Transcript Functions of Random Variables

Chapter 3
DeGroot & Schervish
Functions of a Random Variable
 the distribution of some function of X
 suppose X is the rate at which customers are served in
a queue
 then 1/X is the average waiting time
 If we have the distribution of X, we should be able to:
 determine the distribution of 1/X
 or of any other function of X
Random Variable with a Discrete
Distribution
 Distance from the Middle example
 Let X have the uniform distribution on the integers 1,
2, . . . , 9.
 Suppose that we are interested in how far X is from the
middle of the distribution, namely, 5.
 We could define Y = |X − 5| and compute probabilities
such as
 Pr(Y = 1) = Pr(X ∈ {4, 6}) = 2/9.
Function of a Discrete Random
Variable
 Let X have a discrete distribution with p.f. f ,
 let Y = r(X) for some function of r defined on the set of
possible values of X
 For each possible value y of Y , the p.f. g of Y is
Distance from the Middle
 The possible values of Y in the previous example are 0,
1, 2, 3, and 4.
 We see that Y = 0 if and only if X = 5
 g(0) = f (5) = 1/9.
 For all other values of Y , there are two values of X that
give that value of Y . For example,
 {Y = 4} = {X = 1} ∪ {X = 9}.
 So, g(y) = 2/9 for y = 1, 2, 3, 4.
Random Variable with a
Continuous Distribution
 If a random variable X has a continuous distribution,
then the procedure for deriving the probability
distribution of a function of X differs from that given
for a discrete distribution.
 One way to proceed is by direct calculation
Average Waiting Time
 Let Z be the rate at which customers are served in a
queue,
 suppose that Z has a continuous c.d.f. F.
 The average waiting time is Y = 1/Z.
 If we want to find the c.d.f. G of Y , we can write
Random Variable with a
Continuous Distribution
 In general, suppose that the p.d.f. of X is f and that
another random variable is defined as Y = r(X).
 For each real number y, the c.d.f. G(y) of Y can be
derived as follows:
 If the random variable Y also has a continuous
distribution, its p.d.f. g can be obtained from the
relation
Direct Derivation of the p.d.f.
 Let r be a differentiable one-to-one function on the open
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interval (a, b).
Then r is either strictly increasing or strictly decreasing.
Because r is also continuous, it will map the interval (a, b)
to another open interval (α, β), called the image of (a, b)
under r.
That is, for each x ∈ (a, b), r(x) ∈ (α, β), and for each y ∈ (α,
β) there is x ∈ (a, b) such that y = r(x) and this y is unique
because r is one-to-one.
So the inverse s of r will exist on the interval (α, β),
meaning that for x ∈ (a, b) and y ∈ (α, β) we have r(x) = y if
and only if s(y) = x.
Theorem
 Let X be a random variable for which the p.d.f. is f and for which
Pr(a <X<b) = 1.
 Here, a and/or b can be either finite or infinite.
 Let Y = r(X), and suppose that r(x) is differentiable and one-to-
one for a <x <b.
 Let (α, β) be the image of the interval (a, b) under the function r.
 Let s(y) be the inverse function of r(x) for α <y <β.
 Then the p.d.f. g of Y is
Proof
 If r is increasing, then s is increasing, and for each y ∈ (α, β)
 Because s is increasing, ds(y)/dy is positive; hence, it equals |ds(y)/dy|
and this equation implies the theorem.
 Similarly, if r is decreasing, then s is decreasing, and for each y ∈ (α, β),
 Since s is strictly decreasing, ds(y)/dy is negative so that −ds(y)/dy
equals |ds(y)/dy|. It follows that the equation implies the theorem.
The Probability Integral
Transformation
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Let X be a continuous random variable
The p.d.f. f (x) = exp(−x) for x >0 and 0 otherwise.
The c.d.f. of X is F(x) = 1− exp(−x) for x >0 and 0 otherwise.
If we let F be the function r, we can find the distribution of
Y = F(X).
 The c.d.f. or Y is, for 0 < y <1,
 which is the c.d.f. of the uniform distribution on the
interval [0, 1]. It follows that Y has the uniform distribution
on the interval [0, 1].
Theorem
 Let X have a continuous c.d.f. F,
 let Y = F(X).
