FZ(a) = P(Z≤a) = P(max{X1, . . . , Xn} ≤a) = P(X1≤a, . . . , Xn≤a). Now suppose that the events {Xi ≤ ai} are independent for every choice of the ai. In this case we call the random variables independent (see also Chapter 9, where we study independence of random variables). In particular, the events{Xi≤a} are independent for alla. It then follows that
FZ(a) = P(X1≤a, . . . , Xn≤a) = P(X1≤a)· · ·P(Xn≤a).
Hence, if all random variables have the same distribution function F, then the following result holds.
The distribution of the maximum. LetX1, X2, . . . , Xn be n
independent random variables with the same distribution function F, and letZ= max{X1, X2, . . . , Xn}. Then
FZ(a) = (F(a))n.
Quick exercise 8.5 Let X1, X2, . . . , Xn be independent random variables,
all with aU(0,1) distribution. LetZ = max{X1, . . . , Xn}. Compute the dis- tribution function and the probability density function ofZ.
What can we say about the distribution of the minimum? Let V = min{X1, X2, . . . , Xn}.
We can now find the distribution function FV of V by observing that the minimum of theXi islarger than a numberaif and only if allXi arelarger
thana. The trick is to switch to the complement of the event{V ≤a}: FV(a) = P(V ≤a) = 1−P(V > a) = 1−P(min{X1, . . . , Xn}> a)
= 1−P(X1> a, . . . , Xn > a).
So using independence and switching back again, we obtain
FV(a) = 1−P(X1> a, . . . , Xn> a) = 1−P(X1> a)· · ·P(Xn> a) = 1−(1−P(X1≤a))· · ·(1−P(Xn≤a)).
We have found the following result for the minimum.
The distribution of the minimum. Let X1, X2, . . . , Xn be n
independent random variables with the same distribution function F, and letV = min{X1, X2, . . . , Xn}. Then
FV(a) = 1−(1−F(a))n.
Quick exercise 8.6 Let X1, X2, . . . , Xn be independent random variables,
all with aU(0,1) distribution. LetV = min{X1, . . . , Xn}. Compute the dis- tribution function and the probability density function ofV.
110 8 Computations with random variables
8.5 Solutions to the quick exercises
8.1 Clearly Z can take the values 1, . . . ,150. The value 150 is special: the plane is full if 150 or more people buy a ticket. Hence P(Z = 150) = P(X≥150) = 51/200.For the other values we have P(Z =i) = P(X =i) = 1/200, fori= 1, . . . ,149. Clearly, hereg(x) = min{150, x}.
8.2 The probability density ofY = 1/X is fY(y) = 1 y2 1 π(1 + (1y)2)= 1 π(1 +y2).
We see that 1/X has the same distribution asX! (This distribution is called the standard Cauchy distribution, it will be introduced in Chapter 11.)
8.3 First defineZ = (X−4)/5, which has anN(0,1) distribution. Then from Table B.1 P(X ≤5) = P Z≤ 5−4 5 = P(Z≤0.20) = 1−0.4207 = 0.5793. Similarly, using the symmetry of the normal distribution,
P(X ≥2) = P Z≥ 2−4 5 = P(Z≥ −0.40) = P(Z≤0.40) = 0.6554.
8.4 Ifg(x) = e−x, theng(x) = e−x>0; henceg is strictly convex. It follows from Jensen’s inequality that
e−E[X] ≤Ee−X. Moreover, if Var(X)>0, then the inequality is strict.
8.5 The distribution function of theXi is given byF(x) =xon [0,1]. There- fore the distribution function FZ of the maximum Z is equal to FZ(a) = (F(a))n =an.Its probability density function is
fZ(z) = d
dzFZ(z) =nz
n−1 for 0≤z≤1.
8.6 The distribution function of theXi is given byF(x) =xon [0,1]. There- fore the distribution function FV of the minimum V is equal to FV(a) = 1−(1−a)n.Its probability density function is
fV(v) = d
dvFV(v) =n(1−v)
8.6 Exercises 111
8.6 Exercises
8.1Often one is interested in the distribution of the deviation of a random variable X from its mean µ = E [X]. LetX take the values 80,90,100,110, and 120, all with probability 0.2; then E [X] = µ= 100. Determine the dis- tribution ofY =|X−µ|. That is, specify the valuesY can take and give the corresponding probabilities.
8.2Suppose X has a uniform distribution over the points {1,2,3,4,5,6} and thatg(x) = sin(π2x).
a. Determine the distribution of Y =g(X) = sin(π
2X), that is, specify the valuesY can take and give the corresponding probabilities.
b. LetZ= cos(π
2X). Determine the distribution ofZ.
c. Determine the distribution of W = Y2+Z2. Warning: in this example there is a very special dependency betweenY andZ, and in general it is much harder to determine the distribution of a random variable that is a function oftwoother random variables. This is the subject of Chapter 11.
8.3The continuous random variableU is uniformly distributed over [0,1].
a. Determine the distribution function ofV = 2U + 7. What kind of distri- bution doesV have?
b. Determine the distribution function of V =rU +s for all real numbers r >0 ands. See Exercise 8.9 for what happens for negativer.
8.4Transforming exponential distributions.
a. LetX have anExp(12) distribution. Determine the distribution function of 1
2X. What kind of distribution does 1
2X have?
b. LetX have an Exp(λ) distribution. Determine the distribution function ofλX. What kind of distribution doesλX have?
8.5LetX be a continuous random variable with probability density func- tion fX(x) = 3 4x(2−x) for 0≤x≤2 0 elsewhere.
a. Determine the distribution functionFX.
b. LetY =√X. Determine the distribution functionFY.
c. Determine the probability density ofY.
