Download Continuous Random Variables Chapter 5

Continuous Random Variables
Chapter 5
Nutan S. Mishra
Department of Mathematics and
Statistics
University of South Alabama
Continuous Random Variable
When random variable X takes values on an
interval
For example GPA of students X  [0, 4]
High day temperature in Mobile X  ( 20,∞)
Recall in case of discrete variables a simple event
was described as (X = k) and then we can
compute P(X = k) which is called probability
mass function
In case of continuous variable we make a change
in the definition of an event.
Continuous Random Variable
Let X  [0,4], then there are infinite number of
values which x may take. If we assign probability
to each value then
P(X=k)  0 for a continuous variable
In this case we define an event as
(x-x ≤ X ≤ x+x ) where x is a very tiny
increment in x. And thus we assign the
probability to this event
P(x-x ≤ X ≤ x+x ) = f(x) dx
f(x) is called probability density function (pdf)
Properties of pdf
f ( x)  0
upper lim

f ( x)dx  1
lower lim
(cumulative) Distribution Function
The cumulative distribution function of a
continuous random variable is
a
F ( a )  P( X  a ) 
 f ( x)dx
lower limit
Where f(x) is the probability density function of x.
b
P(a  x  b)  P(a  x  b)   f ( x)dx
a
 F (b)  F (a)
Relation between f(x) and F(x)
x
F ( x) 

f (t )dt

dF ( x)
 f ( x)
dx
Mean and Variance
upper lim
 xf ( x)dx

lower lim
upper lim
 
2
 (x  )
2
f ( x)dx
lower lim
 
2
upper lim
2
x
f
(
x
)
dx



lower lim
2
Exercise 5.2
To find the value of k
1
3
kx
 1
0
1
4
x
1
L.H .S .  k  x 3  k[ ]10  k[ ]  1
4
4
0
k 4
Thus f(x) = 4x3 for 0<x<1
3/ 4
3
4
x
P(1/4<x<3/4) =  dx =
1/ 4
1
3
4
x
P(x>2/3) =  dx =
2/3
Exercise 5.7
P( x  3)  F (3)  1  4 / 9  5 / 9
P(4  x  5)  F (5)  F (4) 
Exercise 5.13
1
1
1
   x * 4 x dx   4 x dx  4 x dx 
3
4
0
4
0
0
1
1
   x * 4 x dx    4  x dx   
2
2
0
3
2
5
0
2
Probability and density curves
• P (a<Y<b): P(100<Y<150)=0.42
Useful link:
http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/cprob/cprob2.html
Normal Distribution
X = normal random variate with parameters µ and
σ if its probability density function is given by
1
2
2
f ( x) 
e  ( x   ) / 2    x  
2
µ and σ are called parameters of the normal
distribution
http://www.willamette.edu/~mjaneba/help/normalcu
rve.html
Standard Normal Distribution
The distribution of a normal random
variable with mean 0 and variance 1 is
called a standard normal distribution.
-4
-3
-2
-1
0
1
2
3
4
Standard Normal Distribution
• The letter Z is traditionally used to
represent a standard normal random
variable.
• z is used to represent a particular value of
Z.
• The standard normal distribution has been
tabularized.
Standard Normal Distribution
Given a standard normal distribution, find
the area under the curve
(a) to the left of z = -1.85
(b) to the left of z = 2.01
(c) to the right of z = –0.99
(d) to right of z = 1.50
(e) between z = -1.66 and z = 0.58
Standard Normal Distribution
Given a standard normal distribution, find
the value of k such that
(a) P(Z < k) = .1271
(b) P(Z < k) = .9495
(c) P(Z > k) = .8186
(d) P(Z > k) = .0073
(e) P( 0.90 < Z < k) = .1806
(f) P( k < Z < 1.02) = .1464
Normal Distribution
• Any normal random variable, X, can be
converted to a standard normal random variable:
z = (x – μx)/x
Useful link: (pictures of normal curves borrowed from:
http://www.stat.sc.edu/~lynch/509Spring03/25
Normal Distribution
Given a random Variable X having a normal
distribution with μx = 10 and x = 2, find the
probability that X < 8.
z
-4
x
-3
-2
-1
0
1
2
3
4
6
8
10
12
14
16
4
Relationship between the Normal
and Binomial Distributions
• The normal distribution is often a good
approximation to a discrete distribution when the
discrete distribution takes on a symmetric bell
shape.
• Some distributions converge to the normal as
their parameters approach certain limits.
• Theorem 6.2: If X is a binomial random variable
with mean μ = np and variance 2 = npq, then
the limiting form of the distribution of Z = (X –
np)/(npq).5 as n  , is the standard normal
distribution, n(z;0,1).
Exercise 5.19
-4
-3
-2
-1
0
1
2
3
4
1.5
P( x  1.5)   e

1 / 2 t 2
dt
Uniform distribution
The uniform distribution with parameters α and β has the
density function
1
f ( x) 
for   x  
 
