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Brownian Motion and Battery Storage Homework Assignment

Q. 1 (Gymnastics). Let {Bt} be standard Brownian motion and fix 0 < s < t.

a. Compute E(B7t).

b. Compute E(e−B2t ).

c. Compute E(|Bt|).

d. Compute the joint distribution P{Bs ∈ dx, Bt ∈ dy}.

e. Compute the conditional distribution P{Bt ∈ dy|Bs = x}.

Q. 2 (Brownian transformations). Let {Bt} be standard Brownian motion.

a. Let s > 0 and define B˜t := Bs+t − Bs. Show that {B˜t} is also Brownian motion.

b. Let c > 0 and define Bˆt := cBt/c2 . Show that {Bˆt} is also Brownian motion.

One way of interpreting parts a. and b. is that if we shift or zoom in on the path of a Brownian motion, it always still looks like Brownian motion no matter how far we zoom. This shows that Brownian motion is self-similar or, more colloquially, a fractal.

Q. 3 (Brownian motion with drift). Let Sn = X1 + X2 + · · · + Xn define a biased simple random walk with P{Xk = 1} = p and P{Xk = −1} = 1 − p. This describes the motion through water of a pollen particle that has a propellor attached to it which is pushing it in a particular direction. (While you should never take homework problems too seriously, this is not as crazy as it may seem: some bacteria have propellors attached to them, called flagella.) Even though the random walk is biased, there is still a sensible continuous time limit. To obtain it, we define as before the process

C Nt = εX1 + εX2 + · · · + εXNt, t ≥ 0.

In order for the continuous time limit to make sense, both the step size ε and the bias p must be scaled as functions of the number of steps per unit time N.

a. In what way(s) can you scale ε and p to get a nontrivial continuous time limit?

b. Express the continuous time limit {Ct} of the biased random walk (called Brownian motion with drift) in terms of a standard Brownian motion {Bt}.

Q. 4 (Battery storage). Every second, the battery on your laptop loses a fraction 0 < 1 − α < 1 of its power. However, in every second your battery also gets recharged. Because you have switched to all-wind power, however, the amount of power you get tends to be a bit spotty: sometimes it is a lot, sometimes not so much, and sometimes you actually drain your laptop battery a little more when there is no wind and you want to run your electric toothbrush. Assume that the amount of extra charge you get in the kth second is a random variable Zk ∼ N(0, 1), and that Z1, Z2, . . . are independent.1

Suppose we start initially with a healthy charge of X0 = 100 bars. If Xn is the total

charge in your batter after n seconds, then we can express our model as follows:

Xn = αXn−1 + Zn, n ≥ 1.

a. Find a formula that expresses Xn in terms of X0 and Z1, . . . , Zn only.

b. What is the distribution of the random variable Xn?

c. What happens to this distribution after a long time as n → ∞?

5.Q. 5 (SETI). Aliens transmit a binary message S to earth: the message is either +a (“we come in peace”) or −a (“we will conquer your planet and turn you into slaves”) with probabilities P{S = +a} = p, P{S = −a} = 1−p. Unfortunately, the message is transmitted from the aliens’ home planet Qzur0rp, which is 12 lightyears away, and the signal received on earth has been corrupted by a lot of noise: the received signal is R = S + X, where the noise X ∼ N(0, σ2) is independent of the message S. Concerned astronomers are trying to guess the message S on the basis of the signal R. Of course, the larger R, the more likely it is that the message was +a, while the smaller R, the more likely it is that the message was −a. The astronomers therefore settle on the following ad-hoc decision scheme. They fix a number x, and decide that the message must have been +a if R > x, while the message must have been −a if R ≤ x. The probability that the astronomers have incorrectly decoded the message is

?(x) = P{S = +a, R ≤ x} + P{S = −a, R > x}.

a. Compute ?(x) for any x.

b. Find the critical number x that minimizes the probability of error ?(x).

The decision scheme used by the astronomers is criticized by NASA as being too adhoc. In response, they come up with the following alternative scheme. Given the observed value of the signal R = r, the probability that the message was in fact +a is

ψ(r) = P{S = +a|R = r}.

The astronomers conclude that if ψ(R) >1/2 , the message is most likely to have been +a, while if ψ(R) ≤1/2 , the message is most likely to have been −a.

c. Compute the conditional probability ψ(r) = P{S = +a|R = r}.

d. How does the alternative decision scheme compare to the scheme of part b.?

The aliens have decided to invite the leaders of earth to discuss their plans for invasion. In order to arrange a meeting, they must transmit to earth the location Y of their space ship, which has the distribution N(µ, τ 2 ). The message received on earth is M = Y + X, where the noise X ∼ N(0, σ2 ) is independent of the message Y . The unfortunate astronomers are again called in to infer the location. e. The astronomers decide to compute the expected location of the aliens’ spaceship given the received message: E(Y |M = m). Help the astronomers (who did not take ORF 309) compute this quantity. The world’s future rests on your shoulders.