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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