# Expected value of the sum

Let $$S = \sum_{i=1}^{N} X_i.$$

$$\mathbb EX = 1, \mathbb DX = 2.$$

$$N$$ has a negative binomial distribution with parameters $$k=80$$ and $$p=0.4.$$

$$\mathbb P(N=l)=\begin{pmatrix} k+l-1 \\ l \\ \end{pmatrix} p^kq^l, l=0,1,2,.. .$$ Find $$\mathbb ES = (\mathbb EN)( \mathbb EX).$$

How to find $$(\mathbb EN)$$?
• You could simply recall the formula for the expected value of a Negative Binomial random variable, which is $\color{blue}{kq/p}$ here. This gives $80\times 0.6/0.4 = 120$. Or are you asking for a proof of the formula? – Minus One-Twelfth Apr 5 at 19:34
• What does $\Bbb D$ denote? – J.G. Apr 5 at 19:34
• $\newcommand{\E}{\mathbb{E}}\newcommand{\P}{\mathbb{P}}$If you're asking for a proof of the formula for the expected value of a Negative Binomial random variable, perhaps the easiest way is to recall that such a Negative Binomial random variable $X$ is equal in distribution to a sum of $k$ Geometric$(p)$ random variables, so the expected value is $\E[ X] = k\times \E[Z]$ where $Z\sim\mathsf{Geom}(p)$ (i.e. $\P(Z=l) = q^l p$ for $l=0,1,2,\ldots$). Then recall or prove that $\E[Z] = q/p$. – Minus One-Twelfth Apr 5 at 19:41
• Oh, I forgot you used $N$ for the Negative Binomial random variable. You can replace the $X$ above with $N$ if you want. – Minus One-Twelfth Apr 5 at 19:47