This lesson is concerned with linear combinations or if you would like linear transformations of the variables. Mathematically linear combinations can be expressed as shown in the expression below:
\(Y = c_1X_1 +c_2X_2 +\dots + c_pX_p = \sum_{j=1}^{p}c_jX_j = \mathbf{c}'\mathbf{X}\).
Here what we have is a set of coefficients c_{1} through c_{p} that are multiplied by corresponding variables X_{1} through X_{p}. So, in the first term we have c_{1} times X_{1} which is added to c_{2} times X_{2} and so on up to the variable X_{p}. Mathematically this is expressed as the sum of j = 1, ... , p of the terms c_{j} times X_{j}. The random variables X_{1} through X_{p} are collected into a column vector X and the coefficient c_{1} to c_{p} are collected into a column vector c. Hence, the linear combination can be expressed as c′X.
The selection of the coefficients c_{1} through c_{p} are very much dependent on the application of interest and what kinds of scientific questions we would like to address.
Upon completion of this lesson, you should be able to do the following:
Later on in this course, when we learn about multivariate data reduction techniques, interpretation of linear combinations will be of great importance.
If you look at the Women's Nutrition data you might recall that we have following observations:
In addition to addressing questions about the individual nutritional component, we may wish to address questions about certain combinations of these components. For instance, we might want to ask what is the total intake of vitamins A and C (in mg). We note that in this case Vitamin A is measuring in micrograms while Vitamin C is measured in milligrams. There are a thousand micrograms per milligram so the total intake of the two vitamins, Y, can be expressed as the following:
Y = 0.001X_{4} + X_{5}
In this case, our coefficients c_{1} , c_{2} and c_{3} are all equal to 0 since the variables X_{1}, X_{2} and X_{3} do not appear in this expression. In addition, c_{4} is equal to 0.001 since each microgram of vitamin A is equal to 0.001 milligrams of vitamin A. In summary, we have
c_{1} = c_{2} = c_{3} = 0, c_{4} = 0.001, c_{5} = 1
Another example where we might be interested in linear combinations is in the Monthly Employment Data. Here we have observations on 6 variables:
Net employment decrease:
In looking at the net job increase, which is equal to the number of jobs created, minus the number of jobs lost.
Y = X_{4}  X_{1}  X_{2}  X_{3}
In this case we have the number of jobs created, (X_{4}), minus the number of people laid off or fired, (X_{1}), minus the number of people resigning, (X_{2}), minus the number of people retired, (X_{3}). These are all of the people that have left their jobs for whatever reason.
In this case
c_{1} = 1 = c_{2} = c_{3}. c_{4} = 1
and since variables 5 and 6 are not included in this expression,
c_{5} = c_{6} = 0.
Net employment increase:
In a similar fashion, net employment increase is equal to the number of people hired, (X_{5}), minus the number of people laid off or fired, (X_{1}), minus the number of people resigning, (X_{2}), minus the number of people retired, (X_{3}).
Y = X_{5}  X_{1}  X_{2}  X_{3}
In this case
c_{1} = ... = c_{3} = 1. c_{4} = 0. c_{5} = 1 and c_{6} = 0.
Net unemployment increase:
Net unemployment increase is going to be equal to the number of people laid off or fired, (X_{1}), plus the number of people resigning, (X_{2}), plus the number of people entering the workforce, (X_{6}), minus the number of people hired, (X_{5}).
Y = X_{1} + X_{2} + X_{6}  X_{5}
Unfilled jobs:
Finally, if we wanted to ask about the number of jobs that went unfilled, this is simply equal to the number jobs created, (X4), minus the number of people hired, (X_{5}).
Y = X_{4}  X_{5}
In other applications, of course other linear combinations would be of interest.
Since linear combinations are functions of random quantities, they also are random vectors, and hence have population means and variances. Moreover, if you are looking at several linear combinations, they will have covariances and correlations as well.
Therefore we are interested in knowing:
Population Mean:
Population mean of a linear combination is equal to the same linear combination of the population means of the component variables. If
\[Y = c_1X_1 + c_2X_2 +\dots c_pX_p =\sum_{j=1}^{p}c_jX_j = \mathbf{c}'\mathbf{X}\].
then
\[E(Y) = c_1 \mu_1 +c_2\mu_2 +\dots + c_p\mu_p = \sum_{j=1}^{p}c_j\mu_j = \mathbf{c}'\mathbf{\mu}\].
Mathematically you express this as the sum of j = 1 to p of c_{p} times the corresponding mean of the j^{th} variable. If the coefficient c's are collected into a vector c and the mean μ are collected into a mean vector μ you can express this as c transposed μ.
We can estimate the population mean by replacing the population means with the corresponding sample means; that is replace all of the μ's with \(\bar{x}\)'s so that \(\bar{y}\) equals c_{1} times \(\bar{x}_{1}\) plus c_{2} times \(\bar{x}_{2}\) and so on...
\[\bar{y} = c_1\bar{x}_1 + c_2\bar{x}_2 + \dots + c_p\bar{x}_p = \sum_{j=1}^{p}c_j\bar{x}_j = \mathbf{c}'\mathbf{\bar{x}}\].
The following table shows the sample means for each of our five nutritional components that we computed in the previous lesson.
Variable

