1 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 Program Department of Engineering Management, Information and Systems

2 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

3 Consider the situation described by the figure. Assuming that the double arrow 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 The relationship between x and  is given by x = a tan , where a > 0. Example

4 Hence, 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. 0 x Example

5 If X 1, X 2,..., X n are independent random variables with means  1,  2,...,  n Standard deviations  1,  2,...,  n, respectively if a 1, a 2, … a n are real numbers then the random variable has mean Linear Combinations of Random Variables

6 Referred to as Root Mean Square, RSS If X i ~ N( µ i,σ i ) for i=1,2,…,n, then Y ~ N( µ Y,σ Y )

7 If X 1, X 2,..., X n are mutually independent random variables that have, respectively Chi-squared distributions with ν 1, ν 2,..., ν n degrees of freedom, then the random variable Y = X 1 + X X n has a Chi-squared distribution with ν = ν 1 + ν ,+ ν n 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. Linear Combinations of Random Variables

8 Corollary: If X 1, X 2,..., X n are independent random variables having identical normal distributions with mean  and variance  2, then the random variable has a chi-squared distribution with  = n degrees of freedom, since Linear Combinations of Random Variables

9 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 Z 1, Z 2,..., Z n 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. Linear Combinations of Random Variables

10 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%? Linear Combinations of Random Variables – Example

11 Y ~ N(100, 5) a. P(Y > 110) Linear Combinations of Random Variables – Example

12 b. P(Y > c) Linear Combinations of Random Variables – Example

13 c. P(Y > 110) = 0.05 Linear Combinations of Random Variables – Example

14 Tolerance Limits - example Consider the assembly shown. Suppose that the specifications on this assembly are 6.00 ± 0.06 in. Let each component x 1, x 2, and x 3, be normally and independently distributed with means  1 = 1.00 in.,  2 =3.00 in., and  3 = 2.00 in., respectively. x 1  1 =1.00 x 2  2 =3.00 x 3  3 =2.00 y

15 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.

16 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 ) That is, if  y  , then the number of nonconforming assemblies produced will be less than or equal to 7 ppm.

17 Tolerance Limits - example solution Suppose that the variances of the component lengths are equal.

18 Tolerance Limits - example solution This can be translated into specification limits on the individual components. If we assume natural tolerance limits, then so

19 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 inches and a standard deviation of inches. Also, they measured the inside of several boxes and found the inside lengths to have a mean of inches and a standard deviation of 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.

20 Distribution of Errors - Example Solution Let U = Y - (X 1 + X 2 + X 3 + X 4 ) where U is the clearance inside of the box Y is the inside box length X i 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

21 Distribution of Errors - Example Solution

22 Distribution of Errors - Example Solution So currently almost 98.17% will not fit into the box, which is bad.

23 Distribution of Errors - Example Solution