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Consider a vector $\{v_1,...,v_n\}$, $v_k\in B$, where $B$ consists of $B_1,...,B_m$ disjoint subsets and let \begin{align*} L(v_k) = \begin{cases} 1 \quad \text{, if a specific event occurs for} \quad v_k \newline 0 \quad \text{, else} \end{cases} \end{align*}

The probability that a specific event occurs for $v_k$ is $p_i\in(0,1)$. The probabilities $p_i$, $i\in\{1,...,m\}$, are exposed to random fluctuations $\vartheta_i$, $i\in\{1,...,m\}$. These fluctuations are realizations of random variables $\theta_i$, $i\in\{1,...,m\}$, with properties \begin{align} \mathbb{E}[\theta_i]=1 \quad \text{and} \quad Cov(\theta_i,\theta_j)=\begin{cases} \sigma^2 \quad \text{, if} \quad i=j \newline \rho\sigma^2 \quad \text{, if} \quad i\neq j \end{cases} \end{align} Further, let \begin{align*} S_i:=\sum_{v_k\in B_i} L(v_k) \quad \text{and} \quad S:=\sum_{l=1}^{m} S_l \end{align*}

Let us make the following two assumptions

\begin{align} S_i | (\theta_1,...,\theta_m) \sim Poi(n_i p_i \theta_i)\quad ,n_i \in \mathbb{N}\newline Cov(S_i,S_j | (\theta_1,...,\theta_m)) = 0 \quad \text{, for} \quad i \neq j, \end{align}

Question 1: How can I calculate $Cov(S_i,S)$?

Question 2: Following the calculation provided below by K. A. Buhr for $Cov(S_i,S)$, I now also tried to calculate $Var(S)$ by the law of total variance: \begin{align} Var(S)=\mathbb{E}[Var(S|\theta^*)]+Var(\mathbb{E}[S|\theta^*]) \end{align} The first term is then just $\mathbb{E}[\sum_{l=1}^m n_l p_l \theta_l]=\sum_{l=1}^m n_l p_l$, assuming $\mathbb{E}[\theta_i]=1$. The seond term becomes \begin{align} Var[\sum_{l=1}^m n_l p_l \theta_l] =\sum_{l=1}^m Var(n_l p_l \theta_l)+\sum_{l\neq j} Cov[n_l p_l \theta_l,n_j p_j \theta_j] =\sigma^2(\sum_{l=1}^m n_l^2 p_l^2 + \rho \sum_{l\neq j} n_l p_l n_j p_j) \end{align} Therefore, \begin{align} Var(S)=\sum_{l=1}^m n_l p_l+\sigma^2(\sum_{l=1}^m n_l^2 p_l^2 + \rho \sum_{l\neq j} n_l p_l n_j p_j) \end{align} Are my calculations for $Var(S)$ correct?

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  • $\begingroup$ Can you clarify what you mean by $Poi(n_i p_i\theta)$? (1) What is $\theta$, (2) are the $n_i$ and $p_i$ assumed fixed constants, and (3) how do the $\theta_i$ relate to any of these parameters? $\endgroup$
    – K. A. Buhr
    Dec 19, 2023 at 19:09
  • $\begingroup$ My formulation was indeed not very precice, sorry. I clarified the problem. $\endgroup$ Dec 20, 2023 at 8:22

1 Answer 1

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Note that you have Poisson rates $n_i p_i \theta_i \geq 0$ but $E(\theta_i)=0$. I'll assume you actually mean $E(\theta_i)=1$ (or equivalently, rates $n_i p_i (1+\theta_i)$ with $E(\theta_i)=0$).

Anyway, you should be able to apply the "law of total covariance": $$\mathrm{Cov}(X,Y)=E[\mathrm{Cov}(X,Y\mid Z)] + \mathrm{Cov}(E[X\mid Z], E[Y\mid Z])$$

Writing $\theta_\star:=(\theta_1,\dots,\theta_n)$, by the conditional (lack of) correlation and marginal distributions of the $S_i$, you have: $$\mathrm{Cov}(S_i,S\mid\theta_\star) = \mathrm{Var}(S_i\mid\theta_\star) = n_i p_i \theta_i$$ and so $E[\mathrm{Cov}(S_i,S\mid\theta_\star)] = n_i p_i$.

You also have $E[S_i\mid\theta_\star] = n_i p_i \theta_i$ and $E[S\mid\theta_\star] = \sum_j n_j p_j \theta_j$ giving: $$\mathrm{Cov}(E[S_i\mid\theta_\star],E[S\mid\theta_\star]) = \sum_j n_i n_j p_i p_j \mathrm{Cov}(\theta_i,\theta_j) = \sigma^2 n_i p_i \left( n_i p_i + \rho\sum_{j\not=i} n_j p_j \right)$$

The law of total covariance then gives: $$\mathrm{Cov}(S_i,S) = n_i p_i + \sigma^2 n_i p_i \left(n_i p_i + \rho \sum_{j\neq i} n_j p_j\right)$$

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  • $\begingroup$ Thank you very much! I added a second question and there I tried to also calculate $Var(S)$ by applying similar ideas as yours. Are my calculations correct? $\endgroup$ Dec 21, 2023 at 7:19

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