Here is the situation:
We a roll a fair dice $2$ times. ( independent )
$a)$ $X$ denotes the number of the first throw; $Y$ denotes the sum of the two throws.
Calculate $\mathbb{C}ov(X,Y)$. Calculate the correlation coefficient$\phi_{X,Y}$.
$b)$ now $X$ denotes the lower number and $Y$ denotes the higher number. Calculate the covariance and the corelation coefficient.
What have I tried? Success with $a): U,W $~ Laplace({$1,…,6$}). So $X = U$ and $ Y = U + W $. $ Cov(X,Y) = Cov(U,U+W) = Cov(U,U) + Cov(U,W) = Var(U) + 0 = Var(U) = \frac{35}{12} $. For $\phi_{X,Y}$ I used the formula.
But I have a problem with b) and hope that somebody can fix it. I'm very much aware that this it is pretty similar to a). So let's say: $U, W$~ Laplace({$1,…,6$}) now $X$ has to be $min(U,W)$ and $Y=max(U,W)$. Consider $Cov(X,Y) = Cov(min(U,W),max(U,W))$. But what do I now?
Best Answer
I don't immediately see a really quick way of doing this so I would proceed as follows.
The definition of covariance is (equivalent to) $$Cov(X,Y) = E[XY]-E[X]E[Y].$$ Now calculate $E[XY]$, $E[X]$ and $E[Y]$ directly by first finding the marginal distributions of $X$ and $Y$. You should get $E[X]=\frac{91}{36}, \, \, E[Y]= \frac{161}{36}$ and $E[XY]=\frac{441}{36}$.
Maybe the easiest way to find $E[XY]$ is to note that it's equal to $E[UV]$ when $X = min(U,V)$ and $Y = max(U,V)$.