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3.16 Effect of scaling and offset on average and standard deviation. Suppose x is an n-vector and a and ß are scalars. (a) Show that avg(ax +31) = a avg(x) + ß. (b) Show that std(ax + 1) = |a|std(x).

3.16 Effect of scaling and offset on average and standard deviation. Suppose x is an n-vector and a and ß are scalars. (a) Show that avg(ax +31) = a avg(x) + ß. (b) Show that std(ax + 1) = |a|std(x).

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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3.16 Effect of scaling and offset on average and standard deviation. Suppose x is an n-vector and a and ß are scalars. (a) Show that avg(ax +31) = a avg(x) + B. (b) Show that std(ax +31) = |a|std(x).
Transcribed Image Text:3.16 Effect of scaling and offset on average and standard deviation. Suppose x is an n-vector and a and ß are scalars. (a) Show that avg(ax +31) = a avg(x) + B. (b) Show that std(ax +31) = |a|std(x).
Standard deviation of sum. We can derive a formula for the standard deviation of a sum from (3.6): Vstd(a)² + 2p std(a) std(b) + std(b)². (3.9) To derive this from (3.6) we let ã and b denote the de-meaned versions of a and b. Then a + b is the de-meaned version of a + b, and std (a + b)² = ||ã + ỗ||²/n. Now using (3.6) and p = cos / (a, b), we get n std (a + b)² || +ỗ||2 std (a + b) = = = = = ||ã||²+2p||ã|||||| + ||õ|| ² n std(a)² + 2pn std(a) std(b) + n std(b)². Dividing by n and taking the squareroot yields the formula above. If p 1, the standard deviation of the sum of vectors is the sum of their standard deviations, i.e., std(a + b) = std(a) + std(b). As Р decreases, the standard deviation of the sum decreases. When p = 0, i.e., a and b are uncorrelated, the standard deviation of the sum a + b is std (a + b) = √/std(a)² + std(b)², which is smaller than std(a) + std(b) (unless one of them is zero). When p= -1, the standard deviation of the sum is as small as it can be, std (a + b) = | std(a) — std(b)|.
Transcribed Image Text:Standard deviation of sum. We can derive a formula for the standard deviation of a sum from (3.6): Vstd(a)² + 2p std(a) std(b) + std(b)². (3.9) To derive this from (3.6) we let ã and b denote the de-meaned versions of a and b. Then a + b is the de-meaned version of a + b, and std (a + b)² = ||ã + ỗ||²/n. Now using (3.6) and p = cos / (a, b), we get n std (a + b)² || +ỗ||2 std (a + b) = = = = = ||ã||²+2p||ã|||||| + ||õ|| ² n std(a)² + 2pn std(a) std(b) + n std(b)². Dividing by n and taking the squareroot yields the formula above. If p 1, the standard deviation of the sum of vectors is the sum of their standard deviations, i.e., std(a + b) = std(a) + std(b). As Р decreases, the standard deviation of the sum decreases. When p = 0, i.e., a and b are uncorrelated, the standard deviation of the sum a + b is std (a + b) = √/std(a)² + std(b)², which is smaller than std(a) + std(b) (unless one of them is zero). When p= -1, the standard deviation of the sum is as small as it can be, std (a + b) = | std(a) — std(b)|.
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