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Sketch of solutions to Sheet 6 September 30, 2015 • Try not to consult these before you have tried the questions thoroughly. • Very likely the solutions outlined below only represent a tiny subset of all possible ways of solving the problems. You are highly encouraged to explore alternative approaches! 1. X is NOT measurable w.r.t F because, for example, X −1 (1) = {1} ∈ / F. Y is measurable w.r.t F. 2. The range of X consists of three possible values 1, −1 or 0. Check that X −1 (1) = {HH}, X −1 (−1) = {T T } and X −1 (0) = {HT, T H}. Thus σ(X) must at least contain the three events {HH}, {T T } and {HT, T H}. All we need to do is to add in a few suitable extra elements (unions, compliments, the empty set, etc...) to turn it into a σ-algebra. We then obtain σ(X) = {∅, Ω, {HH}, {T T }, {HT, T H}, {HH, T T }, {HT, T H, T T }, {HH, HT, T H}}. 3. A probability mass function (pmf) pX (k)P has to satisfy two conditions: i) pX (k) > 0 for each k belonging to the support of X; ii) k pX (k) = 1. Here the first condition is clearly satisfied. P P∞ k On the second condition, k pX (k) = e−λ k=0 λk! = e−λ eλ = 1 (recall the Taylor’s expank sion of the function eλ ). Hence pX (k) = e−λ λk! is a well defined pmf. P∞ λk−1 P P∞ k −λ −λ λ E(X) = k kpX (k) = e−λ k=0 kλ e = λ, and k=1 (k−1)! = λe k! = λe E(X 2 ) = X k 2 pX (k) = e−λ k = e−λ ∞ X k 2 λk k=0 ∞ X k=0 =e k! (k(k − 1) + k)λk k! ∞ ∞ X X λk−2 λk−1 +λ λ (k − 2)! (k − 1)! k=2 k=1 λ2 eλ + λeλ −λ 2 = e−λ = λ2 + λ. Thus var(X) = E(X 2 ) − (E(X))2 = λ. 1 ! 4. E(Y ) = ∞ X k=0 1 λk e−λ (1 + k) k! ∞ e−λ X λk+1 = λ (k + 1)! e−λ = λ k=0 ∞ X k=0 ! λk −1 k! e−λ λ = (e − 1) λ 1 = (1 − e−λ ). λ 5. We have a sequence of “head” and “tail” after flipping the coin n times. What is the number of such sequence with “head” appearing k times? It is given by Ckn (think of it as selecting k flips from the total n flips such that we assign “head” to each selected flip). The probability of getting any one of these Ckn outcomes is pk (1 − p)n−k . Hence we have P (H = k) = Ckn pk (1 − p)n−k for k = 0, 1, 2..., n. This is a binomial distribution with parameters (n, p). H can be interpreted as a sum of n independent and identically distributed Bernoulli random variables with rate of success p, where each of these Bernoulli random variables has identical mean p and variance p(1 − p). By linearity P of expectation and Pn variance (where the latter n H relies on independence), we have E(H) = E(1 ) = k k=1 k=1 p = np and var(H) = Pn Pn H k=1 var(1k ) = k=1 p(1 − p) = np(1 − p). 6. If we need N = k, then the light bulb does not fail in the first k −1 days which has probability (1 − p)k−1 , and fails on day k which has probability p. Then P(N = k) = (1 − p)k−1 p where k = 1, 2, 3... Here N has a geometric distribution with parameter p. Pk We first work out P(N > k) = 1 − P(N 6 k) = 1 − i=1 (1 − p)i−1 p = (1 − p)k . Then P(survives for extra 5 days|has survived previous 10 days) = P(N > 10 + 5|N > 10) P(N > 15, N > 10) P(N > 10) P(N > 15) = P(N > 10) = = (1 − p)15 /(1 − p)10 = (1 − p)5 . Similarly, P(survives for extra 5 days|has survived previous 100 days) = P(N > 100 + 5|N > 100) = P(N > 105) P(N > 100) = (1 − p)5 , where probability is the same. 7. A probability density function (pdf) f (x) has to satisfy two conditions: i) f (x) > 0 for all x, R and ii) R f (x)dx = 1. R R∞ The first condition is clearly satisfied. On the second condition, R f (x)dx = 0 λe−λx dx = e−λx |0∞ = 1. 