3 (EEE 350 - Fall 2016) Name: Student ID: The grade scale is from 0 to 4: 4 Excellent: solved with no errors, shows perfect patronage of the concepts tested. 3 Good: some errors are present, but overall it shows that the understanding of the subject is good. 2 Sufficient: partially solved with several errors; it shows some clear gaps in understanding. 1 Insufficient: an attempt to solve the problem with insufficient results, indicating serious gaps in understanding the related material. 0 Inadequate: the homework has not been returned, or it is a completely inadequate effort to solve the problem. Reasoning and work must be shown to gain partial/full credit REMEMBER TO WRITE YOUR NAME! 1 Problems Problem 1 - Mean and Variance: A random variable X has possible outcomes X = {0, 3, 5}. You are told that the mean and variance of the random variable are E[X] = 3.5 and V ar(X) = 4. Determine the probability mass function pX (x). Hint: Since there are only 3 possible outcomes you need three equations to solve for the three unknowns in this problem pX (0), pX (3), pX (5). The equations are given by the constraints E[X] = 3.5, V ar(X) = 4 and P (X ) = 1. Problem 2 - Practice on Random variables: You have a Binomial random variable X with parameters p = 0.1 and n = 3. - Calculate the probability that 2 X 4. - Calculate the probability that 2 X 4 or 3 X < 7. - Calculate the probability mass function of Y = 2X + 1. - What is the mean, mean square and variance of X? - What is the expected value of Y = 35X ? - What is the variance of Y = 35X ? Problem 3 - Moment Generating Function As you know, the moment generating function (MGF) of a random variable Z is defined as GK (z) = E[z K ]. - Prove that if Xi , i = 1, . . . , n are independent and identically distributed random variables with a given MGF GXi (z) = GX (z) (i.e. the function is the same irrespective of i), then the random variable K = X1 + . . . + Xn is such that: GK (z) = . . . = GX1 (z)GX2 (z) . . . GXn (z) = (GX (z))n . Hint: Remember that, if the random variables X, Y are independent, the expectation E[g(X)f (Y )] = E[g(X)]E[f (Y )]. This is true for any type of functions g(x) f (y) and for more than two random variables. Also remember that z X+Y = z X z Y . Problem 4 - Two random variables: The number of emails you receive daily is a Poisson random variable with expected value 25. Only 10% of the emails you receive is urgent. Let N be the number of emails you receive on a certain day and K be the number of urgent emails among them. Derive: - The conditional probability mass function - The joint probability mass function pN K (n, k) of N, K. - The expectation and variance of K. - The expectation of Z = N K. - The covariance of N, K. - Let me tell you the following: Fact: Note that K = X1 + . . . + XN where N is also random and it is a Poisson random variable with mean E[N ] = and Xi are independent Bernoulli random variables with pX (1) = p. Recall that GN (z) = e(z1) , GX (z) = p(z 1) + 1 Using the theorem of total expectation one can show that in this case: GK (z) = E[z K ] = EN [(GX (z))N ] = e((p(z1)+1)1) = e(z1)p this is the MGF of a Poisson random variable with parameter p. It is a fairly remarkable and non obvious result because it shows that the marginal distribution of the random variable K in this problem is actually also a Poisson random variable, but with a smaller mean, equal to p, compared to N . Knowing this fact, it follows that the marginal distribution pK (k) in the problem is actually a Poisson random variable with parameter 25 0.1 = 2.5 (this is not entirely obvious if you try to saturate the random variable n from the joint distribution). Question: Keeping this in mind write the expression of pN |k (n|k). Problem 5 - Capital: You have c0 = $20K and you want to invest them for your retirement. At this point in your life the number of years until your retirement is a Geometric random variable Y with expectation E[Y ] = 45. the rate of return for year i = 1, . . . , Y is also random, and it is a discrete random variable Ri independent from the rate of return for other years and has the same Probability Mass Function (PMF): pRi (0.3) = 0.1, pRi (0.1) = 0.1, pRi (0) = 0.7, pRi (0.05) = 0.1. The way your capital grows is as follows. The capital at year Y is: cY = c0 (1 + R1 )(1 + R2 ) (1 + RY ). - Calculate the expectation E[Ri ] (this is the expected yearly return). - Calculate the probability that after the first year you lost at least $1000 (i.e. P (c1 $19K). - Calculate the expectation of your capital after exactly 45 years (i.e. c45 ). Hint: You need again to use the fact that if the random variables X, Y are independent, the expectation E[g(X)f (Y )] = E[g(X)]E[f (Y )]. This is true for more than two random variables, which in this case are the factors 1 + Ri . - Calculate your expected capital when you retire. Hint: You need to apply the theorem of total expectation here. In the style of the previous question you can derive E[cY |Y = y] first and then average over the distribution of y