Section 3.7
(phone calls)
(one hour)
Suppose the number of occurrences in a “unit” interval follows a Poisson
distribution with mean .
(10 phone calls per hour)
Recall that for w > 0,
P(interval length to obtain the first occurrence  w) =
1 – P(interval length to obtain the first occurrence > w) =
1 – P(no occurrences in interval of length w) = 1 – e–w .
Consider a random variable
W = interval length to obtain the first occurrence
with distribution function G(w) = P(W  w) = 1 – e–H(w) for w > 0 , where
w
H(w) =
0
(t) dt
.
If the function (t) is a constant, say , then H(w) = w , so that the
p.d.f. of W is
g(w) = e–w if w > 0 ,
1
) distribution.
which is the p.d.f. for an exponential(—

A non-constant function (t) implies that the mean of the Poisson
process changes over an interval. For instance, the number of
occurrences in a “unit” interval may steadily increase (or decrease).
(phone calls per hour)
(increase as time goes on)
We observe that in general, H /(w) = (w) , so that the p.d.f. of W is
g(w) = H /(w)e–H(w) = (w)e–H(w) if w > 0 .
When W represents length of life (for an item or living organism), (w)
is called the failure rate or force of mortality, and we find that
Consider (t) = 10 versus
g(w)
g(w)
= ———— .
(t) = 5t / 2 when W =
(w) = ——
–H(w)
e
1 – G(w)
hours until first phone call
w 
If H(w) = — for  > 0 and  > 0, we say W has a Weibull distribution.

For W = hours until first phone call, compare
P(W > 1/10) when (t) = 10 and (t) = 5t / 2.
When (t) = 10, the p.d.f. of W is 10e–10w if w > 0 .
We can say that W has a Weibull
distribution with  = 1 and  = 1/10
P(W > 1/10) = e–1 = 0.3679
2
–(5/4)w
(5w / 2)e
if w > 0 .
When (t) = 5t / 2, the p.d.f. of W is
We can say that W has a Weibull
distribution with  = 2 and  = 2/5
P(W > 1/10) = e–1/80 = 0. 9876
1. A random variable W has p.d.f. g(w) = H /(w)e–H(w) if w > 0 , where
H(w) = 243w5/3. Find the values of  and  so that we may write
w 
H(w) = — in order to demonstrate that W has a Weibull

distribution.
 = 5/3
 = 1/27
2. Suppose the random variable W represents length of life and has
p.d.f. g(w) = H /(w)e–H(w) if w > 0 , with an exponential force of
mortality
g(w)
(w) = ———— = aebw
for a > 0 and b > 0.
1 – G(w)
(a) Find the distribution function for W.
w
w
H(w) =
0
bw – a
ae
aebt dt = ———
b
(t) dt =
0
The distribution function for W is G(w) =
0
if w  0
a – aebw
1 – exp ———
b
if 0 < w
(b) Find the p.d.f. for W (known as Gompertz law).
bw
a
–
ae
The p.d.f. for W is g(w) = aebw exp ———
b
if w > 0 .
Let X be a random variable with c.d.f. F(x). In general, for any value v,
P(X = v) = P(X  v)  P(X < v) = P(X  v)  Lim P(X  x) =
x  v
F(v)  Lim F(x) .
x  v
Example. Let X be a random variable with
c.d.f.
F(x) =
Suppose v is a value less than 5, say 4.8.
P(X = v) = P(X = 4.8) =
F(4.8)  Lim F(x) = 0  0 = 0
0
1/4
(x  5) / 4
1
if
if
if
if
x<5
5x<6
6x<9
9x
1
x   4.8
3/4
1/2
1/4
5
6
7
8
9
0
Example. Let X be a random variable with
c.d.f.
F(x) =
0
1/4
(x  5) / 4
1
if
if
if
if
Suppose v is 5.
P(X = v) = P(X = 5) =
F(5)  Lim F(x) = 1/4  0 = 1/4
x<5
5x<6
6x<9
9x
1
3/4
x  5
1/2
1/4
5
6
7
8
Suppose v is a value greater than 5 but less than 6, say 5.8.
P(X = v) = P(X = 5.8) =
F(5.8)  Lim F(x) = 1/4  1/4 = 0
x   5.8
9
0
Example. Let X be a random variable with
c.d.f.
F(x) =
0
1/4
(x  5) / 4
1
if
if
if
if
x<5
5x<6
6x<9
9x
Suppose v is a value such that 6  v < 9, say 7.5.
P(X = v) = P(X = 7.5) =
F(7.5)  Lim F(x) =
3/4
5 / 8  (7.5  5) / 4 = 0
1/2
1
x   7.5
1/4
5
6
7
8
Suppose v is a value greater than or equal to 9, say 9.8.