Mean

Calcium 
624.0 mg

Iron 
11.1 mg

Protein 
65.8 mg

Vitamin A 
839.6 μg

Vitamin C 
78.9 mg

If, as previously, we define Y to be the total intake of vitamins A and C (in mg) or :
Y = 0.001X_{4} + X_{5}
Then we can work out the estimated mean intake of the two vitamins as follows:
\(\bar{y}=0.001 \bar{x}_4 +\bar{x}_5 = 0.001 \times 839.6 + 78.9248 = 0.8396 + 78.9248 = 79.7680\) mg.
Linear combinations not only have a population mean but they also have a population variance. The population variance of a linear combination is expressed as the following double sum of j = 1 to p and k = 1 to p over all pairs of variables.
\[var(Y) =\sum_{j=1}^{p}\sum_{k=1}^{p}c_jc_k\sigma_{jk}=\mathbf{c}'\mathbf{\Sigma}\mathbf{c}\].
In each term within the double sum, the product of the paired coefficients c_{j} times c_{k} is multiplied by the covariance between the j^{th} and k^{th} variables. If Σ is the variancecovariance matrix of X, then Var(Y) = c′ Σ c.
Expressions of vectors and matrices of this form are called a quadratic forms.
When using this expression, the covariance between the variables and itself, or σ_{jj} is simply equal to the variance of the j^{th} variable, or σ_{j}^{2 }.
\( \sigma_{jj} = \sigma^2_j\)
The variance of the random variable y can be estimated by the sample variances, or s squared Y. This is obtained by substituting in the sample variances and covariances for the population variances and covariances as shown in the expression below.
\[s^2_Y = \sum_{j=1}^{p}\sum_{k=1}^{p}c_jc_ks_{jk} = \mathbf{c}'\mathbf{S}\mathbf{c}\]
A simplified calculation can be found below. This involves two terms.
\[s^2_Y = \sum_{j=1}^{p}c^2_j \sigma^2_j +2\sum_{j<k}c_jc_ks_{jk}\]
The first term involves summing over all the variables. Here we take the squared coefficients and multiply them by their respective variances. In the second term, we sum over all unique pairs of variables j less than k. Again take the product of c_{j} times c_{k} times the covariances between variables j and k. Since each unique pair appears twice in the original expression, we must multiply the sum by 2.
Looking at the Women's Nutrition survey data we obtained the following variance/covariance matrix as shown below from the previous lesson.
\(S = \left(\begin{array}{RRRRR}157829.4 & 940.1 & 6075.8 & 102411.1 & 6701.6 \\ 940.1 & 35.8 & 114.1 & 2383.2 & 137.7 \\ 6075.8 & 114.1 & 934.9 & 7330.1 & 477.2 \\ 102411.1 & 2383.2 & 7330.1 & 2668452.4 & 22063.3 \\ 6701.6 & 137.7 & 477.2 & 22063.3 & 5416.3 \end{array}\right)\).
If we wanted to take a look at the total intake of vitamins A and C (in mg) remember we defined this earlier as:
Y = 0.001X_{4} + X_{5}
Therefore the sample variance of Y is equal to (0.001)^{2} times the variance for X_{4}, plus the variance for X_{5}, plus 2 times 0.001 times the covariance between X_{4} and X_{5}. The next few lines carries out the mathematical calculations using these values.
\(\begin{array}{lll}s^2_Y & = & 0.001^2s^2_4 + s^2_5 + 2 \times 0.001s_{45}\\ & = & 0.000001 \times 2668452.4 + 5416.3 + 0.002 \times 22063.3\\ & = & 2.7 + 5416.3 + 44.1 \\ & = & 5463.1\end{array}\)
Sometimes we are interested in more than one linear combination or variable. In this case we may be interested in the association between those two linear combinations. More specifically, we can consider the covariance between two linear combinations of the data.
Consider the pair of linear combinations:
\[Y_1 = \sum_{j=1}^{p}c_jX_j \text{and} Y_2 = \sum_{k=1}^{p}d_kX_k\].
