Question

Random variables X Poisson~ ( a) ,Y Poisson ~ ( a) . X and Y are independent. If 2 1, 2 1. U =2X+ Y-1, V=2X- Y +1. Find: ) Cov (U ,V ).

Answer 1

By definition of covariance,

`\mathrm{Cov}(U,V)=E[(U-E[U])(V-E[V])]=E[UV-E[U]V-UE[V]+E[U]E[V]]=E[UV]-E[U]E[V]`

Since
`U=2X+Y-1` and
`V=2X-Y+1`, we have

`E[U]=2E[X]+E[Y]-1`

`E[V]=2E[X]-E[Y]+1`

`\implies E[U]E[V]=(2E[X]+E[Y]-1)(2E[X]-(E[Y]-1))=4E[X]^2-(E[Y]-1)^2=4E[X]^2-E[Y]^2+2E[Y]-1`

and

`UV=(2X+Y-1)(2X-(Y-1))=4X^2-(Y-1)^2=4X^2-Y^2+2Y-1`

`\implies E[UV]=4E[X^2]-E[Y^2]+2E[Y]-1`

Putting everything together, we have

`\mathrm{Cov}(U,V)=(4E[X^2]-E[Y^2]+2E[Y]-1)-(4E[X]^2-E[Y]^2+2E[Y]-1)`

`\mathrm{Cov}(U,V)=4(E[X^2]-E[X]^2)-(E[Y^2]-E[Y]^2)`

`\mathrm{Cov}(U,V)=4V[X]-V[Y]=4a-a=\boxed{3a}`