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Continuous Distributions. Continuous random variables. For a continuous random variable X the probability distribution is described by the probability density function f ( x ), which has the following properties :. f ( x ) ≥ 0. - PowerPoint PPT Presentation
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Continuous Distributions
Continuous random variables
For a continuous random variable X the probability distribution is described by the probability density function f(x), which has the following properties :
1. f(x) ≥ 0
2. 1.f x dx
3. .
b
a
P a X b f x dx
Graph: Continuous Random Variableprobability density function, f(x)
1.f x dx
.b
a
P a X b f x dx
The Uniform distribution from a to b
Definition: A random variable , X, is said to have a Uniform distribution from a to b if X is a continuous random variable with probability density function f(x):
1
0 otherwise
a x bf x b a
Graph: the Uniform Distribution(from a to b)
0
0.1
0.2
0.3
0.4
0 5 10 15
1
b a
a b
f x
x
The Cumulative Distribution function, F(x) (Uniform Distribution from a to b)
0
0.1
0.2
0.3
0.4
0 5 10 15
0
1
x a
x aF x P X x a x b
b ax b
a b
f x
x
F x
Cumulative Distribution function, F(x)
0
1
x a
x aF x P X x a x b
b ax b
0
0.5
1
0 5 10 15 a b x
F x
The Normal distribution
Definition: A random variable , X, is said to have a Normal distribution with mean and standard deviation if X is a continuous random variable with probability density function f(x):
2
221
2
x
f x e
Graph: the Normal Distribution(mean , standard deviation )
Note: 2
221
2
x
f x e
2
222
11 0
2
x
f x e x
if 0 or x x
Thus the point is an extremum point of f(x). (In this case a maximum)
Thus the points – , + are points of inflection of f(x)
2
22
3
1
2
xdf x e x
dx
2 2
2 22 22
5 3
1 1
2 2
x x
e x e
2
22
2
23
1 1 0
2
x xe
2
2if 1 or 1
x x
i.e. x
Also 2
221
12
x
f x dx e dx
xz
To evaluate
Proof: 2
221
2
x
e dx
Make the substitution1
dz dx
When , and when , . x z x z
2 22
22 2 21 1 1
2 2 2
x z z
e dx e dz e dz
Consider evaluating
2
2
z
e dz c
Note:2 2 2
22 2 2 2
z z uzc e dz e dz e dz e du
2 2
2 2
22 2
z uz u
e e dudz e dudz
Make the change to polar coordinates (R, )
z = R sin() and u = R cos()
Using
or
2 2 2R z u tanz
u and
2 2R z u 1tan
z
u
and
2 2
22
z u
c e dudz
2
0 0
, sin , cosf z u dzdu f R R RdRd
Hence
2
2
2
2 2 22
00 0 0 0
1 2R
R
e RdRd e d d
and
2
2 2z
c e dz
or 22
2221 1
12 2
xz
e dz e dz
The Exponential distribution
Consider a continuous random variable, X with the following properties:
1. P[X ≥ 0] = 1, and
2. P[X ≥ a + b] = P[X ≥ a] P[X ≥ b] for all a > 0, b > 0.
These two properties are reasonable to assume if X = lifetime of an object that doesn’t age.
The second property implies:
P X a b
P X a b X a P X bP X a
The property:
models the non-aging property
P X a b X a P X b
i.e. Given the object has lived to age a, the probability that is lives a further b units is the same as if it was born at age a.
Let F(x) = P[X ≤ x] and G(x) = P[X ≥ x] .Since X is a continuous RV then G(x) = 1 – F(x)(P[X ≥ x] = 0 for all x.)
1. G(0) = 1, and
2. G(a + b) = G(a) G(b) for all a > 0, b > 0.
The two properties can be written in terms of G(x):
We can show that the only continuous function, G(x), that satisfies 1. and 2. is a exponential function
From property 2 we can conclude
22G a G a a G a G a G a
Using induction
nG na G a a G a G a G a
Hence putting a = 1.
1n
G n G
Also putting a = 1/n.
11n
nG G 1
1 or 1 nnG G
Finally putting a = 1/m.
1
1 1 1n
m mnn
nm mG G G G
Since G(x) is continuous
1x
G x G for all x ≥ 0.
