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This is a simple question:

Do we need mutual independence or only pairwise independence in order to state that

$$\mathrm{Var}\left[\sum_{i=1}^n X_i\right] = \sum_{i=1}^n \mathrm{Var}\left[X_i\right]?$$

As I do not know what uncorrelated means (I know that this is the actual condition), I am not sure whether it is enough for each pair to be independent.

Thanks for your help.

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    $\begingroup$ Pairwise is enough, because uncorrelatedness is enough. Uncorrelated means $\operatorname{Cov}(X_i,X_j) = 0$, and this is a condition on two variables. $\endgroup$ – user191371 Nov 11 '14 at 8:46
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Edit (with mainly the same content and some adaptions)

Pairwise independence is enough here. In general:

$$\text{Var}\left(\sum_{i=1}^{n}X_{i}\right)=\sum_{i=1}^{n}\sum_{j=1}^{n}\text{Cov}\left(X_{i},X_{j}\right)$$

If $X_{i}$ and $X_{j}$ are independent then $\mathbb{E}X_{i}X_{j}=\mathbb{E}X_{i}\mathbb{E}X_{j}$ and consequently $$\text{Cov}\left(X_{i},X_{j}\right):=\mathbb{E}X_{i}X_{j}-\mathbb{E}X_{i}\mathbb{E}X_{j}=0$$ Or in words: $X_{i}$ and $X_{j}$ are uncorrelated. This leads to:

$$\text{Var}\left(\sum_{i=1}^{n}X_{i}\right)=\sum_{i=1}^{n}\text{Var}X_{i}$$

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    $\begingroup$ Why is this downvoted? One could argue that the question should be closed as a multi duplicate, but, if the question stays open... $\endgroup$ – Did Nov 11 '14 at 9:41
  • $\begingroup$ I find that this is not the answer to my question. I asked for mutually or pairwise and explicitly stated, that I know that the actual condition is correlation. He then answers that uncorrelated is enough whih is weaker than mutually independent. BUt this does not say anything on pairwise independent. So actually I do not think that the answer is ok. But the comment of user191371 answered it perfectly. $\endgroup$ – user136457 Nov 11 '14 at 11:48
  • $\begingroup$ "does not say anything on pairwise independent." That is not true. If the variables are pairwise independent then they are uncorrelated (i.e. their covariance equals $0$) so the answer above tells you that pairwise independence is indeed enough. That is the essence of the word 'weaker' here. Can you mention me one thing that you have met in the indeed excellent comment of @user136457 and was not met in my answer? That is a condition on two variables is immediately clear from the definition. $\endgroup$ – drhab Nov 11 '14 at 12:10
  • $\begingroup$ You are saying in your question that you don't know what uncorrelated means. So I explained that in my answer. Maybe it should have been said explicitly that mutual independence implies uncorrelated. I cannot think of any other missing link. Besides that nothing is wrong with it and it answers your question in the same line as the comment of @user136457. $\endgroup$ – drhab Nov 11 '14 at 12:22
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    $\begingroup$ I am glad with your reaction and hope that my edit meets your needs. If not then let me know. Cheers. $\endgroup$ – drhab Nov 11 '14 at 20:09

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