1 functions of random variables dr. jerrell t. stracener, sae fellow leadership in engineering emis...

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Functions of Random Variables

Dr. Jerrell T. Stracener, SAE Fellow

Leadership in Engineering

EMIS 7370/5370 STAT 5340 : PROBABILITY AND STATISTICS FOR SCIENTISTS AND ENGINEERS

Systems Engineering ProgramDepartment of Engineering Management, Information and Systems

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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

Functions of Random Variables

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Consider the situation described by the figure. Assuming that the doublearrow is spun so that has a uniform 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

x axis

a

x

0

1

f 22 ,

The relationship between x and is given by x = a tan , where a > 0.

elsewhere ,

Example

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Hence, 22 xa

a

dx

d

22

1

xa

axg

and 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.

0x

Example

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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 numbersthen the random variable

has mean

n

iiiXaY

1

n

iiiY a

1

Linear Combinations of Random Variables

6

n

iiiY a

1

22


Referred to as Root Mean Square, RSS

If Xi ~ N(µi,σi) for i=1,2,…,n,

then Y ~ N(µY,σY)

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If X1, X2, ..., Xn are mutually independent random variables thathave, 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 offreedom.

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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Corollary:If X1, X2, ..., Xn are independent random variables havingidentical normal distributions with mean and variance 2, thenthe random variable

2n

1i

ixY

has a chi-squared distribution with = n degrees of freedom, since

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2

i2i ~

xZ


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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

has a chi-squared distribution and the single parameter, , the degrees of freedom, is n, the number of standard normal variates.

n

1i

21Z


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Screws are packaged 100 per box. If individuals have weightsthat independently and normally distributed with mean of 1 ounceand 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 screwswill exceed 110 ounces is 5%?

Linear Combinations of Random Variables – Example

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Y ~ N(100, 5)

a. P(Y > 110)

100

1iiXY

100)1(100100 niY

52

122

12 nniY

5

100110ZP

0228.000.2 ZP


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b. P(Y > c) 01.05

100

c

ZP

01.033.2 ZP

33.25

100

c

65.111c


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c. P(Y > 110) = 0.05

05.0645.1 ZP

05.0100110

Y

ZP

645.110

Y

079.6Y

6079.0 6.05.0:


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Tolerance Limits - example

Consider the assembly shown.

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.

x11=1.00

x2

2=3.00

x3

3=2.00

y

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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 ± 4.49y. (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)

That is, if y 0.0134, then the number of

nonconforming assemblies produced will be less than or equal to 7 ppm.

0134.049.4

06.0

yσ

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Suppose that the variances of the component lengths are equal.

00018.0

0134.0σσσ 223

22

21

2

yσ

2σ32yσ

00006.03

00018.03

σ2y

2σ

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This can be translated into specification limits on the individual components. If we assume natural tolerance limits, then

so

3xx

0232.000.100006.0300.1 :x1

0232.000.300006.0300.3 :x2

0232.000.200006.0300.2 :x3

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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 boxY is the inside box lengthXi 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 UU σ,μN~U

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Uμ

752.0

2.5774556.9

μ1μ1μ1μ1μ1

μa

4321 XXXXY

k

1iii

2Uσ

133.0

061.04053.0

σ4σ

σ1σ1σ1σ1σ1

σa

22

2X

2Y

2X

22X

22X

22X

22Y

2

k

1i

2i

2i

4321

Uσ 36.0

22


So currently almost 98.17% will not fit into the box, which is bad.

fittingnot P

9817.

09.2P

0.36

752.00

σ

μUP

)0(

U

U

Z

UP

23


fittingnot P

645.1P Z0.05

05.0

0.36

μ-308.10645.1 Y

90.10μY

1 functions of random variables dr. jerrell t. stracener, sae fellow leadership in engineering emis...

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