Linear Transformations and Combinations

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person buys, X, is a random variable. Suppose we want an expression for the amount of money spent by that particular person. Letting Y denote the amount of money spent, we can find a function relating Y to X. If a person buys X tickets, then they spent 2X dollars on tickets. In addition, they spent $5 to enter the carnival. The total amount spent is Y = 2X + 5. Since Y has the correct form (aX + b, with a = 2, and b = 5), we say Y is a linear transformation of X. If we know the mean and variance of X, then there are simple formulas to compute the mean and variance of a linear transformation Y . We derive the formulas for discrete random variables. We know that E[X] = x xP (X = x) and E[h(X)] = x h(x)P (X = x). A linear transformation is a particular form of h(x), so The formula applies because E[X] = x random variable. E[aX + b] = x (ax + b)P (X = x) = x axP (X = x) + bP (X = x) = x axP (X = x) + x bP (X = x) = a x xP (X = x) + b x P (X = x) = aE[X] + b xP (X = x) by definition and x P (X = x) = 1 for any V [aX + b] = E[((aX + b) − (aE[X] + b)) 2 ] = E[(aX + b − aE[X] − b) 2 ] = E[a 2 (X − E[X]) 2 ] = a 2 E[(X − E[X]) 2 ] = a 2 V [X] These formulas E[aX + b] = aE[X] + b and V [aX + b] = a 2 V [X] apply for any random variable, discrete or continuous (we did not derive the calculus, but integrals have many of the properties of summations, such as factoring out constants and breaking sums into two parts, so it makes sense). The expectation formula is simple to remember, the expectation of a linear transformation is the same linear transformation of the expectation. The variance requires a little more explanation. Variance is a measure of spread. Adding a constant b doesn’t change the spread, so it can be ignored in computed the variance. When we multiply by a, we have to remember the variance is in squared units, so the constant a is squared in the formula. Suppose for our carnival example any particular person buys an average of 5 tickets with a variance of 9 tickets. What is the mean and variance of the amount of money spent by a particular person? We said Y = 2X + 5, so E[Y ] = 2E[X] + 5 = 2(5) + 5 = 15 and V [Y ] = 2 2 (9) = 36. 3 Linear Combinations Often we deal with several random variables at once. A linear combination of two random variables X and Y has the form aX + bY + c, where a, b, and c are fixed constants found from the problem. Linear combinations can involve many random variables. A linear combination Y of n random variables X1, . . . , Xn is Y = a1X1 + a2X2 + . . . + anXn + b where the ai and b coefficients are fixed values which, again, are found in the problem. 2
We will not go through the proofs on how to compute the mean and variance of a linear combination, but they are similar to the formulas for a linear transformation of a single random variable. The mean of a linear combination is E[a1X1 + a2X2 + . . . + anXn + b] = a1E[X1] + a2E[X2] + . . . + anE[Xn] + b so the expectation of a linear combination is the same linear combination of the expectations. If all the random variables in the linear combination are independent (don’t forget this assumption), then the variance of a linear combination is V [a1X1 + a2X2 + . . . + anXn + b] = a 2 1V [X1] + a 2 2V [X2] + . . . + a 2 nV [Xn] Two simple linear combinations of two independent random variables are Z1 = X + Y and Z2 = X − Y , where X and Y are independent. These may be written Z1 = 1X + 1Y + 0 and Z2 = 1X + (−1)Y + 0. Using the formulas, we may derive E[Z1] = E[X] + E[Y ], E[Z2] = E[X] − E[Y ], and V [Z1] = V [X] + V [Y ], V [Z2] = V [X] + V [Y ]. Note that while the variance of a sum is the sum of the variances, the variance of a difference is also the sum of the variances. If we add independent sources of noise into a problem, we increase the overall noise of the system. 4 Examples 4.1 Tables and Chairs At a school, each room contains only tables and chairs. Suppose that, on average, each room contains 50 chairs with a standard deviation of 20 chairs. Suppose further that, on average, each room contains 5 tables with a standard deviation of 2 tables. Each chair weighs 10 pounds and each table weighs 30 pounds. You select a random room and place all the items in the room into a 10000 pound truck. a) Write a formula for Z, the total weight of the truck and its contents after the truck has been loaded with the contents of the room. Z = 10C + 30T + 10000 b) What are the mean and variance of Z? E[Z] = 10E[C] + 30E[T ] + 10000 = 10(50) + 30(5) + 10000 = 10650. V [Z] = 10 2 V [C] + 30 2 V [T ] = 10 2 (20 2 ) + 30 2 (2 2 ) = 43600 4.2 Christmas Donations Each year at Christmas, charity donations are accepted at a local grocery chain. Suppose the chain donates an initial $100 and then customers donate either in Lexington or in Nicholasville. In Nicholasville, customers donate an average of $2100 each year with a standard deviation of $100, while in Lexington customers donate an average of $5500 with a variance of 90000. Suppose further that the local grocery chain matches the donations. For every dollar donated in Nicholasville, the local grocery chain gives an additional $0.25 and for every dollar donated in Lexington the local grocery chain gives an additional $0.50. What are the mean, variance, and standard deviation of the total amount of donations to the charity in a given year? When we look at this problem, the mean and variance are given for two random quantities. These are the amount donated by customers in Nicholasville (N) and the amount donated by 3
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Page 5: V [Ct] + V [Cv] = 15000 + 2100 = 17

Linear Transformations and Combinations

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