Markus Petty

2022-07-20

Suppose ${X}_{1},\dots ,{X}_{m}$ and ${Y}_{1},\dots ,{Y}_{n}$ are random variables with the property that $Cov\left({X}_{i},{Y}_{j}\right)<\mathrm{\infty }$ for all i and j. Show that, for any constants ${a}_{1},\dots ,{a}_{m}$ and ${b}_{1},\dots ,{b}_{n}$,
$Cov\left(\sum _{i=1}^{m}{a}_{i}{X}_{i},\sum _{j=1}^{n}{b}_{j}{Y}_{j}\right)=\sum _{i=1}^{m}\sum _{j=1}^{n}{a}_{i}{b}_{j}Cov\left({X}_{i},{Y}_{j}\right)$

suchonos6r

Expert

$\to {x}_{1},\dots ,{x}_{m}$ and ${y}_{1},\dots ,{y}_{n}$ are random variable with the property that _
$cov\left({x}_{i},{y}_{j}\right)<\mathrm{\infty }$
for all i and j. Show that _
for any constants
$cov\left(\sum _{i=1}^{m}{a}_{i}{x}_{i},\sum _{j=1}^{n}{b}_{j}{Y}_{j}\right)=\sum _{i=1}^{m}\sum _{j=1}^{n}{a}_{i}{b}_{j}cov\left({x}_{i},{Y}_{j}\right)$
$cov\left(\sum _{i=1}^{m}{a}_{i}{x}_{i},\sum _{j=1}^{n}{b}_{j}{Y}_{j}\right)=\sum _{i=1}^{m}\sum _{j=1}^{n}{a}_{i}{b}_{j}\left(cov\left({a}_{i}{x}_{i},{b}_{j}{Y}_{j}\right)\right)$
$=\sum _{i=1}^{m}\sum _{j=1}^{n}{a}_{i}{b}_{j}cov\left({x}_{i},{Y}_{j}\right)$
Hence, show that for any constant
$cov\left(\sum _{i=1}^{m}{a}_{i}{x}_{i},\sum _{j=1}^{m}{b}_{j}{Y}_{j}\right)=\sum _{i=1}^{m}\sum _{j=1}^{n}{a}_{i}{b}_{j}cov\left({X}_{i},{Y}_{j}\right)$

Do you have a similar question?