ECS 315: Probability and Random Processes 2015/1

HW Solution 11 — Due: Nov 25, 9:19 AM

Lecturer: Prapun Suksompong, Ph.D.

Problem 1. Suppose that the time to failure (in hours) of fans in a personal computer canbe modeled by an exponential distribution with λ = 0.0003.

(a) What proportion of the fans will last at least 10,000 hours?

(b) What proportion of the fans will last at most 7000 hours?

[Montgomery and Runger, 2010, Q4-97]

Solution : See handwritten solution

Problem 2. Consider each random variable X defined below. Let Y = 1 + 2X. (i) Findand sketch the pdf of Y and (ii) Does Y belong to any of the (popular) families discussed inclass? If so, state the name of the family and find the corresponding parameters.

(a) X ∼ U(0, 1)

(b) X ∼ E(1)

(c) X ∼ N (0, 1)

Solution : See handwritten solution

Problem 3. Consider each random variable X defined below. Let Y = 1 − 2X. (i) Findand sketch the pdf of Y and (ii) Does Y belong to any of the (popular) families discussed inclass? If so, state the name of the family and find the corresponding parameters.

(a) X ∼ U(0, 1)

(b) X ∼ E(1)

(c) X ∼ N (0, 1)

Solution : See handwritten solution

11-1

HW11 Q1: Exponential RVFriday, November 21, 2014 9:02 AM

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HW11 Q2 Affine TransformationTuesday, November 11, 2014 4:15 PM

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HW11 Q3 Affine TransformationTuesday, November 11, 2014 4:15 PM

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ECS 315 HW Solution 11 — Due: Nov 25, 9:19 AM 2015/1

Problem 4. Let X ∼ E(5) and Y = 2/X.(a) Check that Y is still a continuous random variable.

(b) Find FY (y).

(c) Find fY (y).

(d) (optional) Find EY .

Hint: Because ddye−

10y = 10

y2e−

10y > 0 for y 6= 0. We know that e−

10y is an increasing

function on our range of integration. In particular, consider y > 10/ ln(2). Then,

e−10y > 1

2. Hence,

∞∫0

10

ye−

10y dy >

∞∫10/ ln 2

10

ye−

10y dy >

∞∫10/ ln 2

10

y

1

2dy =

∞∫10/ ln 2

5

ydy.

Remark: To be technically correct, we should be a little more careful when writingY = 2

Xbecause it is undefined when X = 0. Of course, this happens with 0 probability;

so it won’t create any serious problem. However, to avoid the problem, we may defineY by

Y =

{2/X, X 6= 0,0, X = 0.

Solution : In this question, we have Y = g(X) where the function g is defined byg(x) = 2

x.

(a) First, we evaluate P [Y = y] = P [g(X) = y].

• For each value of y 6= 0, there is only one x value that satisfies y = g(x). (Thatx value is x = 2

y.)

• When y = 0, there is no x value that satisfies y = g(x).

In both cases, for each value of y, the number of solutions for y = g(x) is countable.Therefore, we can write

P [Y = y] =∑

x:g(x)=y

P [X = x] .

Here, X ∼ E(5). Therefore, X is a continuous random variable and P [X = x] = 0 forany x. Hence,

P [Y = y] =∑

x:g(x)=y

0 = 0.

Because P [Y = y] = 0 for all y, we conclude that Y is a continuous random variable.

11-2

ECS 315 HW Solution 11 — Due: Nov 25, 9:19 AM 2015/1

(b) We consider two cases: “y ≤ 0” and “y > 0”.

(i) Because X > 0, we know that Y = 2X

must be > 0 and hence, FY (y) = 0 fory ≤ 0. Note that Y can not be = 0. We need X = ∞ or −∞ to make Y = 0.However, ±∞ are not real numbers therefore they are not possible X values.

(ii) For y > 0,

FY (y) = P [Y ≤ y] = P[

2

X≤ y]

= P

[X ≥ 2

y

].

Note that, for the last equality, we can freely move X and y without worry-ing about “flipping the inequality” or “division by zero” because both X and yconsidered here are strictly positive. Now, for X ∼ E(λ) and x > 0, we have

P [X ≥ x] =∞∫x

λe−λtdt = −e−λt∣∣∞x

= e−λx

Therefore,

FY (y) = e−5( 2y ) = e

−10y .

Combining the two cases above we have

FY (y) =

{e−

10y , y > 0

0, y ≤ 0

(c) Because we have already derived the cdf in the previous part, we can find the pdf viathe cdf by fY (y) =

ddyFY (y). This gives fY at all points except at y = 0 which we will

set fY to be 0 there. (This arbitrary assignment works for continuous RV. This is whywe need to check first that the random variable is actually continuous.) Hence,

fY (y) =

{10y2e−

10y , y > 0

0, y ≤ 0.

