chapter 5a:functions of random variables [email protected] yang zhenlin

Chapter 5a: Functions of RandomVariables

[email protected]://www.mysmu.edu/faculty/zlyang/

Yang ZhenlinYang Zhenlin

STAT306, Term II, 09/10

Chapter 5a

STAT151, Term I 2015-16 © Zhenlin Yang, SMU

Chapter 5a Contents

Functions of One Random Variable

--- change of variable technique

Functions of Two Random Variables

--- change of variable technique

Sum of Independent Random variables

--- The Moment Generating Function Technique

2

The main purpose of this chapter: Introducing methods for finding the distribution of a function of the random variable(s).

The main purpose of this chapter: Introducing methods for finding the distribution of a function of the random variable(s).


Chapter 5a

STAT151, Term I 2015-16 © Zhenlin Yang, SMU3


Definition 5a.1 (Change-of-Variable Technique) Let X be a continuous type random variable with pdf f(x). Let Y = u(X) be a one-to-one transformation of X with inverse function X = v(Y). Then the pdf of Y is given by

g(y) = f[v(y)] |v(y)|,

where v(y) is the derivative of v(y). If the possible values of X are c1< x <c2, then the possible values of Y are u(c1)< y <u(c2).

Definition 5a.1 (Change-of-Variable Technique) Let X be a continuous type random variable with pdf f(x). Let Y = u(X) be a one-to-one transformation of X with inverse function X = v(Y). Then the pdf of Y is given by

g(y) = f[v(y)] |v(y)|,

where v(y) is the derivative of v(y). If the possible values of X are c1< x <c2, then the possible values of Y are u(c1)< y <u(c2).

Example 5a.1. Let X have a gamma distribution with pdf

Let Y = eX. Find the pdf of Y.

.0,

)(

1)( 1

xexxf x

The case of a continuous random variable.The case of a continuous random variable.


Chapter 5a



Solution: Since the inverse function is X= v(Y) = log (Y), v(y) = 1/y. Thus, by Definition 5.1, the pdf of Y is given by

Since the support of X is (0, ), the support of Y is (1, ). The pdf of Y is thus,

11

1ln1 )(ln

)(

11)(ln

)(

1 |)(|)]([)(

y

y

yeyyvyvfyg y

.1,)(ln

)(

1)(

11

1

yy

yyg

The way to see the change-of-variable technique is through CDF:

G(y) = P{Y y} = P{X v(y)} = F[v(y)].

Taking derivatives leads to g(y) = f[v(y)] |v(y)|. So, the change-of-variable technique is essentially the CDF technique.

The way to see the change-of-variable technique is through CDF:

G(y) = P{Y y} = P{X v(y)} = F[v(y)].

Taking derivatives leads to g(y) = f[v(y)] |v(y)|. So, the change-of-variable technique is essentially the CDF technique.


Chapter 5a



The Change-of-variable technique can be applied to a random variable X of the discrete type, but there is major difference: pmf p(x) = P{X = x} represents probability, but pdf f(x) does not.

For a one-to-one transformation, Y = u(X), with inverse X = v(Y), we can easily see that the pmf g(y) of Y is

g(y) = P{Y = y} = P{X = v(y)} = p[v(y)]

The possible values of Y are found directly from the possible values of X through the functional relation Y = u(X).

The case of a discrete random variable.The case of a discrete random variable.

Example 5a.2. Let X have a Poisson distribution with = 4. Find the pmf of Y = X1/2. Since X = Y2, we have,

...,3,2,1,0,

)!(

)4()(and...,2,1,0,

!

4)(

2

44 2

yy

eygx

x

exp

yx


Chapter 5a



The case of a non-one-to-one function of a continuous r.v.The case of a non-one-to-one function of a continuous r.v.

The change-of-variable technique requires that the function is one-to-one, thus cannot be applied when the function is not one-to-one.

However, as noted earlier, the distribution of functions of a random variable are essentially developed from the CDF. Thus, the distribution of a non-one-to-one function can still be derived from the CDF!

We will demonstrate this idea by showing an important result:

The square of a standard normal random variable is a gamma r.v. with parameters (1/2, 2), this special gamma r.v. is called the chi-squared random variable with degrees of freedom equal to 1.

The square of a standard normal random variable is a gamma r.v. with parameters (1/2, 2), this special gamma r.v. is called the chi-squared random variable with degrees of freedom equal to 1.


Chapter 5a


Functions of One Random Variable Example 5a.3. Let Z be a standard normal r.v. and let X = Z2.

