Page 211 - Applied statistics and probability for engineers

Section 5-5/Linear Functions of Random Variables 189

Suppose that X is a discrete random variable with probability distribution f (x). Let Y = h(X) be
X
a function of X that deines a one-to-one transformation between the values of X and Y and that we

wish to ind the probability distribution of Y. By a one-to-one transformation, we mean that each

value x is related to one and only one value of y = h(x) and that each value of y is related to one
and only one value of x, say, x = u(y) where u(y) is found by solving y = h(x) for x in terms of y.
Now the random variable Y takes on the value y when X takes on the value u(y). Therefore,
the probability distribution of Y is
f Y ( y) = ( y) = P X = ( ⎤ = f X ( ⎤ ⎦
u y)
u y)
P Y =
⎡
⎡
⎣
⎦
⎣
We may state this result as follows.
General Function of
a Discrete Random Suppose that X is a discrete random variable with probability distribution f (x). Let
X
Variable Y = h(X) deine a one-to-one transformation between the values of X and Y so that

the equation y = h(x) can be solved uniquely for x in terms of y. Let this solution be
x = u(y). Then the probability mass function of the random variable Y is
u y)
⎡
f Y ( y) = f X ( ⎤ ⎦ (5-30)
⎣
Example 5-34 Function of a Discrete Random Variable Let X be a geometric random variable with
probability distribution
f X ( x) = p(1 − p) x −1 , x = 1 , , …
2
Find the probability distribution of Y = X .
2
Because X $ 0, the transformation is one to one; that is, y = x and x = y. Therefore, Equation 5-30 indicates that
2
the distribution of the random variable Y is
f y ( ) = ( ) = p(1 − p) y −1 , y = 1 , , ,16 , …
9
4
y
f
Y
We now consider the situation in which the random variables are continuous. Let Y = h(X)
with X continuous and the transformation one to one.

General Function of a
Continuous Random Suppose that X is a continuous random variable with probability distribution f (x).
X
Variable The function Y = h(X) is a one-to-one transformation between the values of Y and X,
so that the equation y = h(x) can be uniquely solved for x in terms of y. Let this solu-
tion be x = u(y). The probability distribution of Y is
⎡
u y) ⏐⏐
f Y ( y) = f X ( ⎤ ⎦ J (5-31)
⎣
where J = u′(y) is called the Jacobian of the transformation and the absolute value
of J is used.

Equation 5-31 is shown as follows. Let the function y = h(x) be an increasing function of x. Now
(
u a)
⎡
∫
P Y ≤ a) = P X ≤ ( ⎤ = u a ( ) f X ( )
x dx
⎣
⎦
−∞
If we change the variable of integration from x to y by using x = u(y), we obtain dx = u′(y) dy
and then
(
⎡
P Y ≤ a) = ∫ a f X ( )⎤ ′( )
u y u y dy
⎦
⎣
−∞