Page 143 - Probability, Random Variables and Random Processes

CHAP. 4) FUNCTIONS OF RANDOM VARIABLES, EXPECTATION, LIMlT THEOREMS 135

4.11. Let Y = tan X. Find the pdf of Y if X is a uniform r.v. over (-7112, 71/2).

The cdf of X is [Eq. (2.4511
x s -. z/2

(X + n/2) -n/2 < x < 42
x 2 42
Now Fy(y) = P(Y I y) = P(tan X 5 y) = P(X I tan- ' y)
-a <y< co

Then the pdf of Y is given by
d 1
fy(y)=-Fy(y)=- -W<Y<W
dy 41 + Y*)
Note that the r.v. Y is a Cauchy r.v. with parameter 1.

4.12. Let X be a continuous r.v. with the cdf FX(x). Let Y = Fx(X). Show that Y is a uniform r.v. over
(0, 1).

Notice from the properties of a cdf that y = FX(x) is a monotonically nondecreasing function. Since
0 I FX(x) I for all real x, y takes on values only on the interval (0, 1). Using Eq. (4.64) (Prob. 4.2), we
1
have

Hence, Y is a uniform r.v. over (0, 1).

4.13. Let Y be a uniform r.v. over (0, 1). Let F(x) be a function which has the properties of the cdf of a
continuous r.v. with F(a) = 0, F(b) = 1, and F(x) strictly increasing for a < x < b, where a and b
could be - co and oo, respectively. Let X = F-'(Y). Show that the cdf of X is F(x).

FX(x) = P(X 5 X) = P[F-'(Y) 5 X]
Since F(x) is strictly increasing, F-'(Y) 5 x is equivalent to Y I: F(x), and hence

FX(x) = P(X I X) = P[Y S F(x)]
Now Y is a uniform r.v. over (0, l), and by Eq. (2.45),
FAy)=P(YIy)=y O<y<l

and accordingly,
Fx(x) = P(X 5 x) = PLY 5 F(x)] = F(x) 0 < F(x) < 1
Note that this problem is the converse of Prob. 4.12.