The CDF of X is

dze

dze

xZxP

xXPxG

zx

zx

x

2

0

2

2

2

2

12

2

1

}{

}{)(

0,

2

1)(

becomes CDF the

,)2/(1,then

,Lettting

2

0

xdyey

xG

dyydz

yz

yx

Taking derivative with respect to x, we obtain the pdf of X:

.0,2

1

2

1)( 21212 xexe

xxg xx

Recognizing that , the above is the pdf of a gamma r.v. with parameters (1/2, 2), called the chi-squared with 1 d.f.

)( 21


Chapter 5a



The above change-of-variable technique can be extended to the case of joint distributions involving two or more random variables. Many interesting problems solved.

Definition 5a.2. (Change-of-Variable Technique) Let X1 and X2 be two continuous type random variables with joint pdf f(x1, x2). Let Y1 = u1(X1, X2) and Y2 = u2(X1, X2) be two continuous functions, which have single-valued inverse: X1 = v1(Y1, Y2) and X2 = v2(Y1, Y2). Then the joint pdf of Y1 and Y2 is

where J, called the Jacobian, is the determinant of the matrixof partial derivatives:

Definition 5a.2. (Change-of-Variable Technique) Let X1 and X2 be two continuous type random variables with joint pdf f(x1, x2). Let Y1 = u1(X1, X2) and Y2 = u2(X1, X2) be two continuous functions, which have single-valued inverse: X1 = v1(Y1, Y2) and X2 = v2(Y1, Y2). Then the joint pdf of Y1 and Y2 is

where J, called the Jacobian, is the determinant of the matrixof partial derivatives:

||)],(),,([),( 21221121 Jyyvyyvfyyg

2

212

1

212

2

211

1

211

),(),(

),(),(

y

yyv

y

yyvy

yyv

y

yyv

J


Chapter 5a


Functions of Two Random Variables Example 5a.4. Let X1 and X2 be two independent r.v.s, each with

pdf f(x) = ex, 0 < x < . Consider Y1 = X1 X2, and Y2 = X1 + X2.

(a)Find the joint pdf of Y1 and Y2

(b)Find the marginal pdfs of Y1 and Y2, respectively.Solution: (a) Since X1 and X2 are independent and have the same distribution,

f(x1, x2) = f(x1) f(x2) = )( 21 xxe , 0 < x1 < , 0 < x2 <

From Y1 = X1 X2, and Y2 = X1 + X2, we obtain

2

211

YYX

and

212

2

YYX

, with J =

21212121

=

2

1

The joint pdf of Y1 and Y2 is

g(y1, y2) = 2

12ye , with y2 < y1 < y2, 0 < y2 < , ???


Chapter 5a



where the possible values of Y1 and Y2 can be found as follows:

•Y2 = X1 + X2 implies 0 < Y2 < ;

•X1 = (Y1 + Y2)/2 > 0 implies Y1 > Y2;

•X2 = (Y2 Y1)/2 > 0 implies Y1 < Y2.

y1

y2

y2 = y1 y2 = -y1

The region of (Y1, Y2) values


Chapter 5a


The latter expression can simply be written as

That is called the double exponential pdf.

11


(b) The marginal pdf of Y2:

22122 0,2

1)( 22

2

2

yeydyeyg yyy

y

.02

1

2

1

,0,2

1

2

1

)(

12

12

111

1

2

1

1

2

yedye

yedyeyg

y

y

y

y

y

y

1

||11 ,

2

1)( 1 yeyg y

The marginal pdf of Y1:


Chapter 5a



Definition 5a.3. Let X and Y be jointly distributed r.v.s with joint pmf p(x, y), or a joint pdf f(x, y). Let u(X, Y) be a continuous function of X and Y. Then, u(X, Y) is also a random variable. If X and Y are both discrete,

And if X and Y are both continuous,

Definition 5a.3. Let X and Y be jointly distributed r.v.s with joint pmf p(x, y), or a joint pdf f(x, y). Let u(X, Y) be a continuous function of X and Y. Then, u(X, Y) is also a random variable. If X and Y are both discrete,

And if X and Y are both continuous,

x y

yxpyxuYXu ),(),()],([E

dxdyyxfyxuYXu ),(),()],([E

If X and Y are jointly distributed r.v.s, then,

Var(X + Y) = Var(X) + Var(Y) + 2Cov(X, Y)

If further X and Y are independent, then

Var(X + Y) = Var(X) + Var(Y)

If X and Y are jointly distributed r.v.s, then,

Var(X + Y) = Var(X) + Var(Y) + 2Cov(X, Y)

If further X and Y are independent, then

Var(X + Y) = Var(X) + Var(Y)


Chapter 5a



Example 5a.5. The joint probability distribution of variables X and Y is shown in the table below,

Y

1X2 3

1 0.20 0.18 0.122 0.15 0.09 0.063 0.07 0.03 0.10

(a) Determine the marginal probability distributions of X and Y.(b) Are X and Y independent? Explain.(c) Find the probability mass function of X+Y.(d) Find the probability of P(X Y).

