Prokis Solution Manual

4) 1 < X < 1) = FX (1−) − 1 = 1 − 1 = 1
P( FX ( 2 ) 2 4 4
5)
2
Problem 4.9
1) P (X > 2) = 1 − P (X ≤ 2) = 1 − FX (2) = 1 − 1 = 0

x < −1 ⇒ FX (x) = 0

−1 ≤ x ≤ 0 ⇒ FX (x) = x (v + 1)dv = ( 1 v2 + v) x = 1 x2 + x + 1
2 −1 2 2
−1

0≤x≤1 ⇒ FX (x) = 0 x + 1)dv = − 1 x2 + x + 1

(v + 1)dv + (−v
0 22
−1

1 ≤ x ⇒ FX (x) = 1

2)

p(X > 1 = 1 − 1 = 1 − 7 = 1
) FX ( 2 ) 8 8

2

3)

p(X > 0|X < 1) = p(X > 0, X < 1 ) = FX ( 1 ) − FX (0) = 3
2 2 2 7

p(X < 1 ) 1 − p(X > 1 )
2 2

4) We find first the CDF

FX (x|X > 1) = p(X ≤ x|X > 1) = p(X ≤ x, X > 1 )
2 2 2

p(X > 1 )
2

If x ≤ 1 then p(X ≤ x|X > 1 ) = 0 since the events E1 = {X ≤ 1 } and E1 = {X > 1 } are disjoint.
2 2 2 2
1 ≤ x|X 1 − 1
If x > 2 then p(X > 2 ) = FX (x) FX ( 2 ) so that

FX (x|X > 1 = FX (x) − FX ( 1 )
) 2

2 1 − FX ( 1 )
2

Differentiating this equation with respect to x we obtain

1 fX (x) x > 1
) 1 2
fX (x|X > = 1−FX ( 2 )
2
0 x ≤ 1
2

5)

∞

E[X|X > 1/2] = xfX (x|X > 1/2)dx

−∞

= 1∞ xfX (x)dx

1 − FX (1/2) 1
2

= 8 ∞ x(−x + 1)dx = 8(− 1 x3 + 1 x2) 1
1 3 21

22

2
=

3

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Problem 4.10

1) The random variable X is Gaussian with zero mean and variance σ2 = 10−8. Thus p(X > x) =

Q( x ) and
σ

p(X > 10−4) = Q 10−4 = Q(1) = .159
10−4

p(X > 4 × 10−4) = Q 4 × 10−4 = Q(4) = 3.17 × 10−5
10−4

p(−2 × 10−4 < X ≤ 10−4) = 1 − Q(1) − Q(2) = .8182

2)

p(X > 10−4|X > 0) = p(X > 10−4, X > 0) = p(X > 10−4) = .159 = .318
p(X > 0) p(X > 0) .5

3) y = g(x) = xu(x). Clearly fY (y) = 0 and FY (y) = 0 for y < 0. If y > 0, then the equation
y = xu(x) has a unique solution x1 = y. Hence, FY (y) = FX (y) and fY (y) = fX (y) for y > 0.
FY (y) is discontinuous at y = 0 and the jump of the discontinuity equals FX (0).

FY (0+) − FY (0−) = FX (0) = 1
2

In summary the PDF fY (y) equals

fY (y) = fX (y)u(y) + 1
δ(y)
2

The general expression for finding fY (y) can not be used because g(x) is constant for some interval
so that there is an uncountable number of solutions for x in this interval.

4)

∞

E[Y ] = yfY (y)dy

−∞

∞1
= y fX (y)u(y) + δ(y) dy
2
−∞

= √1 ∞ ye− y2 √σ
2σ2
dy =
2πσ2 0 2π

5) y = g(x) = |x|. For a given y > 0 there are two solutions to the equation y = g(x) = |x|, that is
x1,2 = ±y. Hence for y > 0

fY (y) = fX (x1) + fX (x2) = fX (y) + fX (−y)
|sgn(x1)| |sgn(x2)|

= √2 e− y2
2σ2

2πσ2

For y < 0 there are no solutions to the equation y = |x| and fY (y) = 0.

E[Y ] = √ 2 ∞ ye− y2 √2σ
2σ2
dy =
2πσ2 0 2π

75

Problem 4.11
1) y = g(x) = ax2. Assume without loss of generality that a > 0. Then, if y < 0 the equation
y = ax2 has no real solutions and fY (y) = 0. If y > 0 there are two solutions to the system, namely

x1,2 = y/a. Hence,

fY (y) = fX (x1) + fX (x2)
|g (x1)| |g (x2)|

= fX ( y/a) + fX (− y/a)
2a y/a 2a y/a

= √ay √1 e− y
2π 2aσ2

σ 2

2) The equation y = g(x) has no solutions if y < −b. Thus FY (y) and fY (y) are zero for y < −b. If
−b ≤ y ≤ b, then for a fixed y, g(x) < y if x < y; hence FY (y) = FX (y). If y > b then g(x) ≤ b < y
for every x; hence FY (y) = 1. At the points y = ±b, FY (y) is discontinuous and the discontinuities
equal to

FY (−b+) − FY (−b−) = FX (−b)

and
FY (b+) − FY (b−) = 1 − FX (b)

The PDF of y = g(x) is

fY (y) = FX (−b)δ(y + b) + (1 − FX (b))δ(y − b) + fX (y)[u−1(y + b) − u−1(y − b)]

= Q b (δ(y + b) + δ(y − b)) + √1 e− y2 [u−1(y + b) − u−1(y − b)]
σ 2πσ2 2σ2

3) In the case of the hard limiter

1
p(Y = b) = p(X < 0) = FX (0) = 2

p(Y = a) = p(X > 0) = 1 − FX (0) = 1
2

Thus FY (y) is a staircase function and

fY (y) = FX (0)δ(y − b) + (1 − FX (0))δ(y − a)

4) The random variable y = g(x) takes the values yn = xn with probability

p(Y = yn) = p(an ≤ X ≤ an+1) = FX (an+1) − FX (an)

Thus, FY (y) is a staircase function with FY (y) = 0 if y < x1 and FY (y) = 1 if y > xN . The PDF
is a sequence of impulse functions, that is

N

fY (y) = [FX (ai+1) − FX (ai)] δ(y − xi)

i=1

= N Q ai − Q ai+1 δ(y − xi)
i=1 σ σ

76