EE 131A Homework Set 7Probability Wednesday, November 26, 2014Instructor: Professor Roychowdhury Wednesday, December 3, 2014

1. Problem 4.61 Let X be an exponential random variable with parameter .

(a) For d > 0 and k a nonnegative integer, find P [kd < X < (k + 1)d].

(b) Segment the positive real line into four equiprobable disjoint intervals.Solution:

(a) The cdf of an exponential random variable with parameter is described as:

Fx(d) = 1 edThen we have:

P (kd < X < (k + 1)d) = FX((k + 1)d) FX(kd) = ekd(1 ed)

(b) To segment the positive real line into four equiprobable disjoint intervals, wefind three segmentation points xk for k = 1, 2, 3 such that FX(xk) =

k4.

x1 =ln4

3

x2 =ln2

x3 =ln4

2. Problem 4.62 The rth percentile, pi(r), of a random variable X is defined by P [X pi(r)] = r/100.

(a) Find the 90%, 95% and 99% percentiles of the exponential random variablewith parameter .

(b) Repeat part a for the Gaussian random variable with parameters m = 0 and2.Solution:

(a) For exponential random variable, r percentiles are computed as follows:

P [X x] = 1 ex = r100

x = 1

ln(1 r100

)

then

pi(90) 2.3

pi(95)

3

pi(99)

4.6

1

(b) For Gaussian random variable, r percentiles are computed as follows:

P [X x] = 1Q(x

) =r

100

x = Q1(1 r100

)

thenpi(90) = 1.3 pi(95) 1.6 pi(99) 2.3

3. Problem 4.67 A binary transmission system transmits a signal X (-1 to send a 0bit; +1 to send a 1 bit). The received signal is Y = X + N where noise N hasa zero-mean Gaussian distribution with variance 2. Assume that 0 bits are threetimes as likely as 1 bits.

(a) Find the conditional pdf of Y given the input value: fY (y|X = +1) andfY (y|X = 1).

(b) The receiver decides a 0 was transmitted if the observed value of y satisfies

fY (y|X = 1)P [X = 1] > fY (y|X = +1)P [X = +1]

and it decides a 1 was transmitted otherwise. Use the results from part a to showthat this decision rule is equivalent to: If y < T decide 0; if y T decide 1.

(c) What is the probability that the receiver makes an error given that a +1 wastransmitted? a -1 was transmitted? Assume 2 = 1/16.

(d) What is the overall probability of error?Solution:

(a) The conditional pdf of Y given the input is found as follows:

fY (y|X = +1) = fN(1 + n y) = fN(y 1) = 12pi

e(y1)222

fY (y|X = 1) = fN(1 + n y) = fN(y + 1) = 12pi

e(y+1)2

22

(b) The receiver decides 0 was transmitted if the observed value of y satisfies:

fY (y|X = 1)P [X = 1] > fY (y|X = +1)P [X = +1]= 1

2pie

(y+1)2

2234> 1

2pie

(y1)222

14

= 3 > e (y1)2

22+(y+1)2

22

= ln 3 > 2y2

= y < ln 322

Therefore, it is equivalent to: if y < T decide 0; if y T ,decide 1 where T = ln 322.

2

(c) The conditional probability of error is given by:

P (error|X = +1) = P [1 +N < T ] = P [N < T 1] = (T 1

) = 5.6076 105

P (error|X = 1) = P [1 +N > T ] = P [N > T + 1] = 1(T + 1

) = 1.7569 105

(d) The overall probability of error is given by:

P (error) = P (error|X = +1)P (X = +1) + P (error|X = 1)P (X = 1)=

1

4(T 1

) +3

4[1 (T + 1

)]

= 2.7196 105

Note: the error probability is close to zero.

4. Problem 4.69 Passengers arrive at a taxi stand at an airport at a rate of one passengerper minutes. The taxi driver will not leave until seven passengers arrive to fill his van.Suppose that passenger interarrival times are exponential random variable, and let Xbe the time to fill a van. Find the probability that more than 10 minutes elapse untilthe van is full.Solution: As we know the time elapsed between two successive occurrences of theevent has an exponential distribution and it is independent of previous occurrences,then the total number of occurrences of the event has a Possion distribution. Then theprobability less than 7 people in 10 minutes is given by:

P [X] = 1 P [X 10]

=6

k=0

(t)k

k!et

=6

k=0

10k

k!e10

= 0.1301

5. Problem 4.74

(a) Find the cdf of the m-Erlang random variable by integration of the pdf. Hint:Use integration by parts.

(b) Show that the derivative of the cdf given by

FSm(t) = 1m1k=0

(t)k

k!et

gives the pdf of an m-Erlang random variable.Solution:

3

(a) The cdf is given by the integral

FX(x) =m1

(m 1)! x0

ym1eydy

Since (m) = (m 1)! Integrate by parts using u = ym1and dv = eydy so thatdu = (m 1)ym2dy and v = ey:

FX(x) =m1

(m 1)!{ym1ey|x0 +

x0

(m 1)ym2eydy}

= (x)m1

(m 1)!ex +

m2

(m 2)! x0

ym2eydy

The integral on the right hand side is identical to the equation for FX(x) with m 1replaced by m 2 . We can therefore repeatly perform integration by parts to obtainthe cdf of X:

FX(x) = (x)m1

(m 1)!ex (x)

m2

(m 2)!ex ...+

x0

eydy

= 1m1k=0

(x)k

k!ex

(b) From part (a), we know

FX(x) = 1m1k=0

(x)k

k!ex =

m1

(m 1)! x0

ym1eydy

therefore:

fX(x) =dFX(x)

dx=

m1

(m 1)!ym1ey

6. Problem 4.90 A voltage X is a Gaussian random variable with mean 1 and variance 2.Find the pdf of the power dissipated by an R-ohm resistor P = RX2

Solution: Note X = P/R and dXdP

= 12

1RP

fP (p) = [fX(x) + fX(x)]|dXdP|

= [fX(P/R) + fX(

P/R)]

1

2RP

=1

2pi

[e(

PR1)2/4 + e(

PR1)2/4]

1

2RP

=1

4PRpi

[e(

PR1)2/4 + e(

PR+1)2/4]

4

7. Let Y = eX , find the pdf of Y when X is Gaussian random variable. In this case Y issaid to be a lognormal random variable.Solution: For y > 0, P [Y y] = P [eX y] = P [X ln y] = FX(ln y)

FY (y) =

{0 y 0FX(ln y) y 0

Therefore, for y > 0 we have

fY (y) =d

dyFY (y) =

1

yfx(ln y)

If X is a Gaussian random variable, then

fY (y) =

{0 y 0e(ln ym)

2/22

y2pi

y 0

5