Techniques for finding the distribution of a transformation ...

TRANSFORMATIONS OF RANDOM VARIABLES

1. INTRODUCTION

1.1. Definition. We are often interested in the probability distributions or densities of functions of one or more random variables. Suppose we have a set of random variables, X1, X2, X3, . . . Xn, with a known joint probability and/or density function. We may want to know the distribution of some function of these random variables Y = (X1, X2, X3, . . . Xn). Realized values of y will be related to realized values of the X's as follows

y = (x1, x2, x3, . . . , xn)

(1)

A simple example might be a single random variable x with transformation

y = (x) = log (x)

(2)

1.2. Techniques for finding the distribution of a transformation of random variables.

1.2.1. Distribution function technique. We find the region in x1, x2, x3, . . . xn space such that (x1, x2, . . . xn) . We can then find the probability that (x1, x2, . . . xn) , i.e., P[ (x1, x2, . . . xn) ] by integrating the density function f(x1, x2, . . . xn) over this region. Of course, F() is just P[ ]. Once we have F(), we can find the density by integration.

1.2.2. Method of transformations (inverse mappings). Suppose we know the density function of x. Also suppose that the function y = (x) is differentiable and monotonic for values within its range for which the density f(x) =0. This means that we can solve the equation y = (x) for x as a function of y. We can then use this inverse mapping to find the density function of y. We can do a similar thing when there is more than one variable X and then there is more than one mapping .

1.2.3. Method of moment generating functions. There is a theorem (Casella [2, p. 65] ) stating that if two random variables have identical moment generating functions, then they possess the same probability distribution. The procedure is to find the moment generating function for and then compare it to any and all known ones to see if there is a match. This is most commonly done to see if a distribution approaches the normal distribution as the sample size goes to infinity. The theorem is presented here for completeness.

Theorem 1. Let FX (x) and FY (y) be two cumulative distribution functions all of whose moments exist. Then

a: If X and Y have bounded support, then FX(u) = FY (u) for all u if and only if E Xr = E Yr for all integers r = 0,1,2, . . . .

b: If the moment generating functions exist and MX (t) = MY (t) for all t in some neighborhood of 0, then FX (u) = FY (u) for all u.

For further discussion, see Billingsley [1, ch. 21-22] .

Date: November 17, 2005. 1

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TRANSFORMATIONS OF RANDOM VARIABLES

2. DISTRIBUTION FUNCTION TECHNIQUE

2.1. Procedure for using the Distribution Function Technique. As stated earlier, we find the region in the x1, x2, x3, . . . xn space such that (x1, x2, . . . xn) . We can then find the probability that (x1, x2, . . . xn) , i.e., P[ (x1, x2, . . . xn) ] by integrating the density function f(x1, x2, . . . xn) over this region. Of course, F() is just P[ ]. Once we have F(), we can find the density by differentiation.

2.2. Example 1. Let the probability density function of X be given by

6 x (1 - x) 0 < x < 1

f(x) =

(3)

0

otherwise

Now find the probability density of Y = X3.

Let G(y) denote the value of the distribution function of Y at y and write

G( y ) = P ( Y y ) = P ( X3 y )

= P X y1/3

y1/3

=

0

6x (1 - x)dx

(4)

y1/3

=

0

6 x - 6 x2 d x

=

3 x2 - 2 x3

|y1/3

0

= 3 y2/3 - 2y

Now differentiate G(y) to obtain the density function g(y)

d G (y) g(y) =

dy

d =

3 y2/3 - 2 y

dy

(5)

= 2 y- 1/3 - 2

= 2 ( y-1/3 - 1 ), 0 < y < 1

2.3. Example 2. Let the probability density function of x1 and of x2 be given by

2 e- x1 - 2 x2 x1 > 0 , x2 > 0

f (x1, x2) =

(6)

0

otherwise

Now find the probability density of Y = X1 + X2 or X1 = Y - X2. Given that Y is a linear function of X1 and X2, we can easily find F(y) as follows.

