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EE 5375/7375 Random Processes September 16, 2003
Homework #3 Solutions
Problem 1. textbook problem 3. The exam grades in a certain class have a Gaussian (normal) pdf with mean m and standard deviation σ. Find the constants a and b so that the random variable Y = aX + b has a Gaussian pdf with mean m′^ and standard deviation σ′.
The linear function Y = aX + b has a simple effect on the mean and variance of X. First, the mean of Y is
m′^ = E(Y ) = E(aX + b) = aE(X) + b = am + b
Second, the variance of Y is
σ′^2 = var(Y ) = var(aX + b) = a^2 var(X) = a^2 σ^2
Thus we have 2 equations for 2 unknowns. The second equation gives
a =
σ′ σ
Substituting into the first equation, we get
b = m′^ − am = m′^ −
σ′ σ
m
Problem 2. textbook problem 3. Suppose that a voltage X is a zero-mean Gaussian random variable. Find the pdf of the power dissipated by an R-ohm resistor P = RX^2.
The pdf of X is the Gaussian density function:
fX (x) =
2 πσ^2
e−x
(^2) / 2 σ 2
The PDF of the power P is
FP (y) = P (RX^2 ≤ y) = P (−
y/R ≤ X ≤
y/R) = FX (
y/R) − FX (−
y/R)
The pdf is found by differentiating with respect to y:
fP (y) =
fX (
y/R) 2
y/R
R
fX (−
y/R) − 2
y/R
R
fX (
y/R) 2
yR
fX (−
y/R) 2
yR
=
yR
2 πσ^2
e−(y/R)/^2 σ
2
yR
2 πσ^2
e−(y/R)/^2 σ
2
2 πσ^2 Ry
e−y/^2 σ
(^2) R
Problem 3. textbook problem 3.
1
Let Y = eX^. (a) Find the PDF and pdf of Y in terms of the PDF and pdf of X. (b) Find the pdf of Y when X is a Gaussian random variable. In this case, Y is said to have a lognormal pdf.
(a) The PDF of Y is FY (y) = P (Y ≤ y) = P (eX^ ≤ y) = P (X ≤ ln y) = FX (ln y)
The pdf is found by differentiating with respect to y:
fY (y) =
d dy FX (ln y) = fX (ln y)
d dy ln y =
y fX (ln y)
(b) The pdf of X is the Gaussian density function:
fX (x) =
2 πσ^2
e−(x−μ)
(^2) / 2 σ 2
Then the pdf of Y is
fY (x) =
y
2 πσ^2
e−(ln^ y−μ)
(^2) / 2 σ 2
Problem 4. textbook problem 3. An urn contains 90 $1 bills, 9 $5 bills, and 1 $50 bill. Let the random variable X be the denomination of a bill that is selected at random from the urn. Find the mean m = E(X). In what sense is the mean the break-even price of a ticket for the right to draw a single bill from the urn?
Since each bill is equally likely, the probability of each denomination is the same as its relative frequency. That is, the probability of $1 is 0.9, probability of $5 is 0.09, and probability of $50 is 0.01. The mean is therefore E(X) = (1)(0.9) + (5)(0.09) + (50)(0.01) = 1. 85
The relative frequency interpretation implies that 1.85 is the long-term expected payoff, if the drawing is repeated independently many times.
Problem 5. textbook problem 3. (a) Suppose a coin is tossed n times. Each coin toss costs d and the reward in obtaining X heads is aX^2 +bX. Find the expected value of the net reward. (b) Suppose that the reward in obtaining X heads is aX^ , where a > 0. Find the expected value of the reward.