(d) We can find EY from fY (y) found in the previous part or we can even use fX(x)Method 1:

EY =∫ ∞−∞

yfY (y) =

∞∫0

y10

y2e−

10y dy =

∞∫0

10

ye−

10y dy

11-3

ECS 315 HW Solution 11 — Due: Nov 25, 9:19 AM 2015/1

From the hint, we have

EY >∞∫

10/ ln 2

10

ye−

10y dy >

∞∫10/ ln 2

10

y

1

2dy =

∞∫10/ ln 2

5

ydy

= 5 ln y|∞10/ ln 2 =∞.

Therefore, EY = ∞ .Method 2:

EY = E[

1

X

]=

∞∫−∞

1

xfX (x) dx =

∞∫0

1

xλe−λxdx >

1∫0

1

xλe−λxdx

>

1∫0

1

xλe−λdx = λe−λ

1∫0

1

xdx = λe−λ lnx|10 =∞,

where the second inequality above comes from the fact that for x ∈ (0, 1), e−λx > e−λ.

Problem 5. In wireless communications systems, fading is sometimes modeled by lognor-mal random variables. We say that a positive random variable Y is lognormal if lnY is anormal random variable (say, with expected value m and variance σ2).

(a) Check that Y is still a continuous random variable.

(b) Find the pdf of Y .

Hint: First, recall that the ln is the natural log function (log base e). Let X = lnY .Then, because Y is lognormal, we know that X ∼ N (m,σ2). Next, write Y as a function ofX.

Solution :Because X = ln(Y ), we have Y = eX . So, here, we consider Y = g(X) where the function

g is defined by g(x) = ex.

(a) First, we evaluate P [Y = y] = P [g(X) = y]. Note that for each value of y, there isonly one x value that satisfies y = g(x). (That x value is x = ln(y).) So, the numberof solutions for y = g(x) is countable. Therefore, we can write

P [Y = y] =∑

x:g(x)=y

P [X = x] .

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ECS 315 HW Solution 11 — Due: Nov 25, 9:19 AM 2015/1

Here, X ∼ N (m,σ2). Therefore, X is a continuous random variable and P [X = x] = 0for any x. Hence,

P [Y = y] =∑

x:g(x)=y

0 = 0.

Because P [Y = y] = 0 for all y, we conclude that Y is a continuous random variable.

(b) Start with Y = eX . We know that exponential function gives strictly positive number.So, Y is always strictly positive. In particular, FY (y) = 0 for y ≤ 0.Next, for y > 0, by definition, FY (y) = P [Y ≤ y]. Plugging in Y = eX , we have

FY (y) = P[eX ≤ y

].

Because the exponential function is strictly increasing, the event [eX ≤ y] is the sameas the event [X ≤ ln y]. Therefore,

FY (y) = P [X ≤ ln y] = FX (ln y) .

Combining the two cases above, we have

FY (y) =

{FX (ln y) , y > 0,0, y ≤ 0.

Finally, we apply

fY (y) =d

dyFY (y).

For y < 0, we have fY (y) =ddy

0 = 0. For y > 0,

fY (y) =d

dyFY (y) =

d

dyFX (ln y) = fX (ln y)×

d

dyln y =

1

yfX (ln y) . (11.1)

Therefore,

fY (y) =

{1yfX (ln y) , y > 0,

0, y < 0.

At y = 0, because Y is a continuous random variable, we can assign any value, e.g. 0,to fY (0). Then

fY (y) =

{1yfX (ln y) , y > 0,

0, otherwise.

Here, X ∼ N (m,σ2). Therefore,

fX (x) =1√2πσ

e−12(

x−mσ )

2

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ECS 315 HW Solution 11 — Due: Nov 25, 9:19 AM 2015/1

and

fY (y) =

{1√2πσy

e−12(

ln(y)−mσ )

2

, y > 0,

0, otherwise.

Extra Questions

Problem 6. Cholesterol is a fatty substance that is an important part of the outer lining(membrane) of cells in the body of animals. Its normal range for an adult is 120–240 mg/dl.The Food and Nutrition Institute of the Philippines found that the total cholesterol level forFilipino adults has a mean of 159.2 mg/dl and 84.1% of adults have a cholesterol level below200 mg/dl. Suppose that the cholesterol level in the population is normally distributed.

(a) Determine the standard deviation of this distribution.

(b) What is the value of the cholesterol level that exceeds 90% of the population?

(c) An adult is at moderate risk if cholesterol level is more than one but less than twostandard deviations above the mean. What percentage of the population is at moderaterisk according to this criterion?

(d) An adult is thought to be at high risk if his cholesterol level is more than two standarddeviations above the mean. What percentage of the population is at high risk?

Solution : See handwritten solution

Problem 7. Consider a random variable X whose pdf is given by

fX(x) =

{cx2, x ∈ (1, 2) ,0, otherwise.

Let Y = 4 |X − 1.5|.

(a) Find EY .

(b) Find fY (y).

Solution : See handwritten solution

11-6

HW11 Q6: Gaussian RV - CholesterolThursday, November 13, 2014 3:49 PM

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HW11 Q7: SISOSaturday, November 22, 2014 11:54 PM

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