Given standard deviation of two of the variables in a function, calculate the standard deviation of the output
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Given a function: z = 2*(x/(x+y)), assuming standard deviation of both x and y are 20%, find the standard deviation and worst case of z.
I tried the documentation page https://www.mathworks.com/help/matlab/ref/std.html but could not find anything useful.
Thanks in advance.
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John Chilleri
am 26 Jan. 2017
Bearbeitet: John Chilleri
am 26 Jan. 2017
Hello,
Assuming x and y have the same mean (among other things), I'd say about 14.61%. I simulated it as follows:
for i = 1:1000
x = 1+randn(10000,1)*.2;
y = 1+randn(10000,1)*.2;
z = 2*(x./(x+y));
std_Z(i) = std(z);
end
mean(std_Z)
As an explanation, x and y have approximate mean 1 and approximate standard deviation .2 (so 20% standard deviation I guess), and z is also approximate mean 1. With 1000 trials, the average standard deviation of z was 14.61% (.1461).
Hope this helps!
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Walter Roberson
am 26 Jan. 2017
Bearbeitet: Walter Roberson
am 26 Jan. 2017
https://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables tells you had to find the standard deviation of x+y . In short, with two variables with the same stand deviation, the standard deviation of the sum will be sqrt(2) times the individual standard deviations.
http://www.talkstats.com/showthread.php/16499-Standard-deviation-of-the-Ratio-of-two-means and look at #5 to see how to find the standard deviation of ratios.
You can put the two together to figure out the standard deviation of the expression.
The result cannot be expressed numerically just in the information you are given. The result depends upon the mean of x and y, which no information is given about. Telling us that the standard deviation is 20% (does that mean 1.2 ?) does not tell us anything about their means. And it matters a lot, since knowing the means gives you information about the probability that the sum X + Y is 0 for X element of x and Y element of y. Imagine for example that mean of x is -2 and mean of y is +2 then the most probably sum is 0.
The worst case for z is of course +/- infinity . Normal distribution has infinite tails so no matter how large the mean, there is a probability that one will be sufficiently negative to be equal to the negative of other, which gives you an infinity with the sign of x.
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