(a) For n tosses, the reward is aX^2 +bX while the cost is nd. The net reward is therefore Y = aX^2 +bX −nd. Each coin toss is a Bernoulli trial so the number of heads X is a binomial probability with mean np and variance np(1 − p). The expected net reward is
E(Y ) = E(aX^2 + bX − nd) = aE(X^2 ) + bE(X) − nd = a(var(X) + [E(X)]^2 ) + bE(X) − nd = a(np(1 − p) + (np)^2 ) + bnp − nd = anp^2 (n − 1) + (a + b)np − nd
(b) The mean reward is
E(aX^ ) =
∑^ n
k=
ak
n k
pk(1 − p)n−k
∑^ n
k=
n k
apk(1 − p)n−k
= (ap + (1 − p))n
2
The variance can now be found by
var(X) = E(X^2 ) − [E(X)]^2 =
b^2 + ab + a^2 3
b + a 2
(b − a)^2 12
Problem 7. textbook problem 3. Let X be the mixture of two exponential random variables. X is then a hyperexponential random variable with pdf fX (x) = pae−ax^ + (1 − p)be−bx
for positive parameters a, b, and x ≥ 0. Find the Laplace transform of the pdf of X.
From the pdf, X can be recognized as
X =
X 1 with probability p X 2 with probability (1 − p)
where X 1 and X 2 are exponential with parameters a and b, respectively. We know that the Laplace transform of the exponential pdf f (x) = λe−λx^ is X∗(s) = (^) λλ+s. The Laplace transform of fX (x) = pae−ax^ +(1−p)be−bx is therefore
X∗(s) = p
a a + s
b b + s
Problem 8. textbook problem 4. Consider a sequence of n + m independent Bernoulli trials with probability of success p in each trial. Let N be the number of successes in the first n trials and let M be the number of successes in the remaining m trials. (a) Why are N and M independent random variables? (b) Find the joint PMF of N and M and the marginal PMFs of N and M. (c) Find the PMF for the total number of successes in the n + m trials.
(a) We know that Bernoulli trials are independent, so the first n trials are independent of the next m trials. (b) The PMF of N is a binomial(n, p) PMF: PN (x) =
(n x
px(1 − p)n−x. Similarly, the PMF of M is PM (y) =
(m y
py^ (1 − p)m−y^. Since they are independent, the joint PMF is simply the product
PN M (x, y) =
n x
px(1 − p)n−x
m y
py^ (1 − p)m−y
(c) The total number of successes Z in n + m trials has a binomial(n + m, p) PMF:
PZ (z) =
n + m z
pz^ (1 − p)n+m−z
Problem 9. textbook problem 4. A message requires N time units to be transmitted, where N is a geometric random variable with PMF pj = (1 − a)aj−^1 for j = 1, 2 ,... A single new message arrives during a time unit with probability p, and no message arrives with probability 1 − p. Let K be the number of new messages that arrive during the transmission of a single message. (a) Find the PMF of K. Hint:
(1 − β)−(k+1)^ =
∑^ ∞
n=k
n k
βn−k
(b) Find E(K) and var(K) using conditional expectation.
4
(a) Given that a message transmission lasts N = n time units, the number of new messages K that arrive during the n time units will be a binomial(n, p) random variable. This is described by the conditional PMF
P (K = k|N = n) =
n k
pk(1 − p)n−k^ , k = 0, 1 ,... , n
We can find the unconditional PMF of K by unconditioning by the geometric PMF of N but must be a little careful about the range of summation. First consider the case K = 0:
P (K = 0) =
∑^ ∞
n=
P (K = 0|N = n)(1 − a)an−^1
∑^ ∞
n=
n 0
(1 − p)n(1 − a)an−^1
= (1 − a)(1 − p)
∑^ ∞
n=
((1 − p)a)n−^1
(1 − a)(1 − p) 1 − (1 − p)a
Next consider the case K > 1:
P (K = k) =
∑^ ∞
n=k
P (K = k|N = n)(1 − a)an−^1
∑^ ∞
n=k
n k
pk(1 − p)n−k(1 − a)an−^1
(1 − a)pkak a
∑^ ∞
n=k
n k
[(1 − p)a]n−k
(1 − a)pkak a(1 − (1 − p)a)k+
=
(1 − a) a(1 − (1 − p)a)
pa 1 − (1 − p)a
)k
for k = 1, 2 ,.. ..
