I’ve solved a really similar question by doing almost the exact same steps and yet here I for some reason I didn’t get the right answer and I don’t know why. I’d really like it if you could help.
The Question
There are 3 balls in a vase numbered $1,2,3$. $2$ balls are taken out of the vase without returning them. $X$ is the number of the first ball and $Y$ is the number of the second ball. Find $\rho_{X,Y}$
My Attempt:
I started off by mentioning a fact - the probability of taking out a number ball in the second draw without returning is exactly the same as if I had returned it.
(proof of specific case: https://i.stack.imgur.com/x60VP.jpg)
Since expectations and moments and subsequently variances of the two random variables basically differ only with respect to $p_X(x)$ and $p_Y(y)$ which are basically the same then I can conclude $E(X) =E(Y) \quad E(X^2) = E(Y^2) \quad \sigma_X = \sigma_Y$.
all that’s left is to calculate these expressions.
$$E(X) = E(Y) = \frac13(1+2+3) = 2 \\ E(X^2) = E(Y^2) = \frac13(1^2+2^2+3^2) = \frac{14}{3} \\ \sigma_X = \sigma_Y = \sqrt{E(X^2)-E(X^2)} = \sqrt{\frac{14}{3}-4} = \sqrt{\frac{2}{3}}$$
$$\left(E(Y|X) = \frac{1+2+3-x}{3}=2-\frac{x}3\right)$$
$$\sigma_{X,Y} = E(XY)-E(X)E(Y)=E(E(XY|X))-4=E(XE(Y|X))-4=\\=E(2X)-E\left(\frac{X^2}{3}\right)-4=4-\frac{14}9-4=-\frac{14}{9}$$
so
$$\rho_{X,Y} = \frac{\sigma_{X,Y}}{\sigma_{X}\sigma_{Y}}=\frac{-14}{9}:\frac{2}{3} = \frac{-7}{3}$$
The actual solution is $-0.5$.
One minor mistake (otherwise, nice work!) $$\left(E(Y|X) = \frac{1+2+3-x}{3}=2-\frac{x}3\right)$$ is incorrect. It should be $$E(Y\mid X=x) = \sum_y y P(Y =y |X = x) = \sum_{y\neq x} yP(Y = y | X = x) = \frac{1}{2} \sum_{y=x} y = \frac{6-x}{2}.$$ You may have realized something was up, if you substituted $x = 3$ into your answer and got the expected value was 1 and not the intuitive 1.5.