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Let a random variable $X$ be the number of boys out of $n$ children. The probability to have a boy or a girl is $0.5$. Calculate $V(2X-n)$.

I know that $Var(2X-n)=4V(X)$.

$\mathbb{P}(X=k)={1\over 2^n}\binom{n}{k}$. Thus $\mathbb{E}(X)=\sum_{i=1}^n{1\over 2^n}\binom{n}{k}\cdot k$, and $\mathbb{E}(X^2)=\sum_{i=1}^n{1\over 2^n}\binom{n}{k}\cdot k^2$. I'm not sure how to keep on.

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$$V(2X - n) = V(2X) = 2^2V(X)$$ $X$ follows a binomial distribution with $n$ trials and $p=0.5$, which has $V(X) = np(1-p)$ $$V(2X - n) = 4np(1-p)$$

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    $\begingroup$ $V(2X-n)=4np(1-p)=4n\cdot 0.5\cdot 0.5=n$? $\endgroup$ – J. Doe Feb 14 at 15:17
  • $\begingroup$ looks good @J.Doe $\endgroup$ – Ahmad Bazzi Feb 14 at 15:17
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HINT $$ \mathrm{Var}\ X = \mathbb{E}\left[(X-\mathbb{E}[X])^2 \right] = \mathbb{E}\left[X^2 \right] - (\mathbb{E}[X])^2 $$

UPDATE

The sums are handled separately. Note that $$ \begin{split} \mathbb{E}[X] &= 2^{-n} \sum_{k=1}^n k \binom{n}{k} \\ &= 2^{-n} \sum_{k=1}^n k \frac{n (n-1)!}{k (k-1)! (n-k)!} \\ &= \frac{n}{2^n} \sum_{k=1}^n \frac{(n-1)!}{(k-1)! (n-k)!} \\ &= \frac{n}{2^n} \sum_{k=0}^{n-1} \frac{(n-1)!}{k! (n-k)!} \\ &= \frac{n}{2^n} 2^{n-1} = n/2. \end{split} $$

The last sum conversion uses the Binomial Theorem. The square sum is transformed similarly to $n/4$.

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  • $\begingroup$ If I sum up the two sums I have I get a monster sum which I don't know how to handle @gt $\endgroup$ – J. Doe Feb 14 at 15:13
  • $\begingroup$ @J.Doe see the update $\endgroup$ – gt6989b Feb 14 at 15:46

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