(b) The mean of K can be found from the conditional mean of K given N as
E(K) = E[E(K|N )]
∑^ ∞
n=
E(K|N = n)(1 − a)an−^1
∑^ ∞
n=
np(1 − a)an−^1
= pE(N ) =
p 1 − a
5
(b) The usual method is to expand the expression by partial fractions, ie, look for an equivalent form of
φZ (z) =
A
α − jω
B
β − jω
In order to be equivalent, A and B must satisfy
A(β − jω) + B(α − jω) = αβ
Expanding this into real and imaginary parts, A and B must satisfy
Aβ + Bα = αβ
−jω(A + B) = 0
The second equation implies A = −B, and by substitution into the first equation,
A =
αβ β − α
, B =
αβ α − β
The characteristic function can be written as
φZ (z) =
β β − α
α α − jω
α α − β
β β − jω
The inverse can be recognized as
fZ (z) =
β β − α
αe−αz^ +
α α − β
βe−βz
Problem 12. textbook problem 5. Suppose that the number of particle emissions by a radioactive mass in t seconds is a Poisson random variable with mean λt. Use the Chebyshev inequality to obtain a bound for the probability that | N^ t( t)− λ| exceeds .
We apply the Chebyshev inequality to bound:
P
N (t) t
− λ| ≥
= P (|N (t) − λt| ≥ t)
var(N (t)) (t)^2
≤
λt ^2 t^2 ≤ λ ^2 t
Problem 13. textbook problem 5. A fair coin is tossed 1000 times. Estimate the probability that the number of heads is between 400 and 600 (using the central limit theorem). Estimate the probability that the number is between 500 and 550.
Let Sn be the sum of heads in n coin tosses. Each coin toss is represented by Xn which has value 1 with probability 0.5 or value 0 with probability 0.5. Hence S 1000 has mean μ = 500 and variance σ^2 = 1000 · 0. 5 ·
7
0 .5 = 250. According to the central limit theorem, the normalized random variable Z 1000 = S^1000 √ 250 −^500 has a standard normal distribution.
P (400 ≤ S 1000 ≤ 600) = P
S 1000 − 500
= P
≤ Z 1000 ≤
= 1 − 2. 54 × 10 −^10
The other probability is found similarly:
P (500 ≤ S 1000 ≤ 550) = P
≤ Z 1000 ≤
Problem 14. textbook problem 5. A student uses pens whose lifetime is an exponential random variable with mean 1 week. Use the central limit theorem to estimate the minimum number of pens he should buy at the beginning of a 15-week semester, so that with probability 0.99 he does not run out of pens during the semester.
Let Sn represent the total lifetimes of n pens which is the sum of n exponential random variables. Thus Sn has mean μ = n · 1 = n weeks and variance σ^2 = n · 12 = n. Assuming that Sn is the sum of many lifetimes, the central limit theorem says that S√n−nn ≈ N (0, 1). The probability in question is
P (Sn > 15) = P
Sn − n √ n
15 − n √ n
15 − n √ n
From the tables of the standard normal distribution,
15 − n √ n
Solving for n, the answer is n = 27.04, which means the student should buy 28 pens.
Problem 15. textbook problem 5. A binary transmission channel introduces bit errors with probability 0.15. Estimate the probability that there are 20 or fewer errors in 100 bit transmissions (using the central limit theorem).
Let S 100 denote the sum of bit errors in 100 bits. Clearly S 100 is the sum of 100 Bernoulli random variables which take the value 1 with probability 0.15 and vlue 0 with probability 0.85. The mean of S 100 is μ = 100 · 0 .15 = 15 and the variance is 100 · 0. 15 · 0 .85 = 12.75. Using the central limit theorem for a Gaussian approximation,
P (S 100 ≤ 20) = P
S 100 − 15
Problem 16. Matlab (optional for EE 5375) Let us verify the law of large numbers. Generate 10 random samples by typing:
8
y
z
x
Fig. 1 for Problem 10
y = x
y
x
Fig. 2 for Problem 10
y = x y = z - x
z
y
x
Fig. 3 for Problem 10
y = x y = z - x
z/