2 E(X) = R R R∞ λxe−λx dx = xe−λx |0∞ + 0 e−λx dx = 0 + λ1 e−λx |0∞ = λ1 , and Z Z ∞ 2 2 E(X ) = x f (x)dx = λx2 e−λx dx R 0 Z ∞ 2 −λx 0 =x e |∞ + 2xe−λx dx 0 Z 2 ∞ λxe−λx dx =0+ λ 0 2 = E(X) λ 2 = 2. λ xf (x)dx = R∞ 0 Thus var(X) = E(X 2 ) − (E(X))2 = 1 λ2 . R R∞ 8. We find c by using the property of a pdf that R f (x)dx = 1. Then 1 = 0 c(1 + x)−3 dx = c −2 0 |∞ = 2c which gives c = 2. 2 (1 + x) R∞ R∞ E(X) = 0 2x(1 + x)−3 dx = x(1 + x)−2 |0∞ + 0 (1 + x)−2 dx = 0 + (1 + x)−1 |0∞ = 1. The cumulative distribution function of X is given by F (x) = P(X 6 x). Since X only has non-zero density for x > 0, X is a positive random variableR and thus F (x) R x = P(X 6 x) = 0 x for x < 0. Otherwise for x > 0, F (x) = P(X 6 x) = 0 f (u)du = 0 2(1 + u)−3 du = (1 + u)−2 |0x = 1 − (1 + x)−2 . In summary, ( 0, x < 0; F (x) = 1 − (1 + x)−2 , x > 0. 9. Let f be the density function of Y . Then Z ∞ Z y=∞ Z u=∞ Z P(Y > y)dy = f (u)dudy = 0 y=0 u=y u=∞ Z y=u Z u=∞ f (u)dydu = u=0 y=0 uf (u)du = E(Y ). u=0 R∞ (Indeed, the relationship E(Y ) = 0 P(Y > y)dy holds for any positive random variable Y , not just those with absolutely continuous density. The proof of this general result is more subtle since Y may not have a density function.) 10. Denote the cumulative distribution function of Y by FY (x) which we try to work out as follows: ln x − µ ln x − µ X −µ 6 =P Z6 , FY (x) = P(Y 6 x) = P(e 6 x) = P(X 6 ln x) = P σ σ σ , where Φ is the where Z is a standard normal random variable. Hence FY (x) = Φ ln x−µ σ cumulative distribution of a standard normal distribution. The density of Y is then obtained by differentiation 2 ! d ln x − µ 1 0 ln x − µ 1 1 ln x − µ d √ exp − fY (x) = FY (x) = Φ = Φ = . dx dx σ σx σ 2 σ σx 2π X for x > 0, and the density is zero elsewhere. (Recall that Φ0 (x) = variable.) 2 x √1 e− 2 2π which is the density function of a standard normal random 3 The mean of Y can be computed as Z ∞ xfY (x)dx E(Y ) = 0 2 ! ln x − µ = dx σ 0 2 Z ∞ u ln x − µ exp(µ + σu) √ exp − du (By change of variable u = ) = 2 σ 2π −∞ Z ∞ 2 1 1 √ exp − (u − σ)2 du = eµ+σ /2 2 2π −∞ Z ∞ = eµ+σ 1 1 √ exp − 2 σ 2π 2 /2 . Notice that we have performed a completing square trick from the third to the forth line. In the last line we have used the fact that the density of a standard normal variable integrates R∞ u2 to 1, i.e. −∞ √12π e− 2 du = 1. 11. From the question setup, T = min(X, C) where X ∼ Exp(λ). Now we would like to find the CDF of T . Clearly T is non-negative. For t > 0. FT (t) = P(T 6 t) = P(min(X, C) 6 t) = P(X 6 t or C 6 t) = P(X 6 t) + P(C 6 t) − P(X 6 t and C 6 t). Notice that in the above expression C and t are just some non-random numbers. If t > C, then the event “t > C” always happens with probability one. Otherwise if t < C, then “t > C” will be an impossible event. This gives: ( P(X 6 t) + 0 − 0, t < C; FT (t) = P(X 6 t) + 1 − P(X 6 t), t > C. Since X follows Exp(λ), P(X 6 t) = gives Rt 0 λe−λu du = 1 − e−λt . Substitution into the above ( FT (t) = 1 − e−λt , t < C; 1, t > C. A sketch of CDF is given below. There is a discontinuity in the function at t = C. FT (t) 1 C t T is not a continuous random variable. It is because P(T = C) = P(T 6 C) − P(T < C) = 1 − (1 − e−λC ) = e−λC > 0. But a continuous random variable always has zero probability of assuming any particular value, i.e P(T = x) = 0 must hold for any x if T was continuous. So T cannot be continuous. T is also not discrete because the possible range of T is [0, C], which is not a countable set. 4