Transcript functionsofRV - Lyle School of Engineering

Systems Engineering Program
Department of Engineering Management, Information and Systems
EMIS 7370/5370 STAT 5340 :
PROBABILITY AND STATISTICS FOR SCIENTISTS AND ENGINEERS
Functions of Random Variables
Dr. Jerrell T. Stracener, SAE Fellow
Leadership in Engineering
1
Functions of Random Variables
Theorem:
X is a continuous random variable with probability distribution
f(x). Let Y = u(X) define a one-to-one transformation between
the values of X and Y so that the equation y = u(x) can be
uniquely solved for x in terms of y, say x = w(y). Then the
probability distribution of Y is:
g(y) = f[w(y)]|J|
where J = w'(y) and is called the Jacobian of the transformation
2
Example

Consider the situation described by
a
the figure. Assuming that the double
x axis
arrow is spun so that  has a uniform
x
density from -(/2) to /2, suppose we want to find the probability
density of x, the coordinate at the point to which the double arrow
points. We are given
1
f   

0
,

2
 

2
, elsewhere
The relationship between x and  is given by x = a tan ,
where a > 0.
3
Example
Hence,
and
d
a
 2
dx a  x 2
a
g x    2
2
 a x
1
for -  < x < 
where a > 0. The probability density described below is
called the Cauchy distribution. It plays an important role in
illustrating various aspects of statistical theory. For example,
its moments do not exist.
0
x
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Linear Combinations of Random Variables
If X1, X2, ..., Xn are independent random variables
with means 1, 2, ..., n
Standard deviations 1, 2, ..., n, respectively
if a1, a2, … an are real numbers
then the random variable
n
Y   ai X i
i 1
has mean
n
Y   ai  i
i 1
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Linear Combinations of Random Variables
Y 
n
2 2
a
 i i
Referred to as Root Mean Square, RSS
i 1
If Xi ~ N(µi,σi) for i=1,2,…,n,
then Y ~ N(µY,σY)
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Linear Combinations of Random Variables
If X1, X2, ..., Xn are mutually independent random variables that
have, respectively Chi-squared distributions with ν1, ν2, ..., νn
degrees of freedom, then the random variable
Y = X1 + X2 + ... + Xn
has a Chi-squared distribution with ν = ν1+ ν2+ ...,+ νn degrees of
freedom.
Remark:
The Poisson, the Normal and the Chi-squared distributions all
possess a property in that the sum of independent random
variables having any of these distributions is a random
variable that also has the same type of distribution.
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Linear Combinations of Random Variables
Corollary:
If X1, X2, ..., Xn are independent random variables having
identical normal distributions with mean  and variance 2, then
the random variable
 xi   
Y  

 
i 1 
n
2
has a chi-squared distribution with  = n degrees of freedom, since
 xi   
2
Z 
 ~ 1
  
2
2
i
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Linear Combinations of Random Variables
Remark:
This corollary is extremely important because it establishes a
relationship between the important chi-squared and normal
distributions. It also should provide a clear idea of what we
mean by the parameter that we call degrees of freedom.
Note that if Z1, Z2, ..., Zn are independent standard normal
random variables, then n
2
Z
 1
i 1
has a chi-squared distribution and the single parameter, , the
degrees of freedom, is n, the number of standard normal
variates.
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Linear Combinations of Random Variables – Example
Screws are packaged 100 per box. If individuals have weights
that independently and normally distributed with mean of 1 ounce
and standard deviation of 0.5 ounce.
a. What is the probability that a randomly selected box will
weigh more than 110 ounces?
b. What is the box weight for which there is a 1% chance of
exceeding that weight?
c. What would the per screw standard deviation have to be in
order that the probability that a randomly selected box of screws
will exceed 110 ounces is 5%?
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Linear Combinations of Random Variables – Example
100
Y   Xi
i 1
Y  100   i  n  100(1)  100
 Y  
  n 
1
2 2
i
1
2 2
 n  5
Y ~ N(100, 5)
a. P(Y > 110)
110  100 

