Let the number X be chosen at random from among 1,2,3,4 and the number Y be chosen from among those atleast as large as X. Prove that Cov(X,Y) = 5/8.
My attempt: I have calculated P(X=k) which is equal to 1/4. And conditional probability, P(Y|X=x), x=1,2,3,4. Hence, I have calculated P(X=x, Y=y). But my calculation is not yielding Cov(X,Y)= 5/8.
I don't know where the mistake is. But I suspect it's in calculating the mean of Y. How to calculate mean of Y?
It will be helpful if you share the answer. Thanks :)
I suggest you to derive the joint pmf with a contingency table like the following
then calculate cov by its definition
$$\mathbb{Cov}[X,Y]=\mathbb{E}[XY]-\mathbb{E}[X]\cdot\mathbb{E}[Y]=8.75-2.5\times3.25=0.625=5/8$$