Timeline for Confusion related to calculation of variance

6 events

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Dec 21, 2012 at 22:04	comment	added	Dilip Sarwate		$$E[X(X-1)]=\sum_{i=0}^{n-1}i(i-1)\binom{n-1}{i}p^i(1-p)^{n-1-i} = \sum_{i=2}^{n-1}i(i-1)\binom{n-1}{i}p^i(1-p)^{n-1-i}$$ since the first two terms of the middle sum are $0$. Now write $\binom{n-1}{i}=\frac{(n-1)!}{i!(n-1-i)!}$, cancel $i(i-1)$ and simplify.
Dec 21, 2012 at 22:02	history	tweeted			twitter.com/#!/StackStats/status/282244578424266753
Dec 21, 2012 at 21:29	answer	added	Douglas Zare		timeline score: 3
Dec 21, 2012 at 20:58	comment	added	user34790		what is E[X(X-1)]] for a binomial distribution how is it calculated?
Dec 21, 2012 at 16:56	comment	added	Dilip Sarwate		There is most likely an assumption that the events $A_i = \text{"I have an edge with neighbor}~i"$ are $n-1$ mutually independent events, making the degree a binomial random variable with parameters $(n-1,p)$. The result then follows from the properties of binomial random variables. If you want an explicit derivation, use the result that $$\text{var}(X) = E[X(X-1)] + E[X] - (E[X])^2$$ since $E[X(X-1)]$ is easier to compute for a binomial random variable than the $E[X^2]$ needed for applying the standard result $$\text{var}(X) = E[X^2] - (E[X])^2.$$
Dec 21, 2012 at 16:47	history	asked	user34790	CC BY-SA 3.0