Advanced Probability Theory for Biomedical Engineers - John D. Enderle

TRANSFORMATIONS OF RANDOM VARIABLES 73

Drill Problem 6.4.1. Random variables x and y have joint PDF

fx,y (α, β) = e −α−β u(α)u(β).

Random variable z = x and w = xy. Find: (a) fz,w(1, 1), (b) fz,w(−1, 1).

Answers: 0, e −2.

Drill Problem 6.4.2. Random variables x and y have joint PDF

fx,y (α, β) =	4αβ,	0 < α < 1, 0 < β < 1
	0,	otherwise.

Random variable z = x + y and w = y2. Determine: (a) fz,w(1, 1/4), (b) fz(−1/2), (c ) fz(1/2),

(d ) fz(3/2).

Answers: 0, 1/12, 1, 13/12.

Drill Problem 6.4.3. Random variables x and y have joint PDF

fx,y	(α, β)	=	1.2(α2 + β),	0 < α < 1, 0 < β < 1
			0,	otherwise.

Random variable z = x2 y. Determine: (a) fz(−1), (b) fz(1/4), (c ) fz(1/2), (d ) fz(3/2).

Answers: 0.7172, 0, 0, 1.3.

Drill Problem 6.4.4. Random variables x and y have joint PDF

fx,y (α, β) =	2β, 0 < α < 1, 0 < β < 1
		0,	otherwise.
Random variable z = x − y and event		A = {(x, y) : x + y ≤ 1}. Determine: (a) fz| A(−3/

2| A), (b) fz| A(−1/2| A), (c )Fz| A(0| A), (d ) fz| A(1/2| A).

Answers: 0, 3/4, 15/16, 3/16.

6.5SUMMARY

This chapter presented a number of different approaches to find the probability distribution of functions of random variables.

Two methods are presented to find the probability distribution of a function of one random variable, z = g (x). The first, and the most general approach, is called the CDF technique. From the definition of the CDF, we write

Fz(γ ) = P (z ≤ γ ) = P (g (x) ≤ γ ) = P (x A(γ )),

(6.33)

74 ADVANCED PROBABILITY THEORY FOR BIOMEDICAL ENGINEERS

where A(γ ) = {x : g (x) ≤ γ }. By partitioning A(γ ) into disjoint intervals, the CDF Fz may be found using the CDF Fx . The second approach, called the PDF technique, involves evaluating

∞	fx (αi (γ ))
fz(γ ) =		,	(6.34)
	g (1)(αi (γ ))
i=1
where
αi = αi (γ ) = g −1(γ ), i = 1, 2, . . . ,			(6.35)

denote the distinct solutions to g (αi ) = γ . Typically, the PDF technique is much simpler to use than the CDF technique. However, the PDF technique is applicable only when z = g (x) is continuous and does not equal a constant in any interval in which fx is nonzero.

Next, we evaluated the probability distribution of a random variable z = g (x, y) created from jointly distributed random variables x and y using two approaches. The first approach, a CDF technique, involves evaluating

Fz(γ ) = P (z ≤ γ ) = P (g (x, y) ≤ γ ) = P ((x, y) A(γ )),		(6.36)
where
A(γ ) = {(x, y) : g (x, y) ≤ γ }.		(6.37)
The CDF for the RV z can then be found by evaluating the integral
Fz(γ ) =	A(γ ) dFx,y (α, β).	(6.38)

The ease of solution here involves transforming A(γ ) into proper limits of integration. We wish to remind the reader that the special case of a convolution integral is obtained when random variables x and y are independent and z = x + y. The second approach involves introducing an auxiliary random variable and using the PDF technique applied to two functions of two random variables.

To find the joint probability distribution of random variables z = g (x, y) and w = h(x, y) from jointly distributed random variables x and y, a bivariate CDF as well as a joint PDF technique were presented. Using the joint PDF technique, the joint PDF for z and w can be found as

∞	fx,y (αi , βi )
fz,w(γ , ψ ) =		,	(6.39)
	J (αi , βi )
i=1
where
(αi (γ , ψ ), βi (γ , ψ )),	i = 1, 2, . . . ,		(6.40)

TRANSFORMATIONS OF RANDOM VARIABLES 75

are the distinct solutions to the simultaneous equations
	g (αi , βi ) = γ						(6.41)
and
	h(αi , βi ) = ψ.						(6.42)
The Jacobian
		∂g (α, β)		∂g (α, β)
J (α, β) =		∂ α			∂β	,	(6.43)
	∂ h(α, β)		∂ h(α, β)
		∂ α			∂β

where the indicated partial derivatives are assumed to exist. Typically, the most difficult aspect of the joint PDF technique is in solving the simultaneous equations. The joint PDF technique can also be used to find the probability distribution for z = g (x, y) from fx,y by introducing an auxiliary random variable w = h(x, y) to find fz,w, and then integrating to obtain the marginal PDF fz.

