Linear Transformations and Combinations

1 Introduction
Linear Transformations and Combinations
STA 281 Fall 2004
We are often interested in transformations of random variables. A transformation is any function
of the random variable. A simple example is changing units. If X is a random variable measured in
feet, then the random variable Y = 12X would measure the same quantity in inches. Only slightly
more complicated is changing temperature scales from Fahrenheit to Celsius. If F is a temperature
measurement in Fahrenheit, than C = (5/9)(F − 32) is the same measurement in Celsius.
We’ve already discussed computing means and variances for transformations. In particular,
recall the formula
E[h(X)] =
h(x)P (X = x)
x
To illustrate, suppose a store sells cubic fish tanks where 40% are 1 foot on a side, 50% are 2 feet
on a side, and 10% are 3 feet on a side. Let X be the side length for a randomly chosen tank and
let Y be the volume of a tank. Y is a function of X because Y = X 3 . Finding the mean and
variance of Y may be done by converting back to X and computing the expectations with respect
to the distribution of X. Thus
E[Y ] = E[X 3 ] = (1 3 )(0.4) + (2 3 )(0.5) + (3 3 )(0.1) = 7.1
E[Y 2 ] = E[(X 3 ) 2 ] = E[X 6 ] = (1 6 )(0.4) + (2 6 )(0.5) + (3 6 )(0.1) = 105.3
V [Y ] = E[Y 2 ] − E[Y ] 2 = 105.3 − (7.1 2 ) = 54.89
The standard deviation of Y is σY = √ 54.89 = 7.4088.
For many transformations (i.e. functions h(X)) we have to go through this general formula.
However, for a particular set of transformations, called linear transformations, the general formula
simplifies considerable and thus we can acquire simpler formulas for E[h(X)] and V [h(X)]. These
methods apply even if the random variables are continuous.
2 Linear Transformations
Let X be a random variable (discrete or continuous). A linear transformation of X is the quantity
aX + b for some constants a and b. In problems, you will have to determine these constants from
the given situation.
For example, suppose you are running a carnival. You charge $5 for admission and then charge
$2 for tickets that may be used for rides. Suppose further the number of ride tickets a particular
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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.
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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
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customers in Lexington (L). We are trying to find the mean and variance of D, the total amount
of donations. To use the information in the problem, we should determine how D is related to
the quantities L and N we know something about. The local grocery chain donates $100 at first,
then customers donate. In addition to customer donations, the local grocery chain also does partial
matching. From Nicholasville, the charity gains 1.25N (each dollar given by a customer matched
with $0.25 from the grocery chain. Similarly, the charity gains 1.50L from Lexington. The total
amount of donations , D, is the sum of these three quantities,
D = 1.25N + 1.50L + 100
Using the formulas for mean and variance, we find
4.3 TV, VCR Example
µD = 1.25(2100) + 1.50(5500) + 100 = 10975
σ 2 D = 1.252 (10000) + 1.50 2 (90000) = 218125
σD = √ 218125 = 467.0385
Recall our example with a customer buying a TV or a VCR. From a set of probabilities, we
constructed a probability table
T V T V c
V CR 0.10 0.20 0.30
V CR c 0.50 0.20 0.70
0.60 0.40 1.00
Suppose we are given the information that a TV costs $250 and a VCR costs $100. We may also
construct a table showing the costs for each outcome in the probability table.
T V T V c
V CR 350 100
V CR c 250 0
We may compute the mean and variance of C directly from the definitions, since we know the costs
and probabilities associated with the four outcomes. We find
E[C] = 350(0.1) + 250(0.5) + 100(0.2) + 0(0.2) = 180
E[C 2 ] = 350 2 (0.1) + 250 2 (0.5) + 100 2 (0.2) + 0 2 (0.2) = 45500
V [C] = E[C 2 ] − (E[C]) 2 = 45500 − 180 2 = 13100
Notice that the cost is really just the sum of the amount spent on TVs, Ct, and the amount
spent on VCRs, Cv. Note Ct has two possible values, 0 if they didn’t buy a TV, which occurs
with probability 0.4, and 250 is they did buy a TV, which occurs with probability 0.6. We may
find that E[Ct] = 250(0.6) + 0(0.4) = 150 and E[Cv] = 100(0.3) + 0(0.7) = 30. We may also find
V [Ct] = 15000 and V [Cv] = 2100.
We said the total cost is the sum of Ct and Cv, so we may write
C = Ct + Cv
Using the formula for expectation, we find E[C] = E[Ct] + E[Cv] = 150 + 30 = 180, which agrees
with our previous calculation, as expected. However, the variances do not sum to V [C], since
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V [Ct] + V [Cv] = 15000 + 2100 = 17100, not the 13100 we derived previously. Why? The random
variables Ct and Cv are not independent. The events “buy a TV” and “buy a VCR” are
not independent because 0.1 = P (T V V CR) = P (T V )P (V CR) = (0.60)(0.30) = 0.18. The
variance formula only applies to independent random variables.
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