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Continuous Distributions
Beta Distribution
• When modeling probabilities for some proportion Y,
0 < Y < 1, consider the Beta distribution:
 ya 1 (1  y ) b 1
,0  y  1

f ( y )   B (a , b )

0, otherwise

for parameters a, b > 0,
where
(a )( b )
B(a , b ) 
(a  b )
Beta D ens ity f unct ion, dbeta(x , 3, 4)
2
f ( x)
1
0
0
0.5
1
x
1.5
A Useful Identity
• Since this is a density function, we know
1 
ya 1 (1  y) b 1
dy,
B(a , b )
1
0
or equivalently,

1
0
a 1
y
(1  y)
b 1
(a )( b )
dy  B(a , b ) 
(a  b )
which is easy to compute when a and b are integers.
Mean, Variance for Beta
• If Y is a Beta random variable with parameters a
and b, the expected value and variance for Y are
given by
E (Y ) 
a
ab
ab
V (Y ) 
(a  b )2 (a  b  1)
Cumulative?
• For the special case when a and b are integers, it
can be shown that the cumulative probabilities can
be determined using binomial coefficients.
n
F ( y)   Cn,i ( y)i (1  y) ni ,
i a
where n  a  b  1.
Beta and the Binomial
• Show that the following density function is for a
Beta distribution. Determine a and b.
12 y 2 (1  y), 0  y  1
f ( y)  
0, otherwise

• Show that
F (0.4)  0.1792
…using integration or using the binomial probabilities.
The Lognormal Distribution
• Unlike the normal curve, it’s not symmetric.
• May be appropriate for modeling “insurance
claim severity or investment returns”.
• If Y is lognormal, then
ln(Y) has the normal
distribution.
Lognorm al Dens it y f unc tion, dlnorm(x, 4, 1.1)
f ( x) 0.01
0
0
100
x
200
Lognormal Mean, Variance
• If ln(Y) is a normal random variable with mean m
and variance s 2, then Y is lognormal with mean
and variance given by:
E (Y )  e
V (Y )  e
m (s 2 / 2)
2 m s 2
e
s2

1
Lognormal Probabilities
• Suppose Y is lognormal and ln(Y) = X where
X ~ N(m,s). Then the cumulative probability
F (Y )  P(Y  c)  P( X  ln(c))  FX (ln(c))
may be computed using the z-score and the table
of standard normal probabilities.
If claim amounts are modeled using a lognormal
random variable Y = eX, where X ~ N(7, 0.5 ).
Find the probability P( Y < 1400 ).
Pareto Distribution
• For modeling insurance losses, consider the Pareto
distribution with density function
a 1
ab
f ( y)   
b y
, for a  2 and y  b  0
and distribution function
a
b
F ( y )  1    , for a  2 and y  b  0
 y
Suppose there is a deductible on the policy,
so values of y less than b are not filed.
Pareto Distribution
• If Y is a Pareto random variable with parameters a
and b, the mean and variance given by:
ab
E (Y ) 
a 1
2
2
ab
 ab 
V (Y ) 


a  2  a 1 
Weibull Distribution
• When there was a constant failure rate l, we often
model the time between failures using an
exponential distribution with mean 1/ l.
• If the failure rate increases with time or age,
consider using a Weibull distribution
a 1  b xa
f ( y)  ab ( x)
e
,
for x  0, with a  0, b  0
 b xa
F ( y)  1  e
, for x  0.
Weibull Mean, Variance
• If Y is a Weibull random variable with parameters
a , b  0, the mean and variance are given by:
E (Y ) 
V (Y ) 
(1  a1 )
b
1/ a
1
b
2 /a
 (1 
2
a
)  (1  a )
1 2

Weibull Distribution
• Note that for a = 1, the Weibull distribution agrees
with the exponential distribution.
a 1  b xa
f ( y)  ab ( x) e
= be
But when a > 1, the Weibull
distribution yields a higher
failure rate for larger x
(i.e., failure rate increases
with age).
b x
when a  1
W eibull dens it y f unc tion, dweibull(x , 2)
1
f ( x)
0.5
0
0
2
x
4
Failure Rate
• A “failure rate” function is defined by
f (t )
l (t ) 
1  F (t )
For a time interval t < T < t + h , we consider this as
f (t )dt P(t  T  t  h)
l (t )dt 

1  F (t )
P(T  t )
…the probability of failing in the next h time units,
given the part (or person) has survived to time t.
Thus, l(t) is thought of as failures per unit time.
Comparing Failure Rate
• For the exponential distribution f (y) = le-ly, we note
f (t )
le
l (t ) 
  lt  l is constant.
1  F (t ) e
 lt
Comparing this to the Weibull distribution,
a 1  b xa
f (t )
ab x e
l (t ) 

 b xa
1  F (t )
e
 ab x
a 1
…so here the failure rate is increasing with x, if a > 1.
Moment Generating
• As with discrete distributions, we can define
moment generating functions for continuous
random variables,
m(t )  E[e ]   e f ( y)dy
ty
ty
y
such that the moments of Y are given by
k
d [m(t )]
 E (Y k )
dt
t 0
Some common MGFs
• The MGFs for some of the continuous distributions
we’ve seen include:
1
l
1
exponential: m(t ) 

, where b 
1 bt l  t
l
gamma: m(t )  (1  b t ) a
chi-square: m(t )  (1  2t ) / 2
• Note that not all distributions have a MGF that can be
written in a nice and tidy, closed-form expression.