6.6PROBLEMS

1.Let random variable x be uniform between 0 and 2 with z = exp(x). Find Fz using the CDF technique.

2.Given

fx	(α)	=	0.5(1 + α),				−1 < α < 1
				0,			otherwise,
and
		z =		x − 1,		x < 0
				x + 1	,	x	>	0	.
Use the CDF technique to find Fz.
3. Suppose random variable x has the CDF
				0,		α < 0
	Fx (α) = α2,					0 ≤ α < 1
				1,		1 ≤ α

and z = exp(−x). Using the CDF technique, find Fz.

76ADVANCED PROBABILITY THEORY FOR BIOMEDICAL ENGINEERS

4.The PDF of random variable x is

	1		e −	1	(α+2)u(α + 2).
fx (α) =				2
		2
Find Fz using the CDF technique when z = |x|.
5. Suppose random variable x has the PDF
fx (α) =		2α, 0 < α < 1
	0,				otherwise.
Random variable z is defined by
−1,					x ≤ −1
z =	x,				−1 < x < 1
	1,				x ≥ 1.

Using the CDF method, determine Fz.

6.Random variable x represents the input to a half-wave rectifier and z represents the output, so that z = u(x). Given that x is uniformly distributed between −2 and 2, find:

(a)E(z) using fx , (b) Fz using the CDF method, (c) fz, (d) E(z) using fz (compare with the answer to part a).

7.Random variable x represents the input to a full-wave rectifier and z represents the output, so that z = |x|. Given that x is uniformly distributed between −2 and 2, find:

(a)E(z) using fx , (b) Fz using the CDF method, (c) fz, (d) E(z) using fz (compare with the answer to part a).

8.Random variable x has the PDF fx (α) = e −α u(α). Find Fz for z = x2 using the CDF method.

9.Given

fx (α) =	a
	1 + α2
and
−1,	x < −1
z = x,	−1 ≤ x ≤ 2
2,	x > 2.
Determine: (a) a, (b) Fz.

TRANSFORMATIONS OF RANDOM VARIABLES 77

10.Random variables x and z are the input and output of a quantizer. The relationship between them is defined by:

0,	x < 0.5
1,	0.5 ≤ x < 1.5
z = 2,	1.5 ≤ x < 2.5
3,	2.5 ≤ x < 3.5
4,	x ≥ 3.5.

Given that the input follows a Gaussian distribution with x G(2.25, 0.49), find fz using the CDF method.

11.Random variable x has the PDF fx (α) = e −α u(α). With z = 100 − 25x, find: (a) Fz using the CDF technique, (b) Fz using the PDF technique.

12.Random variable x has the following PDF

4α3, 0 < α < 1 fx (α) = 0, otherwise.

Find the PDFs for the following random variables: (a) z = x3, (b) z = (x − 1/2)2, (c)

z = (x − 1/4)2.

13.Random variable x has the CDF

0,	α < −1
Fx (α) = (α2 + 2α + 1)/9, −1 ≤ α < 2
1,	2 ≤ α
and
x,	x < −0.5
z = 0,	−0.5 ≤ x ≤ 0
x,	0 < x.

Determine Fz.

14. Suppose
fx	(α)	=	(1 + α2)/6,	−1 < α < 2
			0,	otherwise.

Let z = 1/x2. Use the CDF technique to determine Fz.

78 ADVANCED PROBABILITY THEORY FOR BIOMEDICAL ENGINEERS

		z = g( x )
		1
−2	−1	1	2	3	x

−1
FIGURE 6.12: Plot for Problem 15.
15. Random variable x has the CDF
0,			α < −1
Fx (α) = 0.25(α + 1),			−1 ≤ α < 3
1,			3 ≤ α.

Find the CDF Fz, with RV z = g (x), and g (x) shown in Figure 6.12. Assume that g (x) is a second degree polynomial for x ≥ 0.

16.The voltage x in Figure 6.13 is a random variable which is uniformly distributed from −1 to 2. Find the PDF fz. Assume the diode is ideal.

17.The voltage x in Figure 6.14 is a Gaussian random variable with mean ηx = 0 and standard deviation σx = 3. Find the PDF fz. Assume the diodes are ideal.

18.The voltage x in Figure 6.15 is a Gaussian random variable with mean ηx = 1 and standard deviation σx = 1. Find the PDF fz. Assume the diode and the operational amplifier are ideal.

19.Random variable x is Gaussian with mean ηx = 0 and standard deviation σx = 1. With z = g (x) shown in Figure 6.16, find the PDF fz.

+	1K Ω	+
x	3K Ω	z
−		−

FIGURE 6.13: Circuit for Problem 16.

TRANSFORMATIONS OF RANDOM VARIABLES 79

+	R	+
x		z
	2 V	3 V
−		−

FIGURE 6.14: Circuit for Problem 17.

