# Statistics 100A Homework 4 Solutions

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1 Problem 1 For a discrete random variable X, Statistics 100A Homework 4 Solutions Ryan Rosario Note that all of the problems below as you to prove the statement. We are proving the properties of epectation and variance, thus you cannot just use them right off the bat. (1 Prove E(aX + b = ae(x + b. Proof. By definition, E(X = p( where f is a discrete probability distribution. Thus, E(aX + b = = (a + bp( ap( + bp( = a p( + b p( Note that E(X = p( and p( = 1. = ae(x + b This proves linearity of epectation with respect to one variable. (2 Prove Var(aX + b = a 2 Var(X. First, prove a lemma: Var(X = E(X 2 E(X 2. Proof. By definition, Var(X = E [ (X µ 2]. Var(X = E [ (X µ 2] = E [ X 2 + µ 2 2µX ] Recall that µ = E(X. = E [ X 2 µ 2] By part 1, linearity: = E(X 2 µ 2 = E(X 2 E(X 2 1

2 Then we can work with Var(aX + b = a 2 Var(X. Proof. We can use the result from part 1 and the lemma to solve this problem. Var(aX + b = E [{(ax + b (aµ b} 2] = E [ a 2 (X µ 2] = a 2 ( E [ X 2 + µ 2 2µX ] = a 2 ( E(X 2 µ 2 = a 2 ( E(X 2 E(X 2 = a 2 Var(X Or, we can start with basic principles. Let Z = ax + b. Proof. Var(Z = E(Z 2 E(Z 2 ( = (ax + b 2 f( (ax + b 2 = ( [ (ax 2 + b 2 + 2abX ] f( [ (ae(x 2 + b 2 + 2abE(X ] ( = a 2 X 2 f( + b 2 f( + 2ab f( [ a 2 E(X 2 + b 2 + 2abE(X ] = a 2 E(X 2 + b 2 + 2abE(X a 2 E(X 2 b 2 2abE(X = a 2 E(X 2 a 2 E(X 2 = a 2 ( E(X 2 E(X 2 = a 2 Var(X 2

3 (3 Let µ = E(X and σ 2 = Var(X. Let Z = X µ, calculate E(Z and Var(Z. σ We want to find the epected value and variance of Z where Z = X µ σ. This is called a z score, and is the primary metric associated with the normal distribution. The z scores measures how far from the mean, in standard deviations, a particular value is. ( X µ E(Z = E σ By linearity (part 1, we get: ( X ( µ = E E σ σ But since µ and σ are constants, they can be pulled out of the epectation by linearity (part 1. = 1 σ E(X µ σ But since E(X = µ we get = µ σ µ σ = 0 By definition, Var(X = E [ (X µ 2]. Var(Z = E [ (Z µ Z 2] Where µ Z = E(Z = 0 from the previous part. = E [ Z 2] [ (X ] µ 2 = E σ [ X 2 + µ 2 ] 2µX = E σ 2 = 1 σ 2 [ E(X 2 + µ 2 2µE(X ] But E(X = µ so = 1 σ 2 [ E(X 2 µ 2] From the lemma proven in part 2, = 1 σ 2 var(x = σ2 σ 2 = 1 and since σ 2 = Var(X, We have proven that Z N(0, 1. N(0, 1 is called the standard normal distribution. (4 Prove Var(X = E(X 2 E(X 2. This was the lemma that we proved in part 2. 3

4 Problem 2 Suppose in the popultion of voters, the proportion of those who would vote for a certain candidate is 20%. (1 If we randomly sample 5 people from the population of voters, what is the probability that we get at least 2 supporters of this candidate? Each person s support is independent. The probability that a person supports this candidate is 0.2. Let X be the number of people that support this candidate, n = 5 and p = 0.2. Then, X Binomial(n, p P (X 2 = 1 P (X 1 = 1 [P (X = 0 + P (X = 1] [( ( 5 = 1 (0.2 0 ( = 1 ( ( (0.2 1 (0.8 4 = (0.2 1 (0.8 4 ] Of course, computing P (X 2 brute force is also acceptable. (2 If we randomly sample 100 people from the population of voters. Let X be the number of people among these 100 people who support the candidate. What is the probability that X > 28? You only need to write down this probability. You do not need to calculate it. Now we have n = 100 and p = 0.2. So, P (X > 28 = 100 k=28 ( 5 k 0.2 k k Aside: We can calculate this on a computer in software such as R: n <- 100 p <- 0.2 k <- seq(29,100 sum(choose(n,k*(p**k*((1-p**(n - k [1] Thus, the eact answer is P (X > Aside: There is a way to compute this directly. Since n is large, and p is not too close to 0 or 1 (or, from Stats 10, since np > 10 and n(1 p > 10 we can use the normal distribution to approimate the normal distribution where µ = np = = 20 and σ = np(1 p = 16 = 4. 4

