If $X \sim Bin(n,p)$, using $E(X(X-1)) = g''(1) = n(n-1)p^2$ show that Var$(X) = npg$.

Question

If $X \sim Bin(n,p)$ using $E(X(X-1)) = g''(1) = n(n-1)p^2$ show that Var$(X) = npg$.

I understand that g is the generating function $g_{x}(t) = \sum_{k=0}^{\infty} p_{k}t^{k}$, and I know that the mean, E(X) = $g''(1) = n(n-1)p^2$. But does E(X(X-1) represent? And how does it prove that the Var(X) = npq?

Answer 1

Observing that $$\begin{align} g'(1) &=\Bbb E(X)\\ g''(1)&= \Bbb E(X(X − 1)) \end{align} $$ so you have $$ \begin{align} g''(1)+g'(1)-[g'(1)]^2 &=\Bbb E(X(X − 1))+\Bbb E(X)-[\Bbb E(X)]^2\\ &= \Bbb E(X^2) -\Bbb E(X)+\Bbb E(X)-[\Bbb E(X)]^2\\ &= \Bbb E(X^2) -[\Bbb E(X)]^2\\ n(n-1)p^2+np-(np)^2&=\mathrm{Var}(X)=npq \end{align} $$

Answer 2

\begin{align*} \operatorname{Var}(X)&=\mathbb{E}[X^2]-\mathbb{E}[X]^2\\ &= \mathbb{E}[X(X-1)]+\mathbb{E}[X]-\mathbb{E}[X]^2 & \text{linearity}\\ &= n(n-1)p^2 + np - (np)^2 & \mathbb{E}[X]=np\\ &= np[(n-1)p + 1 - np]\\ &= np(1-p)\\ &= npq \end{align*}

Answer 3

Note for pgf the general differentiation rule for m th derivative( repeated derivative m times), when s = 1 is given by

g^m (1) = E( X(X - 1) ......... ( X - m + 1))

and so it follows g^2(1) = n(n - 1 )p^2 for the Binomial R.V. X