20. Random variable x has the following PDF

fx	(α)	=	α + 0.5δ(α − 0.5),	0 < α < 1
			0,	otherwise.

Find Fz if z = x2.

21.Random variable x is uniform in the interval 0 to 12. Random variable z = 4x + 2. Find fz using the PDF technique.

22.Find fz if z = 1/x2 and x is uniform on −1 to 2.

23.Random variable x has the PDF

fx	(α)	=	2(α + 1)/9,	−1 < α < 2
			0,	otherwise.

Find the PDF of z = 2x2 using the PDF technique.

24. Let z = cos(x). Find fz if:

(a)	fx	(α)	=	1/π,	|α| < π/2
				0,	otherwise.

1 K Ω

x

5 K Ω

z

FIGURE 6.15: Circuit for Problem 18.

80 ADVANCED PROBABILITY THEORY FOR BIOMEDICAL ENGINEERS

		z = g( x )
		1.4
−2	−1	1	2	x
		−0.5

FIGURE 6.16: Transformation for Problem 19.

(b) fx (α) =	8α/π 2,	0 < α < π/2
	0,	otherwise.
25. Given that x has the CDF
	0,	α < 0
Fx (α) = α,		0 ≤ α < 1
	1,	1 ≤ α.
Find the PDF of z = −2 ln(x) using the PDF technique.
26. Random variable x has the PDF
fx (α) =	2α/9,	0 < α < 3
	0,	otherwise.

Random variable z = (x − 1)2 and event A = {x : x ≥ 1/2}. Find the PDF of random variable z, given event A.

27. Random variable x is uniform between −1 and 1. Random variable

z =	x2,		x < 0
	x	,	x ≥ 0	.

Using the PDF technique, find fz.

28.A voltage v is a Gaussian random variable with ηv = 0 and σv = 2. Random variable w = v2/ R represents the power dissipated in a resistor of R with v volts across the resistor. Find (a) fw, (b) fw| A if A = {v ≥ 0}.

TRANSFORMATIONS OF RANDOM VARIABLES 81

29. Random variable x has an exponential distribution with mean one. If the PDF technique to determine: (a) fz| A(γ | A) if A = {x : x > 2}, (b)

B= {z : z < 1/2}.

30.Find fz if z = 1/x and

1

fx (α) = π (α2 + 1) .

z = e −x , use fz|B (γ |B) if

31. Given that random variable x has the PDF

α, 0 < α < 1 fx (α) = 2 − α, 1 < α < 2 0, otherwise.

Using the PDF technique, find the PDF of z = x2.

32. Suppose

2α, 0 < α < 1 fx (α) = 0, otherwise,

and z = 8x3. Determine fz using the PDF technique. 33. Given

9α2,	0 < α < 0.5
fx (α) = 3(1 − α2),	0.5 < α < 1
0,	otherwise,

and z = −2 ln(x). Determine: (a) fz, (b) Fz, (c) E(z), (d) E(z2).

34.Using the PDF technique, find fz if z = |x| and x is a standard Gaussian random variable.

35.Suppose RV x has PDF fx (α) = 0.5α(u(α) − u(α − 2)). Find a transformation g such that z = g (x) has PDF

		fz(γ ) = c γ 2(u(γ ) − u(γ − 1)).
36. Let
fx	(α)	=	0.75(1 − α2),	−1 < α < 1
			0,	otherwise,

and z = x2. Determine fz using the PDF technique.

82ADVANCED PROBABILITY THEORY FOR BIOMEDICAL ENGINEERS

37.Find fz if z = sin(x) and

fx	(α)	=	1/(2π ),	0 ≤ α < 2π
			0,	otherwise.

38. Random variable x has the PDF

0.5,

−1 < α < 0

fx (α) = 0.5 − 0.25α,

0 < α < 2

0,

otherwise.

(a) Find the transformation z = g (x) so that

1 − 0.25γ ,

0 < γ < 1

fz(γ ) = 0.5 − 0.25γ ,

1 < γ < 2

0,

otherwise.

(b) Determine fz if z = (2x + 2)u(x).

39. Let random variable x have the PDF

fx

(α)

=

1/12,

−2 < α < 1

1/4,

1 < α < 2;

in addition, P (x = −2) = P (x = 2) = 0.25. If z = 1/x, find fz.

40. Random variable x has PDF

1

1

fx (α) =

δ(α +

1) +

δ(α)

+ (u(α) − u(α − 2)).

4

4

Random variable z = g (x), with g shown in Figure 6.17. Find the PDF fz.

41. Random variable x has PDF

α2

fx (α) =

(u(α + 1) − u(α − 2)).

3

2

z = g( x )

1

−

1

0

1

2 x

FIGURE 6.17: Transformation for Problem 40.