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University of Illinois Spring 2007 ECE 413: Solutions to problem set 9
- Moments of the exponential distribution
(a) E[Xn] =
0 x
nμe−μxdx = −nxn− (^1) e−μx|∞ 0 +^
0 x
n− (^1) ne−μxdx = 0 + nE[Xn−^1 ] μ. (b) E[Xn] = nE[X
n− (^1) ] μ =^
n(n−1)E[Xn−^2 ] μ^2 =^ · · ·^ =^
n! μn^ E[X
(^0) ] = n! μn (c) E[eαX^ ] =
0 e
αX (^) μe−μxdx = μ ∫^ ∞ 0 e
−(μ−α)xdx = − μ μ−α e
−(μ−α)x|∞ 0 =^
μ μ−α (d) On one hand, E[eαX^ ] = E[
n=
αnXn n! ] =^
n=
αnE[Xn] n!. On the other hand, (^) μ−μα = (^1) −^1 μ α
n=
αn μn^. These two series are equal by part (c), so the coefficients of αn^ must be equal for all n. Therefore, E[X
n] n! =^
1 μn^ , or^ E[X
n] = n! μn^.
- Splitting property of the Poisson distribution (a) Given there are n flips, N 1 has the binomial distribution with parameters n and p. Thus, P (N 1 = n 1 |N = n) =
( (^) n n 1
pn^1 (1 − p)n−n^1. (b)
P (N 1 = n 1 , N 2 = n 2 ) = P (N 1 = n 1 |N = n 1 + n 2 )pN (n 1 + n 2 )
=
n 1 + n 2 n 1
pn^1 (1 − p)n^2
e−μμn^1 +n^2 (n 1 + n 2 )!
(c) Which can be written as
P (N 1 = n 1 , N 2 = n 2 ) = e−μ^ (μp)n^1 n 1!
(μ(1 − p))n^2 n 2! = pN 1 (n 1 )pN 2 (n 2 )
where pN 1 (n 1 ) = (μp)
n (^1) e−μp n 1! and^ pN^2 (n^2 ) =^
(μ(1−p))n^2 e−μ(1−p) n 1!. That is,^ N^1 has the Poisson distribution with parameter μp, and N 2 has the Poisson distribution with parameter μ(1 − p).
(Remark: Here is another explanation of the fact that N 1 is a Poisson random variable with parameter μp. Based on the way the Poisson distribution arises in practice, we can imagine that N is the number of successes for n independent trials in an experiment with success probability p′, where n is very large, p′^ is very small, and np′^ = μ. For each success, we flip a coin with bias p, and if heads shows we say the trial is a super success. The number of super successes, N 1 , is thus a binomial random variable with parameters n and p′p, where n is very large, p′p is very small (because p′^ is small) and np′p = μp. Thus, N 1 is a Poisson random variable. )
- Arrival times for a Poisson process
(a) P (N (t) = i) = e
− 2 t(2t)i i! for nonnegative integers^ i^ (Poisson distribution, mean 2t) (b) P (E 1 ) = P (N (3.5) = 0) = e
− (^7) (7) 0 0! =^ e
P (E 2 ) = 0.00091 since E 2 is the same event as E 1
P (E 3 ) = P (N (3.5) ≤ 2) = P (N (3.5) = 0) + P (N (3.5) = 1) + P (N (3.5) = 2) = e−^7 (1 + 7 + 7
2 2 ) = 0.^0296 P (E 4 ) = 0.0296 since E 4 is the same event as E 3
P (E 5 ) = P (N (t) ≤ 2) = P (N (t) = 0) + P (N (t) = 1) + P (N (t) = 2) = e−^2 t(1 + 2t + (2t)
2 2 ) P (E 6 ) = 1 − P (E 5 ) = 1 − e−^2 t(1 + 2t + (2t)
2 2 ) (c) Differentiating yields f (t) = e−^2 t(2(1 + 2t + (2t)
2 2 )^ −^2 −^4 t) =^ e
− 2 t (2t)^2 2 which indeed is the gamma density with parameters (i = 3, μ = 2). (d) The expected time between arrivals is 1/μ, so the expected time until the tenth arrival is 10/μ = 5 minutes.
- Poisson process probabilities (a) The numbers of arrivals in the disjoint intervals are independent, Poisson random variables with mean μ. Thus, the probability is (μe−μ)^3 = μ^3 e−^3 μ. (b) We wish to determine the probability of the event A = {N (0, 2] = 2, N (1, 3] = 2}. A is the union of the disjoint events B 020 , B 111 , and B 202 , were Bijk = {N (0, 1] = i, N (1, 2] = j, N (2, 3] = k}. So
P (A) = P (B 020 )+P (B 111 )+P (B 202 ) = e−μ( μ
2 2 e
−μ)e−μ+(μe−μ) (^3) +( μ^2 2 e
−μ)e−μ( μ^2 2 e
−μ) = ( μ^2 2 +μ
(^3) + μ^4 4 )e
− 3 μ.
(c) P (B 020 |A) = P^ ( PB (^020 A)A )= P^ P(B (A^020 ) )= μ
2 2 /(^
μ^2 2 +^ μ
(^3) + μ^4 4 ) = 2/(2 + 4μ^ +^ μ
(^2) ). (We have applied Bayes rule,
which in this case is the definition of the conditional probability P (B 020 |A) and use of the law of total probability for finding P (A).)
- Calculating probabilities of events involving a Gaussian random variable In parts (a)-(c) we use the fact that X− 410 is a N(0,1) random variable.
(a) P (X ≥ 15) = P ( X− 410 ≥ 15 − 4 10 ) = Q( 15 − 4 10 ) = Q(1.25) = 1 − Φ(1.25) = 0. 1056. (b) P (X ≤ 5) = P ( X− 4 10 ≤ 5 − 410 ) = P ( X− 4 10 ≤ − 1 .25) = Q(1.25) = 0.1056. (c) P (X^2 ≥ 400) = P (X ≥ 20) + P (X ≤ −20) = P ( X− 4 10 ≥ 2 .5) + P ( X− 4 10 ≤ − 7 .5) = Q(2.5) + Q(7.5) ≈ Q(2.5) = 1 − Φ(2.5) = 0.0062 (Note: Q(7.2) < 10 −^12 .) (d) P (X = 2) = 0 because X is a continuous type random variable.
- Normal approximation to the binomial distribution Do problem 20, p. 230 of Ross using the binomial approximation The number of people S favoring the proposition is a binomial random variable with parameters (100, 0.65). It has mean μ = 65 and variance σ^2 = 100(0.65)(1 − 0 .65) = 22.75. The DeMoivre-Laplace limit theorem tells us that S is well approximated by a normal random variable with parameters μ and σ^2. (a) P (S ≥ 50) = P (S ≥ 49 .5) = P ( √S 22 −^65. 75 ≥ 49 √. 225 −. 7565 ) ≈ Q(− 3 .25) = Φ(3.25) = 0. 9994
(b) P (60 ≤ S ≤ 70) = P (59. 5 ≤ S ≤ 70 .5) = P ( 59 √. 225 −. 7565 ≤ √S 22 −^65. 75 ≥ 70 √. 225 −. 7565 ) ≈ Φ(1.153) − Φ(− 1 .153) =
2Φ(1.153) − 1 ≈ 0. 751 (c)P (S < 75) = P (S ≤ 74 .5) = P ( √S 22 −^65. 75 ≤ 74 √. 225 −. 7565 ) ≈ Φ(2.003) ≈ 0. 9774
