3 Random Variable Models

Section 3.1: Probability Distribution Function

3.1. Can the following functions in Figures 3.41 (a), (b), (c) represent possible probability distribution

functions? Explain.

Figure 3.41: Possible probability distribution functions.

Solution: Figures (b) and (c) can be possible distributions, but Figure (a) cannot be because F (x) cannot

decrease in value as x increases.

16

3.2. Show that the expected value of a constant equals that constant.

Solution: Let the random variable, X, take a constant value, c. Then, the probability density function is

the Dirac delta function,

f

X

(x) = δ (x − c) .

Then, the expected value is given by

E {X} =

∞

−∞

f

X

(x) xdx

=

∞

−∞

δ (x −c) xdx

= c.

17

3.3. Random variable X is mixed, containing continuous ranges as well as discontinuities. Its cumulative

probability distribution function is

F

X

(x) =

0, x < 0

x/2, 0 ≤ x < 1

2/3, 1 ≤ x < 2

11/12, 2 ≤ x < 3

1, 3 ≤ x.

Sketch the distribution function. Calculate the following probabilities: (a) Pr(X < 3), (b) Pr(X = 1),

(c) Pr(X > 1/2), (d) Pr(2 < X ≤ 4).

Solution:

(a) Pr(X < 3) = F

X

(3

−

) = 11/12

(b) Pr(X = 1) = F

X

(x = 1

+

) −F

X

x = 1

−1

= 2/3 − 1/2 = 1/6

(c) Pr(X > 1/2) = 1 −Pr (X ≤ 1/2) = 1 −F

X

(x = 1/2) = 1 − (1/2)/2 = 3/4

(d) Pr(2 < X ≤ 4) = Pr (X ≤ 4) − Pr (X ≤ 2) = 1 − 11/12 = 1/12

Figure 2: The cumulative density function for Problem 3.3

18

3.4. Suppose X is distributed as indicated in Figure 3.42. All lines are straight except for the exponential

curve in the range 1 < x < 3.

Figure 3.42: Generic probability distribution function F

X

(x).

Express F

X

(x) algebraically and then ﬁnd the following probabilities:

(a) Pr(X = 1/3) (b) Pr(X = 1) (c) Pr(X < 1/3)

(d) Pr(X 1/3) (e) Pr(X < 1) (f) Pr(X ≤ 1)

(g) Pr(1 < X ≤ 2) (h) Pr(1 ≤ X ≤ 2) (i) Pr(X = 1 or 1/4 < X < 1/2).

Solution: F

X

(x) is a straight line except between 1 < x < 3, where it may be assumed to be parabolic.

Then,

F

X

(x) =

1

4

x for 0 ≤ x < 1

1

16

(x − 1)

2

+

1

2

1

8

x +

3

8

1

for 0 ≤ x < 1

for 1 ≤ x < 3

for 3 ≤ x < 5

for x ≥ 5

From the ﬁgure we ﬁnd:

(a) Pr(X = 1/3) = 0,

(b) Pr(X = 1) = 0.5 − 0.25 = 0.25,

(c) Pr(X < 1/3) =

1

3

×0.25 = 0.083,

(d) Pr(X 1/3) = 0

(e) Pr(X < 1) = 0.25,

(f) Pr(X ≤ 1) = 0.5,

(g) Pr(1 < X ≤ 2) = Pr(X ≤ 2) −Pr(X ≤ 1) = F

X

(2) − F

X

(1

+

) = (1/16 + 1/2) −1/2 = 1/16,

(h) Pr(1 ≤ X ≤ 2) = Pr(X ≤ 2) −Pr(X < 1) = F

X

(2) − F

X

(1

−

) = (1/16 + 1/2) −1/4 = 5/16

(i) Pr(X = 1 or 1/4 < X < 1/2) = Pr (X = 1) + Pr(X < 1/2) −Pr(X ≤ 1/4) = 0 + F

X

(1/2) −F

X

(1/4) =

1/8 −1/16 = 1/16.

19

Section 3.2: Probability Density Function

3.5. (a) Verify and sketch the following probability density function,

f(x) =

1

2

−

1

4

|x − 3|, 1 ≤ x ≤ 5

0, otherwise.

(b) Find and sketch F (x).

Solution: The probability density function is shown below. It is a valid probability density function because

f (x) is non-negative for all valid x and the area under f (x) is 1. This can be seen in Figure 3(a). To ﬁnd

the functional form of the cumulative density function, we can express the probability density as

f(x) =

1

4

x −

1

4

1 ≤ x < 3

−

1

4

x +

5

4

3 ≤ x ≤ 5

0, otherwise.

Then, the cumulative distribution is the area under the probability density function and is given by

F (x) =

x

0

f (ˆx) dˆx =

x

1

1

4

ˆx −

1

4

dˆx =

1

8

x

2

−

1

4

x +

1

8

1 ≤ x < 3

x

3

−

1

4

x +

5

4

dˆx +

3

1

1

4

ˆx −

1

4

dˆx = −

1

8

x

2

+

5

4

x −

17

8

3 ≤ x < 5

1, x ≥ 5.

.

These are two parabolas with vertices at (1, 0) and (5, 1) as shown in Figure 3(b).

Figure 3: Problem 3.5

20

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