 This transformation from X to Y is called the
probability integral transformation.
 The distribution of Y is the uniform distribution on the
interval [0, 1].
Proof
 First, because F is the c.d.f. of a random variable, then 0 ≤ F(x) ≤ 1 for
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−∞ < x <∞.
Therefore, Pr(Y < 0) = Pr(Y > 1) = 0.
Since F is continuous, the set of x such that F(x) = y is a nonempty
closed and bounded interval [x0, x1] for each y in the interval (0, 1).
Let F−1(y) denote the lower endpoint x0 of this interval, which was
called the y quantile of F.
In this way, Y ≤ y if and only if X ≤ x1.
Let G denote the c.d.f. of Y . Then
 Hence, G(y) = y for 0 < y <1. Because this function is the c.d.f. of the
uniform distribution on the interval [0, 1], this uniform distribution is
the distribution of Y .
Functions of Two or More Random
Variables
 When we observe data consisting of the values of
several random variables, we need to summarize the
observed values in order to be able to focus on the
information in the data.
 Summarizing consists of constructing one or a few
functions of the random variables.
 We now describe the techniques needed to determine
the distribution of a function of two or more random
variables.
Random Variables with a Discrete
Joint Distribution
 Suppose that n random variables X1, . . . , Xn have a
discrete joint distribution for which the joint p.f. is f,
and that m functions Y1, . . . , Ym of these n random
variables are defined as follows:
Y1 = r1(X1, . . . , Xn),
Y2 = r2(X1, . . . , Xn),
...
Ym = rm(X1, . . . , Xn).
Random Variables with a Discrete
Joint Distribution
 For given values y1, . . . , ym of the m random variables Y1, . . . , Ym, let
A denote the set of all points (x1, . . . , xn) such that
r1(x1, . . . , xn) = y1,
r2(x1, . . . , xn) = y2,
...
rm(x1, . . . , xn) = ym.
 Then the value of the joint p.f. g of Y1, . . . , Ym is specified at the point
(y1, . . . , ym) by the relation
Random Variables with a
Continuous Joint Distribution
 Suppose that the joint p.d.f. of X = (X1, . . . , Xn) is f (x)
and that Y = r(X).
 Y = r(X1, . . . , Xn),
 The d.f. of Y can be calculated as follows:
 𝐺 𝑦 = 𝑃 𝑌 ≤ 𝑦 = 𝑃 𝑟 𝑋1, 𝑋2, … , 𝑋𝑛 ≤ 𝑦
= … 𝑓 𝑋1, 𝑋2, … , 𝑋𝑛 𝑑𝑥1 … 𝑑𝑥𝑛
 If Y has a continuous distribution, then the derivation of
G(y) gives the pd.f. of Y.
Direct Transformation of a
Multivariate p.d.f.
 Let X1, . . . , Xn have a continuous joint distribution for which the
joint p.d.f. is f .
 Assume that there is a subset S of Rn such that
 Pr[(X1, . . . , Xn) ∈ S]= 1.
 Define n new random variables Y1, . . . , Yn as follows:
Y1 = r1(X1, . . . , Xn),
Y2 = r2(X1, . . . , Xn),
...
Yn= rn(X1, . . . , Xn),
where we assume that the n functions r1, . . . , rn define a one-toone differentiable transformation of S onto a subset T of Rn.
Direct Transformation of a
Multivariate p.d.f.
 Let the inverse of this transformation be given as
follows:
x1 = s1(y1, . . . , yn),
x2 = s2(y1, . . . , yn),
...
xn = sn(y1, . . . , yn).
Direct Transformation of a
Multivariate p.d.f.
 Then the joint p.d.f. g of Y1, . . . , Yn is
 where J is the determinant and |J | denotes the absolute
value of the determinant J .
 This determinant J is called the Jacobian of the
transformation specified by the equations.
Linear Transformations
 Let X = (X1, . . . , Xn) have a continuous joint
distribution for which the joint p.d.f. is f .
 Define Y = (Y1, . . . , Yn) by
Y = AX,
 where A is a nonsingular n × n matrix. Then Y has a
continuous joint distribution with p.d.f.
 where A−1 is the inverse of A.