Download MTH/STA 561 UNIFORM PROBABILITY DISTRIBUTION Perhaps

MTH/STA 561
UNIFORM PROBABILITY DISTRIBUTION
Perhaps the simplest possible continuous probability distribution is the uniform distribution. We say that a continuous random variable Y is uniformly distributed over the interval
( ; ) if the probability that the observed value for Y falls in any subinterval of length y, in
( ; ), is proportional to y. The probability that the observed value for Y is smaller than
or greater than is zero. Thus, the density for Y is positive only for values between
and , and if the probability that the observed value for Y lies in an interval is proportional
to the length of that interval, the density function f (y) must be constant for < y < .
That is f (y) = c for < y < . Since
1=
it must be true that c = 1= (
Z
f (y) dy =
Z
cdy = c (
);
). Thus, we have established the following distribution.
De…nition 1. A random variable Y is said to have a continuous uniform probability
distribution on the interval ( ; ) if the probability density function of Y is given by
f (y; ; ) =
1= (
0
)
for < y <
elsewhere.
Any continuous random variable that follows a uniform distribution is referred to as the
uniform random variable.
The constants, such as and , that determine the speci…c form of a probability density
function are called parameters of the probability density function. Here we employ the symbol f (y; ; ) to emphasize the fact that the uniform distribution depends upon parameters
and . Some continuous random variables in the physical, management, and biological
sciences have approximately uniform probability distributions. For example, suppose that
we are counting events that have a Poisson distribution, such as telephone calls coming into
a switchboard. If it is known that exactly one such event has occurred in a given interval,
1
say (0; t), then the actual time of occurrence is distributed uniformly over this interval.
Example 1.
Suppose that arrivals of customers at a certain checkout counter follow
a Poisson distribution. It is known that, during a given 30-minute period, one customer
arrived at the counter. What is the probability that the customer arrived during the last 5
minutes of the 30-minute period?
Solution. As mentioned above, the actual time T of arrival will follow a uniform
distribution over the interval (0; 30); that is,
f (t; 0; 30) =
1
30
for 0 t 30
elsewhere.
0
Then the desired probability is
P f25
T
30g =
Z30
1
30 25
5
1
dt =
=
= :
30
30
30
6
25
Theorem 1.
with parameters
Proof. For
The cumulative distribution function for the uniform random variable Y
and is given by
8
for y
< 0
y
for < y <
F (y; ; ) =
:
1
for y
<y< ,
F (y; ; ) =
Zy
1
dt = 0 +
t
t=y
=
y
:
t=
The graph of the distribution function for uniform random variable look like
2
Example 2.
The distribution function of the uniform random variable Y as de…ned
in Example 1 is given by
8
for y 0
< 0
y
for 0 < y < 30
F (y; 0; 30) =
: 30
1
for y 30
Theorem 2.
and are
The mean and variance of the uniform random variable Y with parameters
+
(
)2
2
and
= V ar (Y ) =
;
2
12
respectively. The moment-generating function is
= E (Y ) =
et
t(
mY (t) =
et
for t 6= 0
)
Proof. By de…nition,
E (Y ) =
Z
(
=
1
y
dy =
)( + )
=
2(
)
y=
y2
2
1
2
=
y=
2
2(
)
+
2
and
E Y
2
Z
=
y
1
2
) 2+
+
3(
)
(
=
dy =
y=
y3
3
1
=
y=
2
2
3
=
+
3
3(
+
)
2
:
3
Thus,
2
V ar (Y ) = E Y
=
=
2
4
2
2
[E (Y )] =
+
+
+
12
)2
(
12
2
+
2
3
3 ( + )2
2
2
+
2
2 +
12
=
2
:
It is also easy to see that
tY
mY (t) = E e
=
Z
ty
e
1
dy =
3
1
ety
t
y=
=
y=
et
t(
et
)
:
Example 3.
Example 1 are
The mean and variance of the uniform random variable Y as de…ned in
(30 0)2
0 + 30
2
=
= 15 and
=
= 75;
2
12
respectively, whereas the moment-generating function is
mY (t) =
e30t 1
30t
for t 6= 0
Di¤erentiating mY (t) with respect to t, we get
d
30te30t e30t + 1
mY (t) =
dt
30t2
and
d2
t2 [30 (30te30t + e30t ) 30e30t ] 2t (30te30t
m
(t)
=
Y
dt2
30t4
3 30t
2 30t
900t e
60t e + 2te30t 2t
=
30t4
2 30t
900t e
60te30t + 2e30t 2
=
30t3
e30t + 1)
By L’Hôpital’s rule, we obtain
d
30te30t e30t + 1
mY (t)
= lim
t!0
dt
30t2
t=0
30 (30te30t + e30t ) 30e30t
= lim
t!0
60t
900te30t
30e30t
30
= lim
= lim
=
= 15
t!0
t!0
60t
2
2
E (Y ) =
and
E Y2
d2
900t2 e30t 60te30t + 2e30t 2
m
(t)
=
lim
Y
t!0
dt2
30t3
t=0
900 (30t2 e30t + 2te30t ) 60 (30te30t + e30t ) + 60e30t
= lim
t!0
90t2
2 30t
27000t e
= lim
= lim 300te30t = 300
t!0
t!0
90t2
=
Hence,
V ar (Y ) = E Y 2
[E (Y )]2 = 300
4
152 = 75: