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$\newcommand{\Var}{\operatorname{Var}}$

I guess the answer is extremely easy, but I think I've missed something...

When we have a set of $n$ independent identically distributed variables (with the binomial distribution), we can deduce the variance by : $\Var\left(\sum_{i=1}^n X_i\right) = \sum_{i=1}^n \Var(X_i)$ Then $\Var(X_i)=pq \:\forall i$ (with standard notation), because of the identical distributed variables, hence : $\Var\left(\sum_{i=1}^n X_i\right) = npq$. But by using first the argument of identically distributed variables, we have : \begin{equation}\Var\left(\sum_{i=1}^{n} X_i\right) = \Var(nX_1)=n^2\Var(X_1)=\cdots\end{equation} Where is the error ?

Michael Hardy
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faero
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    Maybe $Var\left(\sum X_i\right) \neq Var(nX_i)$ ? – tenpercent Dec 04 '13 at 15:46
  • I guess this is the flaw, but I don't understand why. I assumed that it is what identically distributed variables mean, because they follow same distribution, and it is equivalent to taking one n times. – faero Dec 04 '13 at 16:08
  • You see, $Var(\sum X_i) = \sum Var(X_i)$ only when $X_i$ are independent. So, I think that might be the flaw in the first equation. I doubt about the equivalence, because the variables are random. – tenpercent Dec 04 '13 at 16:14
  • You think there is a flaw in the first equation ? – faero Dec 04 '13 at 16:25
  • Yes, I think you can't substitute $\sum X_i$ with $nX_1$ in this case. That's because they have different distributions, moments, etc. – tenpercent Dec 04 '13 at 16:30
  • But assuming identical distributions doesn't imply these are the same ? – faero Dec 04 '13 at 16:39

2 Answers2

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In the sum X1+X2+...+Xn the values of the variables might be different, the first equation assumes that the values are equal. So the variances do not coincide.

Peter
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The identity $$ \operatorname{var}(X_1+\cdots+X_n)=\operatorname{var}(X_1)+\cdots+\operatorname{var}(X_n) $$ applies if the random variables are independent or even if they're only uncorrelated. But in the expression $\operatorname{var}(X_1+\cdots+X_1)$ (where they're all $X_1$, rather than $X_1,\ldots,X_n$), they're nowhere near uncorrelated.

Michael Hardy
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