 P Z 

5


 PZ  2.00  0.0228
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Linear Combinations of Random Variables – Example
c  100 

b. P(Y > c)  P Z 
  0.01
5 

PZ  2.33  0.01
c  100
 2.33
5
c  111.65
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Linear Combinations of Random Variables – Example
c. P(Y > 110) = 0.05

110  100 
  0.05
P Z 
Y


PZ  1.645  0.05
10
 1.645
Y
 Y  6.079
 : 0.5  0.6
  0.6079
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Tolerance Limits - example
Consider the assembly shown.
y
x1
1=1.00
x2
2=3.00
x3
3=2.00
Suppose that the specifications on this assembly are
6.00 ± 0.06 in. Let each component x1, x2, and x3,
be normally and independently distributed with
means 1 = 1.00 in., 2 =3.00 in., and 3 = 2.00 in.,
respectively.
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Tolerance Limits - example
Suppose that we want the specifications to fall on
the inside of the natural tolerance limits of the
process for the final assembly such that the
probability of falling outside of the specification
limits is 7ppm.
Establish the specification limits for each component.
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Tolerance Limits - example solution
The length of the final assembly is normally
distributed. Furthermore, if the allowable number of
assemblies nonconforming to specifications is 7ppm,
this implies that the natural limits must be located at
 ± 4.49y. (This value can be found on the normal distribution table in
the resource section on the web site with a z-value of 0.0000035)
σ y  0.06
4.49
 0.0134
That is, if y  0.0134, then the number of
nonconforming assemblies produced will be less than
or equal to 7 ppm.
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Tolerance Limits - example solution
σ 2y  σ12  σ 22  σ 32  0.01342
 0.00018
Suppose that the variances of the component
lengths are equal.
σ 2y  3σ 2
σ 
2
σ 2y
3
0.00018

3
 0.00006
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Tolerance Limits - example solution
This can be translated into specification limits on the
individual components. If we assume natural
tolerance limits, then
x  x  3
so
x1 : 1.00  3 0.00006  1.00  0.0232
x 2 : 3.00  3 0.00006  3.00  0.0232
x 3 : 2.00  3 0.00006  2.00  0.0232
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Distribution of Errors - Example
A company is having a packaging problem. The company purchase cardboard
boxes nominally 9.5 inches in length intended to hold exactly 4 units of a
product they produced, stacked side by side in the boxes. Many of the boxes
were unable to accommodate the full 4 units so an objective was established to
specific target dimension on the boxes.
Their interns measured the thickness of 25 units of product. They found that
these had a mean of 2.577 inches and a standard deviation of 0.061 inches.
Also, they measured the inside of several boxes and found the inside lengths to
have a mean of 9.556 inches and a standard deviation of 0.053 inches.
Find a new target dimension for the inside length of the boxes ordered by the
company so that only about 5% of the time 4 units cannot be packaged.
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Distribution of Errors - Example Solution
Let U = Y - (X1 + X2 + X3 + X4)
where
U is the clearance inside of the box
Y is the inside box length
Xi is the length of one unit for i = 1,2,3,4
Analysis of the measurement data indicates that the normal distribution
provides a “good” fir to the data.
Then
U ~ Nμ U , σ U 
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Distribution of Errors - Example Solution
k
μU  a μ
 i i
i 1
 1μ Y   1μ X1   1μ X 2   1μ X 3   1μ X 4
 9.556  42.577 
 0.752
k
σ   a i2 σ i2
2
U
i 1
 12 σ 2Y   1 σ 2X1   1 σ 2X 2   1 σ 2X 3   1 σ 2X 4
2
2
2
2
 σ 2Y  4σ 2X
 0.053  40.061
2
2
 0.133
σ U  0.36
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Distribution of Errors - Example Solution
Pnot fitting
  P(U  0)
 U  μ U 0  0.752 

 P

0.36 
 σU
 PZ  2.09
 .9817
So currently almost 98.17% will not fit into the box, which is bad.
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Distribution of Errors - Example Solution
Pnot fitting   0.05
0.05  PZ  1.645
10.308 - μ Y
 1.645 
0.36
μ Y  10.90
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