Now 0 1 1G
If G(1) = 0 then G(x) = 0 for all x > 0 and G(0) = 0 if G is continuous. (a contradiction)
Also 0 1 and 0.G G
If G(1) = 1 then G(x) = 1 for all x > 0 and G() = 1 if G is continuous. (a contradiction)
Thus G(1) ≠ 0, 1 and 0 < G(1) < 1
Let = - ln(G(1)) then G(1) = e-
Thus 1xx xG x G e e
0
0 0
xe xf x F x
x
To find the density of X we use:
A continuous random variable with this density function is said to have the exponential distribution with parameter .
0 0Finally 1
1 0x
xF x G x
e x
0
0.5
1
-2 0 2 4 6
Graphs of f(x) and F(x)
0
0.5
1
-2 0 2 4 6
f(x)
F(x)
Consider a continuous random variable, X with the following properties:
1. P[X ≥ 0] = 1, and
2. P[x ≤ X ≤ x + dx|X ≥ x] = dx
for all x > 0 and small dx.
These two properties are reasonable to assume if X = lifetime of an object that doesn’t age.
The second property implies that if the object has lived up to time x, the chance that it dies in the small interval x to x + dx depends only on the length of that interval, dx, and not on its age x.
Another derivation of the Exponential distribution
Let F (x ) = P[X ≤ x] = the cumulative distribution function of the random variable, X .
Then P[X ≥ 0] = 1 implies that F(0) = 0.
Determination of the distribution of X
Also P[x ≤ X ≤ x + dx|X ≥ x] = dx implies
P x X x dxP x X x dx X x
P X x
1
F x dx F xdx
F x
or 1
F x dx F x dF xF x
dx dx
We can now solve the differential equation
for the unknown F.
1
dFF
dx
1 1
dF dxF
1 or
1dF dx
F
and ln 1 F x c
ln 1 F x c
and 1 or 1x c x cF e F F x e
Now using the fact that F(0) = 0.
0 1 0 implies 1 and 0c cF e e c
Thus 1 and x xF x e f x F x e
This shows that X has an exponential distribution with parameter .
The Weibull distribution
A model for the lifetime of objects that do age.
Recall the properties of continuous random variable, X, that lead to the Exponential distribution. namely
1. P[X ≥ 0] = 1, and
2. P[x ≤ X ≤ x + dx|X ≥ x] = dx
for all x > 0 and small dx.
Suppose that the second property is replaced by:
The second property now implies that object does age. If it has lived up to time x, the chance that it dies in the small interval x to x + dx depends both on the length of that interval, dx, and its age x.
2. P[x ≤ X ≤ x + dx|X ≥ x] = x)dx = x – 1dx
for all x > 0 and small dx.
A continuous random variable, X satisfies the following properties:
1. P[X ≥ 0] = 1, and
2. P[x ≤ X ≤ x + dx|X ≥ x] = x-1dx
for all x > 0 and small dx.
Derivation of the Weibulll distribution
Let F (x ) = P[X ≤ x] = the cumulative distribution function of the random variable, X .
Then P[X ≥ 0] = 1 implies that F(0) = 0.
Also P[x ≤ X ≤ x + dx|X ≥ x] = x-1dx implies
P x X x dxP x X x dx X x
P X x
1
1
F x dx F xx dx
F x
1 or 1F x dx F x dF x
x F xdx dx
We can now solve the differential equation
for the unknown F.
1 1
dFx F
dx
11 1
dF x dxF
11 or
1dF x dx
F
and ln 1x
F c
ln 1 F x c
1 or 1x c x c
F e F x e
again F(0) = 0 implies c = 0.
The distribution of X is called the Weibull distribution with parameters and.
