Math 489/889
Stochastic Processes and
Advanced Mathematical Finance
Homework

Steve Dunbar

Due Wed, November 10, 2010
  1. If you buy a lottery ticket in 50 independent lotteries, and in each lottery your chance of winning a prize is 1100, write down and evaluate the probability of winning and also approximate the probability using the Central Limit Theorem.
    1. exactly one prize,
    2. at least one prize,
    3. at least two prizes.

    Explain with a reason whether or not you expect the approximation to be a good approximation.

    Solution: The exact probabilities are easy:

    1. 50 1 1 100 1 99 100 49 0.3055586198

    2. 1 50 0 1 100 0 99 100 50 0.3949939329

    3. 1 50 0 1 100 0 99 100 50 50 1 1 100 2 99 100 49 0.08943531310

    The normal approximations (from the Central Limit Theorem using the half-integer, or histogram area corrections) are

    1. 1250(1100) 9950100 < Z < 3250(1100) 9950100 0.4223907551,
    2. 1250(1100) 9950100 < Z = 12 = 0.5,
    3. 3250(1100) 9950100 < Z 0.0776092449.

    The normal approximations are not good since the rule of thumb npq > 18 is not satisfied, in fact npq 12.

  2. A bank has $1,000,000 available to make for car loans. The loans are in random amounts uniformly distributed from $5,000 to $20,000. Make a model for the total amount that the bank loans out. How many loans can the bank make with 99% confidence that it will have enough money available?

    Solution: Let X1,X2,X3 be a sequence of random variables representing the individual loan amounts. These random variables may reasonably be assumed to be independent, and of course are identically distributed random variables on the interval [5000, 20000]. Then E Xi = 12500 and Var Xi = 18,750,000 so σ = 4330.127020. Then the total loan amount is SN = X1 + + ̇Xn. We seek Sn > 1000000 0.01. This is approximately the probability Z > 100000012500n 4330.127020n . Note (for example from tables) this requires

    1000000 12500n 4330.127020n > 2.326347874

    or n < 73.10947801 so the bank can expect to make about 73 loans.

  3. Let W(t) be standard Brownian motion.
    1. Find the probability that 0 < W(1) < 1.
    2. Find the probability that 0 < W(1) < 1 and 1 < W(2) W(1) < 3.
    3. Find the probability that 0 < W(1) < 1 and 1 < W(2) W(1) < 3 and 0 < W(3) W(2) < 12.

    Solution: First note that 0 < W(1) < 1 = 0.3413, 1 < W(2) W(1) < 3 = 0.1574, and 0 < W(3) W(2) < 12 = 0.1915. Then

    1. 0.3413
    2. By independence of increments: 0.3413 0.1574 = 0.05372.
    3. By independence of increments: 0.3413 0.1574 0.1915 = 0.01029.
  4. Let W(t) be standard Brownian motion.
    1. Evaluate the probability that W(5) 3 given that W(1) = 1.
    2. Find the number c such that W(9) > c|W(1) = 1 = 0.10.

    Solution:

    1. Since W(5) W(1) N(0, 4), the required probability is W(5) > 3|W(1) = 1 = W(5) W(1) > 3 1 = (W(5) W(1))2 > 1 = 0.1586552539

    2. Since W(9) W(1) N(0, 8), the required value can be deduced from W(9) > c|W(1) = 1 = W(9) W(1) > c 1 = (W(9) W(1))(22) > (c 1)(22) = 0.10

      Then (c 1)(22) = 1.281551566 and c = 4.624775211.

  5. Use your Homework 5 record of a 100-flip coin flip sequence. Scoring Y i = +1 for each Head and Y i = 1 for each Tail on each flip, keep track of the accumulated sum Tn = i=1nY i for n = 1,, 100 representing the net fortune at any time. Plot the resulting Tn versus n on the interval [0, 100]. Finally, using N = 10 plot the rescaled approximation W10(t) = (110)S(10t) on the interval [0, 10] on the same set of axes.
  6. Let Z be a single normally distributed random variable, with mean 0 and variance 1, i.e. Z N(0, 1). Then consider the continuous time stochastic process X(t) = tZ.
    1. Using a normal random variable generator (from Excel, Maple, Mathematica, Octave, MATLAB, R etc. , all have one and probably the TI-89 or equivalent has one too), find sample values of X(1), X(2), X(4) and X(9).
    2. Explain why the distribution of X(t) is normal with mean 0 with variance t.
    3. Is X(t) a Brownian motion? Explain why or why not.

    Solution: No, X(t) is not Brownian motion for two reasons in spite of the fact that tZ N(0,t) (which follows from being a scalar multiple by t of the N(0, 1) random variable Z.)

    First, for t1 < t2 t3 < t4, X(t2) X(t1) = (t2 t1)Z is not independent of X(t4) X(t3) = (t4 t3)Z since both are multiples of the same sample value Z drawn from the N(0, 1) population.

    Second, the distribution of (t2 t1)Z is normal with variance (t2 t1)2t 2 t1.

    Note nevertheless that X(0) = 0 and X(t) is continuous at t = 0.