5 ( P (X > 28 = P Z > 4 ( = P Z > 8 4 = P (Z > 2 = 1 P (Z < 2 Which can be found in a z table or by using R: Note that the normal approimation is fairly close, but with an error of 10 4 which is OK but not great. (3 What is E(X and Var(X? What is the standard deviation of X? There is an ambiguity in this problem. It is not clear which n to use from the problem. In any case, E(X = np and Var(X = np(1 p and σ = np(1 p. n = 5 n = 100 E(X 1 20 Var(X σ 0.8 =

6 Problem 3 For T Geometric(p, calculate E(T. If p =.2, then what is the probability that T > 5? We need to derive E(T from basic principles. Recall that there are two parameterizations for the geometric distributions. (1 If t is the number of failures before the first success, then P (T = t = (1 p t p. (2 If the tth attempt is the first success, then P (T = t = (1 p t 1 p. By definition, E(T = t tp (T = t Using the second parameterization, E(T = = = p tp (T = t t=1 t(1 p t 1 p t=1 t(1 p t 1 t=1 Trick [ here: ] = p (1 p t 1 + (1 p t t=1 t=2 = p [{ 1 + (1 p + (1 p } + { (1 p + (1 p } +... ] Each term is a convergent geometric series: Σ = a 0 [ ] 1 = p 1 (1 p + 1 p 1 (1 p +... [ 1 = p p + 1 p ] +... p = 1 + (1 p + (1 p = = 1 p which is a geometric series itself, so 1 1 (1 p 1 a n+1 an The proof is very similar for the other parameterization, and there are other ways to do this. One common way is by using derivatives. Now to compute P (T > 5. If we assume that t is the number of failures until the first success, then P (T > 5 = (1 p 5 p + (1 p 6 p +... = (1 p5 p = (1 p5 1 (1 p 6

7 Thus, P (T > 5 = ( In general, assuming t is the number of failures until the first success, then P (T > t = (1 p t. Problem 4 Suppose we divide the time ais into small periods (0, t, ( t, 2 t,... Within each period, we flip a coin independently. The probability of getting a head is λ t. (1 Let X be the number of heads within the interval [0, t]. Calculate the limit of P (X = k as t 0, for k = 0, 1, 2,.... Also calculate E(X. Within each subinterval, we flip a coin. The probability of getting a head (a success is λ t and X is the number of heads that we observe in the interval [0, t]. Note that each subinterval forms a Bernoulli trial and the sequence of trials is independent. Start with the binomial distribution. P (X = k = ( n k p k (1 p n k where the probability of success (head is λ t and the number of trials (subintervals is Thus, n = t t, p = λ t. The first trick that we need to use is the following t t. (1 p n k e np Proof. If we take the log of (1 p n k we get log (1 p n k = (n k log (1 p. Note that for p very small, p << 1, we can approimate that log (1 p p. Then, log (1 p n k = (n k log (1 p (n k( p = p(n k Thus... (1 p n k e p(n k e np = e λt Now recall that ( n k = Another trick we can use is that n! (n k! nk. n! k!(n k! 7

8 Starting with the binomial distribution, we have that n! (n k! n k and (1 p n k e np, thus in the limit P (X = k = (λtk e λt (2 Let T be the time until the first head. Calculate the limit of P (T > t as t 0. In both (1 and (2, let us assume that t is a multiple of λt. We want to find the probability that the first success occurs after the kth attempt. This is geometric. Recall that n = t t, p = λt. k! Recall that by definition, ( 1 + n P (T > k = (1 p k = (1 λ t t t n e, so (1 λ t t t e λt (3 Please give a real eample of X and T with concrete value of λ and its unit. What is the meaning of λ. Answers will vary. In general, λ is the rate parameter and specifies the (average number of events that occur per some unit time (seconds, hours, etc. Everything, λ, X, and T must all use this same unit of time. X is a random variable that represents the number of events that happen during some unit time. T is a random variable that represents the time it takes for the first event to occur. Note that X Poisson(λ. Since T is the amount of time for the first event to occur, T Geometric. If T where the amount of time between two events, then T would be eponential. One common eample is a call center. Let s use cell phones since this is In the call center problem, we have that a particular cell tower receives on average 20 phone calls per fifteen minutes. The unit of time here is fifteen minutes. Everything in the problem must then be written in terms of fifteen minutes. Let X be the number of calls received during some arbitrary fifteen minute block of time. Then T represents the amount of time (in fifteen minute blocks until the cell tower receives the first call. Catch: If we wanted to find the probability that the cell tower receives 70 calls in an hour, we need to rewrite it with respect to fifteen minute blocks. This is where λt comes in. The average number of calls per hour is λt = 20 4 = 80 because there are 4 fifteen minute blocks in one hour. Then, P (X = 70 = (8070 e 80 70! =

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