TRANSFORMATIONS OF RANDOM VARIABLES

3

Let FY (y) denote the value of the distribution function of Y at y and write

FY (y) = P (Y y)

y

=

0

y - x2

2 e- x1 - 2 x2 d x1 d x2

0

y

=

- 2 e- x1 - 2 x2 |y0 - x2 d x2

0

y

(7)

=

- 2 e- y + x2 - 2 x2 - -2 e- 2 x2 d x2

0

y

=

- 2 e- y - x2 + 2 e- 2 x2 d x2

0

y

=

2 e- 2 x2 - 2 e- y - x2 d x2

0

Now integrate with respect to x2 as follows

FY (y) = P (Y y)

y

=

2 e- 2 x2 - 2 e- y - x2 d x2

0

= - e- 2 x2 + 2 e- y - x2 |y0

(8)

= - e- 2 y + 2 e- y - y - - e0 + 2 e- y

= e- 2 y - 2 e- y + 1 Now differentiate FY (y) to obtain the density function f(y)

d F (y) fY (y) = d y

d =

e- 2 y - 2 e- y + 1

dy

(9)

= - 2 e- 2 y + 2 e- y

= 2 e- 2 y (- 1 + e y)

2.4. Example 3. Let the probability density function of X be given by

1 fX (x) =

?

-1

e2

(

) , x - ? 2

- < x <

2

(10)

= 1 2 2

? exp

( x - ? )2

- 2 2

,

- < x <

Now let Y = (X) = eX. We can then find the distribution of Y by integrating the density function of X over the appropriate area that is defined as a function of y. Let FY (y) denote the value of the distribution function of Y at y and write

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TRANSFORMATIONS OF RANDOM VARIABLES

FY (y) = P ( Y y)

= P eX y = P ( X ln y), y > 0 (11)

ln y

1

( x - ? )2

=

? exp -

-

2 2

2 2

d x, y > 0

Now differentiate FY (y) to obtain the density function f(y). In this case we will need the rules for

differentiating under the integral sign. They are given by theorem 2 which we state below without

proof.

Theorem 2.

Suppose that f and

f x

are continuous in the rectangle

R = { (x, t) : a x b , c t d}

and suppose that u0(x) and u1(x) are continuously differentiable for a x b with the range of u0(x) and u1(x) in (c, d). If is given by

u1 (x)

(x) =

f(x, t) d t

(12)

u0 (x)

then

d

u1 (x)

=

f(x, t) d t

dx x u0(x)

(13)

=

f

(x,

u1(x))

du1(x) dx

-

f (x, u0(x))

du0(x) dx

+

u1(x) f (x, t)

dt

u0 (x)

x

If one of the bounds of integration does not depend on x, then the term involving its derivative will be zero.

For a proof of theorem 2 see (Protter [3, p. 425] ). Applying this to equation 11 where y takes the role of x, ln y takes the role of u1(x), and x takes the role of t in the theorem we obtain

ln y

1

(x - ?)2

FY (y) =

? exp

- 22

-

22

dx, y > 0

FY (y) = fY (y) =

1 22

? exp

(ln y - ?)2 -

22

1 y

(14)

ln y d

1

(x - ?)2

+ - dy

? exp -

22

22

dx

1

(ln y - ?)2

=

? exp -

y 22

22

The last line of equation 14 follows because

d

1

(x - ?)2

? exp -

=0

dy 22

22

TRANSFORMATIONS OF RANDOM VARIABLES

5

3. METHOD OF TRANSFORMATIONS (SINGLE VARIABLE)

3.1. Discrete examples of the method of transformations.

3.1.1. One-to-one function. Find a formula for the probability distribution of the total number of heads obtained in four tosses of a coin where the probability of a head is 0.60.

The sample space, probabilities and the value of the random variable are given in table 1.

TABLE 1. Outcomes, Probabilities and Number of Heads from Tossing a Coin Four Times.

Element of sample space HHHH HHHT HHTH HTHH THHH HHTT HTHT HTTH THHT THTH TTHH HTTT THTT TTHT TTTH TTTT

Probability 81/625 54/625 54/625 54/625 54/625 36/625 36/625 36/625 36/625 36/625 36/625 24/625 24/625 24/625 24/625 16/625

Value of random variable X (x) 4 3 3 3 3 2 2 2 2 2 2 1 1 1 1 0

From the table we can determine the probabilities as

16

96

216

216

81

P (X = 0) = , P (X = 1) = , P (X = 2) = , P (X = 3) = , P (X = 4) =

625

625

625

625

625

We can also compute these probabilities using counting rules. The probability of one head and

then three tails is

3222

5555 or

31 23

24

=

5

5

625

The probability of 3 heads and then one tail is

3332

5555 or

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