P(X = v) = P(X = 9.8) =
F(9.8)  Lim F(x) = 1  1 = 0
x   9.8
9
0
Example. Let X be a random variable with
c.d.f.
F(x) =
0
1/4
(x  5) / 4
1
if
if
if
if
Consequently, we see that
if v  5, then P(X = v) = 0 but P(X = 5) = 1 / 4
x<5
5x<6
6x<9
9x
1
3/4
1/2
1/4
5
6
7
8
We also see that P(6 < X  9) = P(X  9)  P(X  6) =
F(9)  F(6) = 1  1 / 4 = 3 / 4
We see then that the space of X is {5}  {x | 6 < x < 9}
9
0
We see then that the space of X is {5}  {x | 6 < x < 9}
The random variable X is not a discrete type random variable and is not
a continuous type random variable; X is called a random variable of the
mixed type.
What kind of random variable X could have this space?
A dart is thrown at a circular board of radius 9 Yellow Area Score = 6
inches. If the dart lands in the area between 3
and 9 inches from the center, a score of 6 is
assigned; if the dart lands within 3 inches of
the center, then a score of 9  x is assigned
where x is the distance from the center.
Blue Area Score = 9  x where
x = inches from the center.
Let F(x) be the distribution function for a random variable X, then
(1) F(x) must be an increasing function, (2) as x  –, F(x)  0 ,
and (3) as x  +, F(x)  1 .
If there exists a discrete set of points at which F(x) is discontinuous
with F /(x) = 0 at all other points, then X must be a discrete type
random variable whose space is the set of points of discontinuity. (The
probability of observing a given value in the space of X is equal to the
“length of the jump in F(x)” at the given value.)
If F(x) is continuous everywhere, then X is a continuous type random
variable whose space is the set of points where F /(x) > 0 (i.e., the set of
points where F(x) is strictly increasing).
If there exists a discrete set of points at which F(x) is discontinuous and
also one or more intervals where F /(x) > 0, then X is called a random
variable of the mixed type. Such a random variable is neither of the
discrete type or the continuous type and has no p.m.f. or p.d.f. The
space of X consists of the union of the set of points of discontinuity and
the intervals where F /(x) > 0.
3. For each distribution function, (1) sketch a graph, and (2) either find
the corresponding p.m.f. or p.d.f., or state why no p.m.f. or p.d.f.
can be found.
1
(a) X1 has distribution function F1(x) = 1 – ——— if –  < x <  .
(1 + ex)
1
1/2
0
X1 is a continuous type random variable with p.d.f. f1(x) = F1(x) =
ex
———
(1 + ex)2
if –  < x <  .
if x < – 3
0
(b) X2 has distribution function F2(x) =
(x3 + 27) / 35 if – 3  x < 2
if 2  x
1
1
–3
0
2
X2 is a continuous type random variable with p.d.f. f2(x) = F2(x) =
3x2
—
35
if – 3  x < 2 .
(c) X3 has distribution function
F3(x) =
11/12 1
0
1/6
1/3
2/3
11/12
1
if x < – 2
if – 2  x < 0
if 0  x < 2
if 2  x < 4
if 4  x < 6
if 6  x
2/3
1/3
1/6
–2
0
2
4
6
X3 is a discrete type random variable. The space
of X is {– 2, 0, 2, 4, 6}. The p.m.f. of X is f3(x) =
1/6
1/6
1/3
1/4
1/12
if x = – 2
if x = 0
if x = 2
if x = 4
if x = 6
(d) X4 has distribution function F4(x) =
0
if x < – 2
(x + 2) / 12 if – 2  x < 0
(2 + 3x ) / 12 if 0  x < 4
11 / 12 if 4  x < 5
1
if 5  x
11/12 1
2/3
1/6
–2
0
2
4
5
X4 has a distribution of the mixed type. X4 has no p.m.f. or p.d.f.
The space of X4 is {x | – 2 < x < 4}  {4,5}.
(e) X5 has distribution function F5(x) =
0
if x < – 2
(x + 2) / 12 if – 2  x < 0
(5 + 3x ) / 12 if 0  x < 4
11 / 12 if 4  x < 5
1
if 5  x
11/12 1
5/12
1/6
–2
0
2
4
5
X5 has a distribution of the mixed type. X5 has no p.m.f. or p.d.f.
The space of X5 is {x | – 2 < x < 0}  {0}  {x | 0 < x < 4}  {5}.
4. An ordinary, fair, six-sided die is rolled, and
each of the two spinners displayed on the right
are spun, with the random variables Y1 and Y2
respectively being set equal to the numbers
selected from Spinner #1 and Spinner #2.
Note that Y1 and Y2 have respective U(2,2)
and U(3,5) distributions. The random variable
X is defined as follows:
X=
Y1
if the die roll results in 1
2
if the die roll results in 2
Y2
if the die roll results in 3, 4, or 5
6
if the die roll results in 6
(a) Find the distribution function of X.
