Question

The joint p.m.f. of x and y is f(x,y) = 1/6, $0 \leq x+y \leq 2$, where x and y are nonnegative integers. compute cov(x, y). g

Answer 1

Definition of covariance:


`\mathrm{Cov}(X,Y)=\mathbb E\bigg[(X-\mathbb E[X])(Y-\mathbb E[Y])\bigg]=\displaystyle{\sum\sum}\limits_((x,y))(x-\mathbb E[X])(y-\mathbb E[Y])f_(X,Y)(x,y)`

We're given that
`f_(X,Y)(x,y)=\frac16` for all
`(x,y)`.

We can compute the expectation of
`X` and
`Y`:


`f_X(x)=\displaystyle\sum_yf_(X,Y)(x,y)`

`f_X(x)=\mathbb P(X=x\mid Y=0)\mathbb P(Y=0)+\mathbb P(X=x\mid Y=1)\mathbb P(Y=1)+\mathbb P(X=x\mid Y=2)\mathbb P(Y=2)`

`\implies f_X(x)=\begin{cases}\frac12&\text{for }x=0\\\\\frac13&\text{for }x=1\\\\\frac16&\text{for }x=2\\\\0&\text{otherwise}\end{cases}`

We'd find an identical PMF for
`Y`. So we have


`\mathbb E[X]=\displaystyle\sum_xxf_X(x)=0\cdot\frac12+1\cdot13+2\cdot16=\frac23`

and similarly
`\mathbb E[Y]=\frac23`.

So the covariance is


`\mathrm{Cov}(X,Y)=\displaystyle\frac16{\sum\sum}\limits_((x,y))\left(x-\frac23\right)\left(y-\frac23\right)`



`\implies\mathrm{Cov}(X,Y)=-\frac5{18}`