Families of Continuous Distributions - University of Arizona
Families of Continuous Distributions
October 6, 2009
We shall, in general, denote the density of a parametric family of discrete distributions by fX (x|) for the distribution depending on the parameter . Some of the mystery surrounding these densities will be solved when we begin to look at multiple random variables.
1 Uniform Distributions
A continuous uniform distribution on [a, b] has density
1
fX (x|a, b) =
b-a
0
if a x b, otherwise.
a+b
EX =
,
2
(b - a)2
Var(X) =
,
12
MX (t)
=
(b
1 (etb - a)t
-
eta).
2 Exponential Distributions
An exponential distribution with parameter has density
fX (x|a, b) =
1
e-t/
0
if x 0, otherwise.
EX = ,
Var(X) = 2,
1
MX (t)
=
1-
. t
The exponential distribution also has the memoryless property. In symbols,
P {X > t + s|X > t} = P {X > s}.
The probability that an individual survives s additional time units does not depend on the present age of the individual.
Exercise 1. Prove the memoryless property of the exponential random variable.
The double exponential distribution can be obtained by shifting and symmetrizing the exponential distribution. The density function is
1
|x - ?|
fX (x|?, ) = 2 exp - ,
By symmetry EX = ?. Check that Var(X) = 22.
x R.
1
3 Gamma Distributions
The Gamma function is defined for > 0 by
() =
te-t
dt .
0
t
Using integration by parts, one can show that ( + 1) = (). Note that (1) = 1. For the case that is a non-negative integer, we can use induction to see that ( + 1) = !
Because the integrand in the gamma function is non-negative, we have a density function for a continuous random variable,
fT (t|) =
1 ()
t-1e-t
0
Let X = T to obtain the two parameter family
for 0 t, otherwise
fX (x|, ) =
1 ()
x-1
e-x/
0
for 0 x, otherwise.
The gamma densities with = 1, 2, ? ? ? , 6, and = 1 are graphed using R in Figure 1 using
> curve(dgamma(x,1,1),0,10) > for (i in 2:6){curve(dgamma(x,i,1),0,10,add=TRUE)}
The parameter determine the sharpness of the peak and is called the shapeness parameter. The parameter determine the spread of the distribution and is called the scale parameter.
Thus,
1 ET =
te-t dt
=
( + 1)
=
()
=
.
() 0
()
()
EX = E[T ] = .
In addition, Thus,
ET2 = 1
t+1e-t dt
=
( + 2)
=
( +
1)()
=
( + 1).
() 0
()
()
Var(T) = ET 2 - (ET )2 = ( + 1) - 2 =
and Var(X) = Var(T ) = 2Var(T ) = 2.
The moment generating function for T , change variables, setting u = t(1 - s)
1 MT (s) = ()
te-test dt = 1 (1 - s)-
0
t ()
ue-u
du
=
(1 - s)-.
0
u
Then,
MX (r) = E[exp rX] = E[exp rT ] = MT (r) = (1 - r)-.
2
dgamma(x, 0.5, 1) 0.0 0.2 0.4 0.6 0.8 1.0
0
2
4
6
8
Figure 1: Gamma density with x= 1, 2, 3, 4, 5, 6 and = 1.
4 Chi-square Distributions
The chi-square densities result from taking = p/2 for some non-negative integer p (called the degrees of freedom) and = 2. Then, the density becomes
fX (x|p) =
1 (p/2)2p/2
x(p/2)-1
e-x/2
0
for 0 x, otherwise.
EX = p/2, Var(X) = 2p, and MX (t) = (1 - 2t)-p/2. We shall later learn the relationship between the chi-square distribution and the normal distribution.
5 Beta Distributions
The beta distribution has two parameters and . It is based on the Beta function
1
B(, ) = t-1(1 - t)-1 dt
0
. The density is concentrated on [0, 1]. f (x|, ) = 1 x-1(1 - x)-1. B(, )
3
8 10
6
dbeta(x, 1, 11)
4
2
0
0.0
0.2
0.4
0.6
0.8
1.0
x Figure 2: Beta density with = 0, 1, ? ? ? , 10 and = 12 - .
EX =
,
+
Var(X )
=
(
+
)2(
+
+
. 1)
The beta densities with + = 12 are graphed using R in Figure 2 using
> curve(dbeta(x,1,11),0,1) > for (i in 2:11){curve(dbeta(x,i,12-i),0,1,add=TRUE)}
6 Normal Distributions
X is called a normal random variable if it is a linear function of Z a standard normal random variable.
X = Z + ?,
X -? Z=
Its density
1
(x - ?)2
fX (x)
=
22
exp -
22
.
If Z has moment generating function MZ then
MX (t) = E[exp tX] = E[exp t(Z + ?)] = e?tMZ (t).
Thus, we have that
EX = ?,
Var(X) = 2,
MX (t) = exp
2 t2 + ?t . 2
4
0.4
0.3
0.2
dnorm(x, 0, 1)
0.1
0.0
!3 !2 !1 0 1 2 3 x
Figure 3: Standard normal density from R command curve(dnorm(x,0,1),-3.5,3.5)
7 Cauchy Distributions
Recall that
d dx
tan-1(x -
)
=
1+
1 (x - )2 .
Thus,
1
1
fX (x|) = 1 + (x - )2 , x R
is a valid probability density function. The absolute mean E|X| does not exist for the Cauchy distribution. To see this note that
1
b 0
1+
x (x - )2
dx
=
1 2
log(1
+ (x - )2)
b 0
=
1 (log(1 + (b
2
- )2)
- log(1 +
2))
as b .
5
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