X) | F_{t}] is a function of X_{t} only. This us supposed to be intuitively clear as a consequence of the Markov property. (iii) Together with question 1, part b is a justification for the backward equation for expected values of final payouts.

4. Suppose we have a 3 state Markov chain with transition matrix

P = ( \begin{matrix}.6 & .2 & .2 \\ .3 & .5 & .2 \\ .1 & .2 & .7 \end{matrix} )

and suppose that X_{1} = 1.

a. Show that the probability distribution of the first t steps conditioned on G_{t+1} is the same as that conditioned on H_{t+1}. This is a kind of backwards Markov property: a forward Markov chain is a backward Markov chain also.

b. Calculate P (X_{2} = 2 | G_{3}). This consists of 3 numbers.