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I have a normal distribution with mean m and standard deviation d.

How can I calculate what will be the deviation if I add N such distributions? As I understand, mean will be m*N. Will standard deviation be d*N?

P.S. It looks like I should be using standard variances vs standard deviation.

Sum of Gauss(Mean1, Standard Variance1) + Gauss(Mean2, Standard Variance2) = Gauss (Mean1 + Mean2, Standard Variance1 + Standard Variance 2).

This way standard variance for N will be v*N, which mean that d=sqrt(N)*d

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closed as off-topic by Andy Ross, Karsten 7., Michael E2, Teake Nutma, m_goldberg Sep 11 '14 at 8:31

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This question appears to be off-topic because it is about statistics. It belongs on CrossValidated –  Andy Ross Sep 11 '14 at 0:21

1 Answer 1

up vote 1 down vote accepted

EDIT: Correction to erroneous first response.


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where the x[k] are independent identically distributed normal variables each with distribution NormalDistribution[m, d]. Assume that y[n] is distributed NormalDistribution[n*m, Sqrt[n]*d] then the distribution for y[n+1] = (y[n] + x[n+1]) is

dist = Assuming[{d > 0, n > 0}, TransformedDistribution[ y + x[n + 1], {y \[Distributed] NormalDistribution[n*m, Sqrt[n]*d], x[n + 1] \[Distributed] NormalDistribution[m, d]}] == NormalDistribution[(n + 1)*m, Sqrt[n + 1]*d] // Simplify]


Since the assumption is true for y[1] = x[1] and true for y[n+1] given that it is true for y[n], then by induction it is true for all positive integers, n.

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Frankly, I am not sure how to read this notation. –  Victor Ronin Sep 10 '14 at 23:50
Bob, I appreciate your help, but my mathematical knowledge is super rusty :) I touched all these last time about 10 years ago in university. So, what is the bottom line? Will deviation be N*d or sqrt(N)*d? –  Victor Ronin Sep 11 '14 at 0:32
The standard deviation for y[n] is Sqrt[n]*d –  Bob Hanlon Sep 11 '14 at 0:35

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