0
elsewhere
Exponential Distribution: Basic Facts
• Density
e  x , x  0
f  x  
,  0
0, x  0
• CDF
1  e x , x  0
F  x  
0, x  0
1
• Mean E  X  

• Variance Var  X   1
2
Key Property: Memorylessness
P  X  s  t X  t  P  X  s for all s, t  0
• Reliability: Amount of time a component has been in
service has no effect on the amount of time until it fails
• Inter-event times: Amount of time since the last event
contains no information about the amount of time until
the next event
• Service times: Amount of remaining service time is
independent of the amount of service time elapsed so far
Exponential Distribution
The exponential distribution is a very commonly used distribution in reliability engineering.
Due to its simplicity, it has been widely employed even in cases to which it does not apply.
The exponential distribution is used to describe units that have a constant failure rate.
The single-parameter exponential pdf is given by:
where:
·λ
·
·
·
= constant failure rate, in failures per unit of measurement, e.g. failures per hour, per
cycle, etc.
λ = .1/m
m = mean time between failures, or to a failure.
T = operating time, life or age, in hours, cycles, miles, actuations, etc.
This distribution requires the estimation of only one parameter, , for its application.
Joint probabilities
For discrete joint probability density function (joint pdf) of a
k-dimensional discrete random variable X = (X1, X2, …,Xk) is
defined to be f(x1,x2,…,xk) = P(X1 = x1, X2 = x2 , …,Xk = xk) for
all possible values x = (x1,x2,…,xk) in X.
Let (X, Y) have the joint probability function specified in the
following table
x/y
0
2
1/24
3
3/24
4
1/24
5
1/24
6/24
1
2/24
2/24
6/24
2/24
12/24
2
2/24
1/24
2/24
1/24
6/24
5/24
6/24
9/24
4/24
24/24
Joint distribution
Consider f (0,2)  1 / 24  .042
P( x  1, y  2)  1
24
2
24
P( x  2, y  3)  5
P( x  1, y  4)  2
f (1,4)  6 / 24  0.25
24
2
P( x  0, y  2)  2
6
24
24
24
24
6
2
P( y  4 | x  2)  2
12
P( y  4 | x  1)  6
12
P( y  2 | x  1)  2
12
24
 11
24
24
12
P( y  3 | x  1)  2
 3
 10
 4
24
24
24
Joint probability distribution
Joint Probability Distribution Function
f(x,y) > 0
 f ( x, y)  1
x
y
P[( X , Y )  A]  
 f ( x, y)
( x , y )A
Marginal pdf of x & y
f X ( x )   f ( x, y )
y
fY ( y)   f ( x, y)
here is an example
x
f ( x, y ) 
x y
21
x =1, 2, 3
y = 1, 2
Marginal pdf of x & y
Consider the following example
x y
y 1 21
2
f X ( x)   f ( x, y)  P( X  x)  
y
x 1 x  2 2x  3



21
21
21
fY ( y )  
x
x = 1,2,3
x y
f ( x, y )  P(Y  y )  
x 1 21
3
1 y 2  y 3  y 6  3y




21
21
21
21
y = 1,2
Independent Random Variables
If
P( X  x  Y  y )  P( X  x) P(Y  y )  independen t
f ( x, y)  f X ( x) fY ( y)
f1 (0) * f 2 (5) 
f1 (0) * f 2 (2)  6
5
24 24
1
1
*
 1
24
4
6
 30
576
 .052  dependent
Properties of expectations
for a discrete pdf, f(x),
The expected value of the function u(x),
E[u(X)] = xS u ( x) f ( x)

Mean =  = E[X] =

xS
x f ( x)
Variance = Var(X) = 2 = x2=E[(X-)2] = E[X2] - 2
For a continuous pdf, f(x)

E(X) = Mean of X = x f ( x)dx
S
E[(X-)2] = E(X2) -[E(X)]2 = Variance of X =
 (x  )
2
S
f ( x)dx
Properties of expectations
E(aX+b) = aE(X) + b
Var (aX+b) = a2var(X)
Mean and variance of Z
Z
Z
x

x

E(Z) = 0 and var(Z) = 1
Is called standardized variable

 x   ax  b


1
Linear combination of two independent
variables
Let x1 and x2 be two Independent random
variables then their linear combination y =
ax1+bx2 is a random variable.
E(y) = aE(x1)+bE(x2)
Var(y) = a2var(x1)+b2var(x2)
Mean and variance of the sample mean
x1, x2,…xn are independent identically distributed
random variables (i.e. a sample coming from a
population) with common mean µ and common
variance σ2
The sample mean is a linear combination of these
i.i.d. variables and hence itself is a random
variable
x1  x2  ...  xn 1
1
1
x
 x1  x2  ...  xn
n
n
n
n
E ( x )  denoted by  x  
var( x ) 
2
n