X 1 2 3 Y 1 2 3

pX(x) 0.42 0.30 0.28 pY(y) 0.50 0.30 0.20

)(xpX )(ypY

Solution: (a) The marginal pmfs are:


Chapter 5a



(b) No. Because p(1, 1) = 0.20, but = 0.420.50 = 0.21.

(c) Let Z = X+Y, then the pmf of Z is

Where, for example, pZ(3) = P(X+Y = 3) = P(X = 1, Y = 2) + P(X = 2, Y = 1)

= 0.15 + 0.18 = 0.33.

(d) P(X Y) = 1 P(X = Y) = 1 P(X = 1, Y = 1) P(X = 2, Y = 2) P(X = 3, Y = 3) = 1 0.20 0.09 0.10 = 0.61

)(xpX )(ypY

Z 2 3 4 5 6

pZ(z) 0.20 0.33 0.28 0.09 0.10

)(zpZ


Chapter 5a


Sum of Independent Random Variables

Using the above property, one can easily see the following results:

Sum of independent binomial r.v.s with the same probability of success is again a binomial r.v.

Sum of independent Poisson r.v.s is again a Poisson r.v.

Sum of independent exponential r.v.s with the same mean is a gamma r.v.

Sum of independent normal r.v.s is again a normal r.v.

And more . . . .

Recall the Uniqueness Property of MGF: The MGF of a r.v. X uniquely determines its distribution, and vise versa, e.g., if the MGF of X is the same of that of a normal r.v., then, X must be normally distributed.

Recall the Uniqueness Property of MGF: The MGF of a r.v. X uniquely determines its distribution, and vise versa, e.g., if the MGF of X is the same of that of a normal r.v., then, X must be normally distributed.


Chapter 5a


To demonstrate the above result using the MGF technique, consider two independent normal random variablesand . Let Y = X1 + X2. The MGF of Y is

16


Sum of independent normal r.v.s is again a normal r.v.Sum of independent normal r.v.s is again a normal r.v.

Recall the MGF of X ~ N(, 2):

If one can show that the MGF of a random variable has the same for as above, then one can conclude that this random variable is normal with mean and variance being, respectively, the quantities in front of ‘t’ and ‘t2’.

Recall the MGF of X ~ N(, 2):

If one can show that the MGF of a random variable has the same for as above, then one can conclude that this random variable is normal with mean and variance being, respectively, the quantities in front of ‘t’ and ‘t2’.

)exp()( 2221 tttM

),(~ 2111 NX

),(~ 2222 NX

)()(][E][E][E][E][E)(21

212121 )( tMtMeeeeeetM XXtXtXtXtXXXttY

Y

Follows from the independence between X1 and X2 Follows from the independence between X1 and X2


Chapter 5a


It follows that

Recognizing that this MGF is in the same form of the MGF of a normal random variable, Y must be normally distributed. In particular,

This result can easily be extended to the case of many normal r.v.s

17


))()exp((

)exp()exp()(22

2212

121

2222

12

2212

11

tt

tttttM Y

),(~ 22

2121 NY


Chapter 5a



Sum of independent binomial r.v.s with the same probability of success is again a binomial r.v.Sum of independent binomial r.v.s with the same probability of success is again a binomial r.v.

To see this, using MGF technique. White board presentation.


Chapter 5a



Sum of independent Poisson r.v.s is again a Poisson r.v.Sum of independent Poisson r.v.s is again a Poisson r.v.



Chapter 5a



Sum of independent exponential r.v.s with the same mean is a gamma r.v.Sum of independent exponential r.v.s with the same mean is a gamma r.v.



Chapter 5a


Functions of Normal R.V.s

In Example 5a.3, we have shown that if Z is a standard normal r.v., then X = Z2 follows a chi-squared distribution with 1 d.f., which is seen to be a special gamma r.v.

We have also shown using the MGF technique that the sum of two independent normal r.v.s is again normally distributed.

There are many other functions of normal r.v.(s) of which the distributions are of interest. In particular in the context of statistical inference, functions of a random sample drawn from a normal population, or functions of two random samples drawn from two independent normal populations, are needed for the purposes of drawing statistical inferences about the normal populations.

We put these into a general topic: “Sampling Distribution”, with details presented in Chapter 5b.

chapter 5a:functions of random variables [email protected] yang zhenlin

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