8.6LetX be a continuous random variable with probability densityfX that takes only positive values and letY = 1/X.
112 8 Computations with random variables a. DetermineFY(y) and show that
fY(y) = 1 y2fX 1 y fory >0.
b. LetZ= 1/Y. Usinga, determine the probability densityfZofZ, in terms offX.
8.7LetX have aPar(α) distribution. Determine the distribution function of lnX. What kind of a distribution does lnX have?
8.8LetXhave anExp(1) distribution, and letαandλbe positive numbers. Determine the distribution function of the random variable
W = X
1/α
λ .
The distribution of the random variableW is called the Weibull distribution with parametersαandλ.
8.9LetX be a continuous random variable. Express the distribution function and probability density of the random variableY =−Xin terms of those ofX.
8.10 Let X be an N(3,4) distributed random variable. Use the rule for normal random variables under change of units and Table B.1 to determine the probabilities P(X ≥3) and P(X ≤1).
8.11LetXbe a random variable, and letgbe a twice differentiable function withg(x)≤0 for all x. Such a function is called aconcave function. Show that for concave functions always
g(E [X])≥E [g(X)].
8.12LetX be a random variable with the following probability mass func- tion:
x 0 1 100 10 000 P(X =x) 14 14 14 14
a. Determine the distribution ofY =√X.
b. Which is larger E
√
X
orE [X]?
Hint:use Exercise 8.11, or start by showing that the functiong(x) =−√x is convex.
c. Compute E [X] and E
√
X
to check your answer (and to see that it makes a big difference!).
8.13Let W have a U(π,2π) distribution. What is larger: E [sin(W)] or sin(E [W])? Check your answer by computing these two numbers.
8.6 Exercises 113 8.14In this exercise we take a look at Jensen’s inequality for the function g(x) =x3 (which is neither convex nor concave on (−∞,∞)).
a. Can you find a (discrete) random variableX with Var(X)>0 such that EX3= (E [X])3?
b. Under what kind of conditions on a random variableX will the inequality EX3>(E [X])3 certainly hold?
8.15LetX1, X2, . . . , Xn be independent random variables, all with aU(0,1) distribution. LetZ= max{X1, . . . , Xn} andV = min{X1, . . . , Xn}.
a. Compute E [max{X1, X2}] and E [min{X1, X2}].
b. Compute E [Z] and E [V] for generaln.
c. Can you argue directly (using the symmetry of the uniform distribu- tion (see Exercise 6.3) and not the result of the computation inb) that 1 − E [max{X1, . . . , Xn}] = E [min{X1, . . . , Xn}]?
8.16In this exercise we derive a kind of Jensen inequality for the minimum.
a. Letaandbbe real numbers. Show that min{a, b}=1
2(a+b− |a−b|).
b. LetX andY be independent random variables with the same distribution and finite expectation. Deduce froma that
E [min{X, Y}] = E [X]−1
2E [|X−Y|].
c. Show that
E [min{X, Y}]≤min{E [X],E [Y]}.
Remark: this is not so interesting, since min{E [X],E [Y]} = E [X] = E [Y], but we will see in the exercises of Chapter 11 that this inequality is also true forX andY,which do not have the same distribution.
8.17LetX1, . . . , Xn be nindependent random variables with the same dis- tribution functionF.
a. Convince yourself that for any numbersx1, . . . , xn it is true that min{x1, . . . , xn}=−max{−x1, . . . ,−xn}.
b. LetZ= max{X1, X2, . . . , Xn}andV = min{X1, X2, . . . , Xn}. Use Exer- cise 8.9 and the observation inato deduce the formula
114 8 Computations with random variables
FV(a) = 1−(1−F(a))n
directly from the formula
FZ(a) = (F(a))n.
8.18 Let X1, X2, . . . , Xn be independent random variables, all with an
Exp(λ) distribution. Let V = min{X1, . . . , Xn}. Determine the distribution function ofV. What kind of distribution is this?
8.19 From the “north pole” N of a circle with diameter 1, a point Q on the circle is mapped to a point t on the line by its projection from N, as illustrated in Figure 8.2. ...... ...... ...... ...... ...... ... ... ... ... ... ... ... ... ... ... ... ... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... ... ... ... ... ... ... ... ... ... ... ... ...... ...... t Q N ϕ • • • ... ...
Fig. 8.2. Mapping the circle to the line.
Suppose that the pointQis uniformly chosen on the circle. This is the same as saying that the angleϕis uniformly chosen from the interval [−π2,π2] (can you see this?). Let X be this angle, so that X is uniformly distributed over the interval [−π2,π2]. This means that P(X ≤ϕ) = 1/2 +ϕ/π (cf. Quick exercise 5.3). What will be the distribution of the projection ofQon the line? Let us call this random variableZ. Then it is clear that the event{Z≤t} is equal to the event{X ≤ϕ}, where t and ϕcorrespond to each other under the projection. This means that tan(ϕ) =t, which is the same as saying that arctan(t) =ϕ.
a. What part of the circle is mapped to the interval [1,∞)?
b. Compute the distribution function ofZusing the correspondence between tandϕ.
c. Compute the probability density function ofZ.
The distribution ofZis called the Cauchy distribution (which will be discussed in Chapter 11).
9
Joint distributions and independence
Random variables related to the same experiment often influence one another. In order to capture this, we introduce the joint distribution of two or more random variables. We also discuss the notion of independence for random variables, which models the situation where random variables do not influence each other. As with single random variables we treat these topics for discrete and continuous random variables separately.