Spinner #1
2
1
1
0
Spinner #2
3
4.5
3.5
4
The space of X is {x | – 2 < x < 2}  {2}  {x | 3 < x < 5}  {6}
P(X  x) = F(x) = 0 if x < – 2
In order for – 2 < X < 2, the die roll must result in 1, so that X = Y1 .
For – 2 < x < 2,
P(X  x) = P({die roll results in 1}  {Y1  x}) =
P(die roll results in 1) P(Y1  x) = (1 / 6) (2 + x) / 4
2+x
P(X  x) = F(x) = ——— if – 2  x < 2
24
In order for X = 2, the die roll must result in 2, and P(X = 2) = 1 / 6 .
We see then that P(X  x) = 1 / 3 for 2  x < 3.
0
2+x
———
24
F(x) =
1
—
3
3x  5
———
12
if x < – 2
if – 2  x < 2
if 2  x < 3
if 3  x < 5
In order for 3 < X < 5, the die roll must result
in 3, 4, or 5, so that X = Y2 . For 3 < x < 5,
P(X  x) = P({X < 3}  {3  X  x}) =
P(X < 3) + P(3  X  x) = 1 / 3 + P(3  X  x) =
1 / 3 + P({die roll results in 3, 4, or 5}  {Y2  x}) =
1 / 3 + P(die roll results in 3, 4, or 5) P(Y2  x) = 1 / 3 + (1/2) (x  3) / 2
It is not possible that 5 < X < 6, and
in order that X = 6, the die roll must
result in 6, and P(X = 6) = 1 / 6 .
0
2+x
———
24
We see then that P(X  x) = 5 / 6 for
5  x < 6.
F(x) =
1
—
3
3x  5
———
12
5
—
6
1
if x < – 2
if – 2  x < 2
if 2  x < 3
if 3  x < 5
if 5  x < 6
if 6  x
(b) Graph the distribution function of X.
0
2+x
———
24
F(x) =
1
5/6
5
—
6
1/3
1
1/6
–2
1
—
3
3x  5
———
12
0
2
4
6
if x < – 2
if – 2  x < 2
if 2  x < 3
if 3  x < 5
if 5  x < 6
if 6  x
When X is a random variable of the mixed type, the expected value of a
function u(X) is calculated by applying the following rules:
(1) Multiply each value u(x) corresponding to the discrete list of values
for x at which F(x) is discontinuous by its probability (i.e., by the
“length of the jump in F(x)” at x).
(2) Integrate u(x)F /(x) over each interval where F /(x) > 0.
(3) Sum the results obtained from (1) and (2).
(c) Find E(X).
0
2+x
———
24
F(x) =
1
E(X) = (2) —
6
+
1
x — dx
24
–2
5
—
6
1
(6) — +
6
2
5
1
1
x — dx
4
+
3
1
—
3
3x  5
———
12
=
1
— + 1 +
3
if x < – 2
if – 2  x < 2
if 2  x < 3
if 3  x < 5
if 5  x < 6
if 6  x
0 + 2 =
10
—
3
5. For each random variable, graph the distribution function, and find
the expected value.
(a) The random variable W = "the length of time in hours a brand W
light bulb will burn" has distribution function
0
if w < 0
F(w) =
1 – e–w/300 if 0  w
F(w)
1
0
Note that W is a continuous type random variable with p.d.f.
e–w/300
f(w) = F (w) = —— if 0 < w < 
300
(an exponential(300) p.d.f.). E(W) = 300
(b) The random variable X = "the length of time in hours a brand X
light bulb will burn" has distribution function
G(x) =
G(x)
1
0
if x < 0
1 – (9/10)e–x/300 if 0  x
1/10
0
X has a distribution of the mixed type. X has no p.m.f. or p.d.f.


1
9e–x/300
9
e–x/300
9
x —— dx = — (300) =
E(X) = (0) — + x ——— dx = —
10
3000
10
300
10
270
0
0
(c) The random variable Y = "the length of time in hours a brand Y
light bulb will burn" has distribution function
0
H(y) =
if y < 0
1 – (9/10)e–y/300 if 0  y < 120
1
if 120  y
H(y)
1
1 – 0.9e–2/5
1/10
0
120
Y has a distribution of the mixed type. Y has no p.m.f. or p.d.f.
120
3e–y/300
y ——— dy + (120) (0.9e–2/5) =
1000
1
E(Y) = (0) — +
10
0
120
– 3(900 + 3y) e–y/300
————————
10
+ 108 e–2/5 =
y=0
– 3(900 + 360) e–2/5
2700
———————— + —— + 108 e–2/5 = 270 – 270 e–2/5
10
10