# Covariance of random variable and sum of random variables

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?

• 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? Dec 19, 2023 at 19:09
• My formulation was indeed not very precice, sorry. I clarified the problem. Dec 20, 2023 at 8:22

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: $$\mathrm{Cov}(S_i,S) = n_i p_i + \sigma^2 n_i p_i \left(n_i p_i + \rho \sum_{j\neq i} n_j p_j\right)$$
• 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? Dec 21, 2023 at 7:19