Compound distributions when $N$ and $\xi_{i}$ are correlated

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I know the following theorem: Let me define $\eta=\sum_{i=1}^{N}\xi_{i}$, where $\left(\xi_{i}\right)_{i}$ variables are $iid$ and they are independent from the non-negative integer random variable $N$, then

$$\mathbb{E}\left(\eta\right)=\mathbb{E}\left(N\right)\cdot\mathbb{E}\left(\xi_{i}\right)$$

and $$\mathbb{D}^{2}\left(\eta\right)=\mathbb{D}^{2}\left(\xi_{i}\right)\mathbb{E}\left(N\right)+\mathbb{E}^{2}\left(\xi_{i}\right)\mathbb{D}^{2}\left(N\right).$$

What happens if $N$ and each $\xi_{i}$ are corraleted?

$$\mathbb{E}\left(\eta\right) =\mathbb{E}\left(\mathbb{E}\left(\eta\mid N=n\right)\right)=\sum_{n=0}^{\infty}\mathbb{E}\left(\sum_{i=1}^{N}\xi_{i}\mid N=n\right)\mathbf{P}\left(N=n\right)$$

In case when $N$ and $\xi_{i}$ are independent, then

$$\mathbb{E}\left(\sum_{i=1}^{N}\xi_{i}\mid N=n\right)=\frac{\mathbb{E}\left(\sum_{i=1}^{N}\xi_{i}\cdot\chi_{\left(N=n\right)}\right)}{\mathbb{E}\left(\chi_{\left(N=n\right)}\right)}=\frac{\mathbb{E}\left(\sum_{i=1}^{n}\xi_{i}\cdot\chi_{\left(N=n\right)}\right)}{\mathbb{E}\left(\chi_{\left(N=n\right)}\right)}=\frac{\mathbb{E}\left(\sum_{i=1}^{n}\xi_{i}\right)\mathbb{E}\left(\chi_{\left(N=n\right)}\right)}{\mathbb{E}\left(\chi_{\left(N=n\right)}\right)}=\mathbb{E}\left(\sum_{i=1}^{n}\xi_{i}\right), $$

but if $N$ and $\xi_{i}$ are not independent, then

$$\mathbb{E}\left(\sum_{i=1}^{N}\xi_{i}\mid N=n\right)=\frac{\mathbb{E}\left(\sum_{i=1}^{N}\xi_{i}\cdot\chi_{\left(N=n\right)}\right)}{\mathbb{E}\left(\chi_{\left(N=n\right)}\right)}=\frac{\mathbb{E}\left(\sum_{i=1}^{n}\xi_{i}\cdot\chi_{\left(N=n\right)}\right)}{\mathbb{E}\left(\chi_{\left(N=n\right)}\right)}=? $$