1

Given (R,X) a multivariate discrete random variable with joint pdf:

$f_{R,X}(r,x)=$ ${x-1}\choose {r-1}$ $ \cdot a^{r}(1-a)^{x-r} \cdot $ ${n}\choose {r} $ $ b^{r}(1-b)^{n-r}$

with support $x=r,r+1,r+2,...$ and $r=0,1,2, ... n$.

The distribution has three parameters: $n \in N^{+}, $ $ a \in (0;1),$ $b \in (0;1). $

Compute $Cov(X,R)$

$Cov(X,R)=E(X\cdot R)-E(X)\cdot E(R) $

First I transformed ${x-1}\choose {r-1}$ $=\frac{(x-1)!}{(x-r)!(r-1)!}=\frac{r\cdot(x)!}{(x-r)!(r)!\cdot x}=\frac{r}{x}\cdot\frac{x!}{(x-r)!\cdot r!}$.

$ E(X\cdot R)=\sum_{r=0}^{n}\sum_{x=r}^{\infty}x\cdot r\cdot\frac{r}{x}\cdot\frac{x!}{(x-r)!\cdot r!}a^{r}(1-a)^{x-r} \cdot $ ${n}\choose {r}$ $ b^{r}(1-b)^{n-r}$

What can I do then?

Hrackadont
  • 581

1 Answers1

1

Building on @StubbornAtom's comment, you have to indeed use that for a random variable $Y$ that follows a negative binomial distribution with parameter $r \in \mathbb{N}_{+}$ (so excluding 0, unlike in the question you stated. This is probably a typo, because if not then for $r = 0$ you get some issues I think), and success probability $p \in (0,1)$, the pmf is given by $$ P(Y = y) = {y-1 \choose r-1}p^{r}(1-p)^{x-r}, y = r, r+1, \dots $$ and as shown in the the other thread, we have $$ E[Y] = \sum_{y=r}^{\infty}y{y-1 \choose r-1}p^{r}(1-p)^{x-r} =\frac{r}{p}. $$

I will show in detail how to compute $E[X]$, the rest is similar.

$$ \begin{align*} E[X] &= \sum_{r=1}^{n}{n\choose r}b^{r}(1-b)^{n-r} \underbrace{\sum_{x=r}^{\infty}x {x-1 \choose r-1}a^{r}(1-a)^{x-r}}_{r/a, \text{ by the above}}\\ &= \frac{1}{a}\underbrace{\sum_{r=1}^{n}r{n\choose r}b^{r}(1-b)^{n-r}}_{\text{Expectation of binomial(n, b) random variable = nb}}\\ &= \frac{nb}{a} \end{align*} $$

Very similarly, you can find $E(R) = nb$, and $E(XR) = \frac{1}{a}(nb(1-b) + n^{2}b^{2})$. Combining all of these yields $cov(X,R) = \frac{nb(1-b)}{a}$.

minginator
  • 441