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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 $\mathbb ES = (\mathbb EN)( \mathbb EX).$

(The answer is 120.)

How to find $(\mathbb EN)$?

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    $\begingroup$ 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? $\endgroup$ – Minus One-Twelfth Apr 5 at 19:34
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    $\begingroup$ What does $\Bbb D$ denote? $\endgroup$ – J.G. Apr 5 at 19:34
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    $\begingroup$ $\newcommand{\E}{\mathbb{E}}\newcommand{\P}{\mathbb{P}}$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 $\E[Z] = q/p$. $\endgroup$ – Minus One-Twelfth Apr 5 at 19:41
  • $\begingroup$ Oh, I forgot you used $N$ for the Negative Binomial random variable. You can replace the $X$ above with $N$ if you want. $\endgroup$ – Minus One-Twelfth Apr 5 at 19:47

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