Continuous Random VariablesChapter 5
Nutan S. MishraDepartment of Mathematics and
StatisticsUniversity 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 VariableLet 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 variableIn 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) dxf(x) is called probability density function (pdf)
Properties of pdf
lim
lim
1)(
0)(upper
lower
dxxf
xf
(cumulative) Distribution Function
The cumulative distribution function of a continuous random variable is
Where f(x) is the probability density function of x.
a
itlower
dxxfaXPaFlim
)()()(
)()(
)()()(
aFbF
dxxfbxaPbxaPb
a
Relation between f(x) and F(x)
)()(
)()(
xfdx
xdF
dttfxFx
Mean and Variance
2lim
lim
22
lim
lim
22
lim
lim
)(
)()(
)(
upper
lower
upper
lower
upper
lower
dxxfx
dxxfx
dxxxf
Exercise 5.2To find the value of k
Thus f(x) = 4x3 for 0<x<1
P(1/4<x<3/4) = =
P(x>2/3) = =
4
1]4
1[]
4[...
1
10
41
0
3
1
0
3
k
kx
kxkSHL
kx
dxx4/3
4/1
34
dxx1
3/2
34
Exercise 5.7
)4()5()54(
9/59/41)3()3(
FFxP
FxP
Exercise 5.13
21
0
521
0
322
1
0
41
0
41
0
3
44*
444*
dxxdxxx
dxxdxxdxxx
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 DistributionX = normal random variate with parameters µ and
σ if its probability density function is given by
µ and σ are called parameters of the normal distribution
http://www.willamette.edu/~mjaneba/help/normalcurve.html
xxexf 22 2/)(2
1)(
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 DistributionGiven a random Variable X having a normal
distribution with μx = 10 and x = 2, find the probability that X < 8.
-4 -3 -2 -1 0 1 2 3 4
4 6 8 10 12 14 16
z
x
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
5.1
2/1 2
)5.1( dtexP t
Uniform distribution
The uniform distribution with parameters α and β has the density function
elsewhere 0
1
)(
xforxf
Exponential Distribution: Basic Facts
• Density
• CDF
• Mean
• Variance
, 0, 0
0, 0
xe xf x
x
1 , 0
0, 0
xe xF x
x
1E X
2
1Var X
Key Property: Memorylessness
• 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
for all , 0P X s t X t P X s s t
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 2 3 4 5
0 1/24 3/24 1/24 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 042.24/1)2,0( f
243
242
241)2,1( yxP
2411
246
245)3,2( yxP
2410
246
242
242)4,1( yxP
244
242
242)2,0( yxP
122)1|2( xyP
122)1|3( xyP
126)1|4( xyP
122)2|4( xyP
25.024/6)4,1( f
Joint probability distribution
Joint Probability Distribution Function f(x,y) > 0
Marginal pdf of x & y
here is an example
x =1, 2, 3 y = 1, 2
x y
yxf 1),(
Ayx
yxfAYXP),(
),(]),[(
y
X yxfxf ),()(
x
Y yxfyf ),()(
21),(
yxyxf
Marginal pdf of x & y
Consider the following example
x = 1,2,3
y = 1,2
x xY
yxyYPyxfyf
3
1 21)(),()(
21
32
21
2
21
1
xxx
y yX
yxxXPyxfxf
2
1 21)(),()(
21
36
21
3
21
2
21
1 yyyy
Independent Random Variables
If
dependentff 052.57630
245
246)2(*)0( 21
tindependen)()()( yYPxXPyYxXP
)()(),( yfxfyxf YX
241
6
1*
4
1)5(*)0( 21 ff
Properties of expectations
for a discrete pdf, f(x), The expected value of the function u(x), E[u(X)] =
Mean = = E[X] =
Variance = Var(X) = 2 = x2=E[(X-)2] = E[X2] - 2
For a continuous pdf, f(x)
E(X) = Mean of X =
E[(X-)2] = E(X2) -[E(X)]2 = Variance of X =
S
dxxfx )(
Sxxfxu )()(
Sxxfx )(
S
dxxfx )()(2
Properties of expectationsE(aX+b) = aE(X) + b
Var (aX+b) = a2var(X)
Mean and variance of Z
x
Z Is called standardized variable
baxxx
Z
1
E(Z) = 0 and var(Z) = 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 meanx1, 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
nn x
nxn
xnn
xxxx
1...
11...21
21
nx
xE x
2
)var(
by denoted )(
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