Thus 1x
F x e
1and 0x
f x F x x e x
The Weibull density, f(x)
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0 1 2 3 4 5
( = 0.5, = 2)
( = 0.7, = 2)
( = 0.9, = 2)
The Gamma distribution
An important family of distibutions
The Gamma Function, (x)
An important function in mathematics
The Gamma function is defined for x ≥ 0 by
1
0
x ux u e du
1x uu e
x
u
Properties of Gamma Function, (x)
1. (1) = 1
0
00
1 1u ue du e e e
2. (x + 1) = x(x)
0
0
1 x u x ux u e du u e du
We will use integration by parts to evaluate x uu e du
x uu e du wdv wv vdw 1where and , hence ,x u x uw u dv e du dw xu dx v e
1Thus x u x u x uu e du u e x u e du
1
00 0
and x u x u x uu e du u e x u e du
or 1 0x x x
3. (n) = (n - 1)! for any positive integer n
using 1 and 1 1x x x
2 1 1 1 1!
3 2 2 2 2!
4 3 3 3 2 3!
121 1
20 0
thus x u ux u e du u e du
4. 12
Recall the density for the normal distribution
2
221
2
x
f x e
If and 1 then
2
21
2
x
f x e
2 2 2
2 2 2
0 0
1 1 2Now 1= 2
2 2
x x x
e dx e dx e dx
2
2
0
Thus =2
x
e dx
Make the substitution 2
2 .xu 121
21
Hence 2 and or .2
ux u du xdx dx du du
x
Also when 0, then 0,x u
122
2 12
0 0
1=
2 2 2
x u ue dx e dx
12and = .
3 1 1 12 2 2 2= .
Using 1 .x x x
5 3 3 3 12 2 2 2 2= .
7 5 5 5 3 12 2 2 2 2 2= .
2 1 2 1 5 31 12 2 2 2 2 2=n nn
12 2
2 !5.
!2 n
nn
n
2
2 ! 2 !
2 !2 !2n n n
n n
n n
Graph: The Gamma Function
0
5
10
15
20
25
0 1 2 3 4 5
x
(x)
The Gamma distribution
Let the continuous random variable X have density function:
1 0
0 0
xx e xf x
x
Then X is said to have a Gamma distribution with parameters and .
Graph: The gamma distribution
0
0.1
0.2
0.3
0.4
0 2 4 6 8 10
( = 2, = 0.9)
( = 2, = 0.6)
( = 3, = 0.6)
Comments
1. The set of gamma distributions is a family of distributions (parameterized by and ).
2. Contained within this family are other distributionsa. The Exponential distribution – in this case = 1, the
gamma distribution becomes the exponential distribution with parameter . The exponential distribution arises if we are measuring the lifetime, X, of an object that does not age. It is also used a distribution for waiting times between events occurring uniformly in time.
b. The Chi-square distribution – in the case = /2 and = ½, the gamma distribution becomes the chi- square (2) distribution with n degrees of freedom. Later we will see that sum of squares of independent standard normal variates have a chi-square distribution, degrees of freedom = the number of independent terms in the sum of squares.
The Exponential distribution
0
0 0
xe xf x
x
The Chi-square (2) distribution with d.f.
21
2 2
112
2
0
0 0
xx e xf x
x
2
2 2
22
10
2
0 0
x
x e x
x
0
0.1
0.2
0 4 8 12 16
Graph: The 2 distribution
( = 4)
( = 5)
( = 6)
Summary
Important continuous distributions
Uniform distribution from a to b
1
0 otherwise
a x bf x b a
0
0.1
0.2
0.3
0.4
0 5 10 15
1
b a
a b
f x
x0
0.5
1
0 5 10 15 a b x
F x
0
1
x a
x aF x P X x a x b
b ax b
Normal distribution mean and standard deviation
2
221
2
x
f x e
0
0 0
xe xf x F x
x
the exponential distribution with parameter
f(x) =
0
0.5
1
-2 0 2 4 6
f(x)
The Weibull distribution - parameters and.
1and 0x
f x F x x e x
f(x) =
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0 1 2 3 4 5
( = 0.5, = 2)
( = 0.7, = 2)
( = 0.9, = 2)
The gamma distribution - parameters and
0
0.1
0.2
0.3
0.4
0 2 4 6 8 10
( = 2, = 0.9)
( = 2, = 0.6)
( = 3, = 0.6)
1 0
0 0
xx e xf x
x
The 2 distribution with degrees of freedom
0
0.1
0.2
0 4 8 12 16
( = 4)
( = 5)
( = 6)
21
2 2
112
2
0
0 0
xx e xf x
x
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