Question

Let $\left(X_i,Y_i\right),i=1,\dots$, be a sequence of independent and identically distributed random vectors. That is, $X_1,Y_1$is independent of, and has the same distribution as, $X_2,Y_2$, and so on. Although $X_i$and $Y_i$can be dependent, $X_i$and $Y_j$are independent when $i\ne j$. Let
${\mu }_x=E\left[X_i\right],{\mu }_y=E\left[Y_i\right],{\sigma }_x^2=Var\left(X_i\right)$

${\sigma }_y^2=Var\left(Y_i\right),\rho =Corr\left(X_i,Y_i\right)$

Find $Corr\left(\sum _{i=1}^n{X}_i,\sum _{j=1}^n{Y}_j\right)$.

Short answer

Therefore,
$Corr\left(\sum _i{X}_i,\sum _j{Y}_j\right)=\rho$

Step-by-step solution

Concept Introduction

Let $\left(X_i,Y_i\right)$for all $i=1,2,\dots$ be a sequence of independent and identically distributed random vectors.

Explanation

Let $\left(X_i,Y_i\right)$for all $i=1,2,\dots$ be a sequence of independent and identically distributed random vectors.
From the given information:

${\mu }_x=E\left(X_i\right)$and ${\mu }_y=E\left(Y_i\right)$

${\sigma }_x^2=Var\left(X_i\right)$and ${\sigma }_y^2=Var\left(Y_i\right)$

Explanation

$\rho =Corr\left(X_i,Y_i\right)$

The objective is to find Corr $\left(\sum _i{X}_i,\sum _j{Y}_j\right)$.

Explanation

From the Correlation properties:
$Corr\left(\sum _i{X}_i,\sum _j{Y}_j\right)=\frac{Cov\left({\sum }_i{X}_i,{\sum }_j{Y}_j\right)}{{\left(Var\left({\sum }_i{X}_i\right)Var\left({\sum }_j{Y}_j\right)\right)}^{\frac{1}{2}}}$

$=\frac{{\sum }_i{\sum }_{j}Cov\left(X_i,Y_j\right)}{{\left(n·{\sigma }_x^2·n·{\sigma }_y^2\right)}^{\frac{1}{2}}}$

Explanation

Simplifying the expression and using the fact that $Cov\left(X_i,Y_i\right)=\rho ·{\sigma }_x·{\sigma }_y$:

$Corr\left(\sum _i{X}_i,\sum _j{Y}_j\right)=\frac{{\sum }_{i}Cov\left(X_i,Y_i\right)+{\sum }_i{\sum }_{j=i}Cov\left(X_i,Y_i\right)}{n·{\sigma }_x·{\sigma }_y}$

$=\frac{n·\rho ·{\sigma }_x·{\sigma }_y}{n·{\sigma }_x·{\sigma }_y}$

$=\rho$

Therefore,
$Corr\left(\sum _i{X}_i,\sum _j{Y}_j\right)=\rho$

Step6:Final Answer

$Corr\left(\sum _i{X}_i,\sum _j{Y}_j\right)=\rho$