📂Probability Distribution

Mean and Variance of the Negative Binomial Distribution

Formula

$X \sim \text{NB}(r, p)$ Plane $$E(X) = {{ r (1-p) } \over { p }} \\ \Var(X) = {{ r (1-p) } \over { p^{2} }}$$

Proof

Strategy: Use the fact that the negative binomial distribution is a generalization of the geometric distribution.

• [b] Generalization of Geometric Distribution: If $Y = X_{1} + \cdots + X_{r}$ and $X_{i} \overset{\text{iid}}{\sim} \text{Geo}(p)$ then $Y \sim \text{NB}(r,p)$

Here, the definition of the geometric distribution is set so that its support is like $\mathcal{S} = \left\{ 0 , 1 , 2, \cdots \right\}$.

Mean and Variance of Geometric Distribution: If $X \sim \text{Geo} (p)$ then $$E(X) = {{ 1-p } \over { p }} \\ \Var(X) = {{ 1-p } \over { p^{2} }}$$

Mean

Since $Y=X_1+X_2+\cdots+X_r$ $$\begin{align*} E(Y) =& E(X_1)+E(X_2)+\cdots+E(X_r) \\ =& \sum_{i=1}^{r} E(X_i) \\ =& { {r(1-p)} \over p } \end{align*}$$ and since $Y \sim \text{NB} (r,p)$, then $\displaystyle E(Y) = { {r(1-p)} \over p }$

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Variance

Since $Y=X_1+X_2+\cdots+X_r$ and $X_1, X_2, \cdots , X_r$ are mutually independent, the covariance is $0$. $$\begin{align*} Var(Y) =& Var(X_1)+Var(X_2)+\cdots+Var(X_r) \\ =& \sum_{i=1}^{r} Var(X_i) \\ =& \frac { r(1-p) }{ { p }^{ 2 } } \end{align*}$$ Since $Y \sim \text{NB} (r,p)$, then $\displaystyle Var(Y) = \frac { r(1-p) }{ { p }^{ 2 } }$

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