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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