Here Y_{1} and Y_{2} are two distinct linear combinations. Both variables Y_{1} and Y_{2} are going to be random and so they will be potentially correlated. We can assess the association between these variables using the covariance as the two vectors c and d are distinct.
The population covariance between Y_{1} and Y_{2} is obtained by summing over all pairs of variables. We then multiply respective coefficients from the two linear combinations as c_{j} times d_{k} times the covariances between j and k.
\[cov(Y_1, Y_2) = \sum_{j=1}^{p}\sum_{k=1}^{p}c_jd_k\sigma_{jk}\]
We can then estimate the population covariance by using the sample covariance. This is obtained by simply substituting the sample covariances between the pairs of variables for the population covariances between the pairs of variables.
\[s_{Y_1,Y_2}= \sum_{j=1}^{p}\sum_{k=1}^{p}c_jd_ks_{jk}\]
The population correlation between variables Y_{1} and Y_{2} can be obtained by using the usual formula of the covariance between Y_{1} and Y_{2} divided by the standard deviation for the two variables as shown below:
\[\rho_{Y_1,Y_2} = \frac{\sigma_{Y_1,Y_2}}{\sigma_{Y_1}\sigma_{Y_2}}\]
This population correlation is estimated by the sample correlation where we simply substitute in the sample quantities for the population quantities as below
\[r_{Y_1,Y_2} = \frac{s_{Y_1, Y_2}}{s_{Y_1}s_{Y_2}}\]
Here is the matrix of the data as was shown previously.
\(S = \left(\begin{array}{RRRRR}157829.4 & 940.1 & 6075.8 & 102411.1 & 6701.6 \\ 940.1 & 35.8 & 114.1 & 2383.2 & 137.7 \\ 6075.8 & 114.1 & 934.9 & 7330.1 & 477.2 \\ 102411.1 & 2383.2 & 7330.1 & 2668452.4 & 22063.3 \\ 6701.6 & 137.7 & 477.2 & 22063.3 & 5416.3 \end{array}\right)\).
We may wish to define the total intake of vitamins A and C in mg as before.
Y_{1} = 0.001X_{4} + X_{5}
and we may also want to take a look at the total intake of calcium and iron:
Y_{2} = X_{1} + X_{2}
Then the sample covariance between Y_{1} and Y_{2} can then be obtained by looking at the covariances between each pair of the component variables time the respective coefficients. So in this case we are looking at pairing X_{1} and X_{4}, X_{1} and X_{5}, X_{2} and X_{4}, and X_{2} and X_{5}. You will notice that in the expression below s_{41}, s_{42}, s_{51} and s_{52} all appear. The variables are taken from the matrix above and substituting them into the expression and the math is carried out below.
\(\begin{array}{lll}s_{Y_1, Y_2} & = & 0.001s_{41} + 0.001s_{42} + s_{51}+s_{52}\\& = & 0.001 \times 102411.1 + 0.001 \times 2383.2 + 6701.6 +137.7\\ & = & 102.4 + 2.4 + 6701.6 + 137.7\\ & = &6944.1 \end{array}\)
You should be able at this point to be able to confirm that the sample variance of Y_{2} is 159,745.4 as shown below:
\(\begin{array}{lll}s^2_{Y_2} & = & s_{11}+s_{22}+2s_{12}\\ & = & 157829.4 + 35.8 + 2 \times 940.1\\ & = & 157829.4 + 35.8 + 1880.2 \\ & = & 159745.4 \end{array}\)
And, if we care to obtain sample correlation between Y_{1} and Y_{2} we take the sample covariance that we just obtained and divide by the square root of the product of the two component variances, 5463.1, for Y_{1} which we obtained earlier and 159745.4 which we just obtained above. Following this math through we end up with a correlation of about 0.235 as shown below.
\[r_{Y_1,Y_2} = \frac{s_{Y_1, Y_2}}{s_{Y_1}s_{Y_2}}= \frac{6944.1}{\sqrt{5463.1 \times 159745.4}}=0.235\]
In this lesson we learned about:
Next, complete the homework problems that will give you a chance to put what you have learned to use...