Probability Theory, Math 170b, Winter 2013, Tonci Antunovic -Homework 1

From the textbook solve the problems 1, 2, 3, 5, 6, 8, 11 and 14 at the endof the Chapter 4.And also the problems below:

Problem 1. Give examples of (not independent) random variables X andY , both of which are uniform in the interval [0, 1] and such that

1) P(X + Y = 1) = 1

2) X + Y is uniform in the interval [0, 2].

Solution:

1) Set X to be uniform in the interval [0, 1] and then set Y = 1 − X.Then we always have X + Y = 1.

2) Set X to be uniform in the interval [0, 1] and then set Y = X. ThenX + Y = 2X which is uniform in the interval [0, 2].

Problem 2. If X is uniform on the interval [0, 2] find the PDF of Y =X2 − 2X.

Solution: The CDF is

FY (y) = P(X2 − 2X ≤ y) = P((X − 1)2 ≤ y + 1).

If y < −1 the above probability is zero, if y > 0 the above probability is oneand if −1 ≤ y ≤ 0 then the above probability is

P(|X − 1| ≤√y + 1) = P(1−

√y + 1 ≤ X ≤ 1 +

√y + 1) =

√y + 1.

Therefore,

FY (y) =

0, if y < −1,√y + 1, if − 1 ≤ y ≤ 0,

1, if y > 0.

and

fY (y) =

0, if y < −1,

12√y+1

, if − 1 ≤ y ≤ 0,

0, if y > 0.

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Problem 3. Two people have arranged to meet at a certain place andtime. However, each of them is coming late and the person that comes firstwill have wait to the other person. The amount of time that the first andthe second person are late is T1 and T2 respectively. Both T1 and T2 areexponential random variables with parameter λ and are independent. Findthe PDF of the amount of time that the person arriving first will have towait for the other person.

Solution: The waiting time is just T = |T1 − T2| and it’s PDF is just

FT (t) = P(|T1 − T2| ≤ t) = P(0 ≤ T2 − T1 ≤ t) + P(0 ≤ T1 − T2 ≤ t).

Since T1 and T2 are independent and they are both exponential with param-eter λ the probabilities on the right hand side are equal so we just computethe first one. We have for t > 0

P(0 ≤ T2 − T1 ≤ t) =

∫ ∞0

∫ x+t

xλe−λxλe−λy dydx

=

∫ ∞0

λe−λx(e−λx − e−λx−λt

)dx =

1

2

(1− e−λt

).

Therefore

FT (t) =

{0, if t < 0,1− e−λt, if t ≥ 0,

and

fT (t) =

{0, if t < 0,λe−λt, if t ≥ 0.

This is an exponential random variable with parameter λ.

Problem 4. Let X be a continuous random variable with the PDF f andthe CDF F . Assume that f(x) > 0 for all real numbers x. Define therandom variable Y = F (X). Show that Y is uniform in the interval [0, 1].

(Hint: Since the derivative of the CDF F is the PDF f which is alwaysstrictly positive, F is monotonically increasing and has an inverse functionh which is monotonically increasing. Now you can try to use the formula wederived with the help of the calculus formula for the derivative of the inversefunction. Alternatively, you can show that P(Y ≤ t) = t for 0 ≤ t ≤ 1 witha little help of the inverse function h.)

Solution: The range of F is the interval (0, 1) so we can use the formula

fY (y) = f(h(y))|dhdy

(y)|,

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for 0 < y < 1. Since F is increasing so is it’s inverse h and so we don’t needthe absolute values. At a point y in (0, 1) (write it as y = F (x)) we have

dh

dy(y) =

1

F ′(x)=

1

f(x)

and so

fY (y) = f(h(y))1

f(x)= f(x)

1

f(x)= 1.

Since the density is 1 on the interval (0, 1) this shows Y is uniform in [0, 1].Alternatively, you can simply compute for 0 < y < 1

FY (y) = P(F (X) ≤ y) = P(X ≤ h(y)) = F (h(y)) = y,

so again the PDF is 1 in the interval (0, 1).

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