Covariance of Geometric Distribution and Negative Binomial Distribution

Question

Let independents $X_1,...X_n \thicksim \text{Bernoulli}(1,p)$ and $W_r$ refer to total number of trial until the $r$-th success, which means $W_r \thicksim \text{NegBin}(r,p)$ then I need to evaluate $Cov(W_1,W_r)$.

First I had checked out whether the $W_1$ and $W_r$ independent or not, but it has been revealed they are dependent since $P(W_1=x_1)\cdot P(W_r=x_r) \neq P(W_1=x_1 , W_r=w_r)$

Then what I might rely on is the definition of Covariance, \begin{align} \text{Cov}(W_1,W_r) &= E\Bigl[ \bigl( W_1 - E[W_1] \bigr) \cdot \bigl( W_r - E[W_2] \bigr) \Bigr] \\ &= E\Bigl[ \bigl( W_1 - \frac1{p} \bigr) \cdot \bigl( W_r - \frac{r}{p} \bigr) \Bigr] \\ & = E\Bigl[ W_1 \cdot W_r - \frac1{p} W_r - \frac{r}{p} W_1 + \frac{r}{p^2} \Bigr] \end{align}

My question is, how could I get the value of $E[W_1\cdot W_r]$?

Answer 1

Hint:$$\mathbb E[W_1W_r\mid W_1=n]=\mathbb E[nW_r\mid W_1=n]=n\mathbb E[W_r\mid W_1=n]=n[n+\mathbb EW_{r-1}]$$

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edit:

So: $$\mathbb E[W_1W_r\mid W_1]=W_1[W_1+\mathbb EW_{r-1}]$$ leading to:

$$\mathbb E[W_1W_r]=\mathbb E[W_1[W_1+\mathbb EW_{r-1}]]=\mathbb EW_1^2+\mathbb EW_1\mathbb EW_{r-1}$$

Then:

$$\text{Covar}(W_1,W_r)=\mathbb EW_1W_r-\mathbb EW_1\mathbb EW_r=\mathbb EW_1^2+\mathbb EW_1\mathbb EW_{r-1}-\mathbb EW_1\mathbb EW_r$$$$=\mathbb EW_1^2-\mathbb EW_1\mathbb E(W_r-W_{r-1})=\mathbb EW_1^2-(\mathbb EW_1)^2=\text{Var}W_1$$

Answer 2

$Cov(W_1,W_r) = E[(W_1-E[W_1])\cdot (W_r - E[W_2])]= E[(W_1-1/p)\cdot(W_r-r/p)]=E[W_1\cdot W_r - 1/pW_r-r/pW_1+r/p^2]\\=E[W_1\cdot W_r]-1/p\cdot r/p-r/p\cdot 1/p+r/p^2\\=E[W_1\cdot (W_{r}-W_1)+ W_1^2]-r/p^2\\=E[W_1](E[W_{r}]-E[W_1])+E[W_1^2] -r/p^2\\=1/p(r/p-1/p)+(2-p)/p^2-r/p^2\\=(1-p)/p^2\\=Var(W_1)$