I'm trying to find the Covariance of $X$ and $Y$.
I see that the $Cov(X,Y) = E[XY] - E[X]E[Y]$.
I know how to find the expectation of single variables. I'm having troubles evaluation $E[XY]$. Can it be $X * E[Y]$?
Thank you!
UPDATE
Smoker Non-SmokerTotal
Heart Attack 0.03 0.03 0.06
No Heart Attack0.44 0.50 0.94
Totals 0.47 0.53
So the table contains all the joint distributions, and marginal distributions. I'm unsure how to plug into the discrete equation.
$\sum_x\sum_y x\, y\; \mathsf P(X=x,Y=y)$
I know that the $P(Smoker)=0.47$ and that $P(Heart Attack)=0.06$ and that $P(Smoker \cap Heart Attack) = 0.03$.
So this mean that $E[Smoker$ $HeartAttack] = 0.47 * 0.03 * 0.06$?
Also, is the $E[Smoker] = P(Smoker)$ (aka is expectation equal to the marginal probability?)
$\endgroup$ 31 Answer
$\begingroup$Nope, $X\,\mathsf E(Y)$ is not a valid option. As André Nicolas commented, that is a random variable while $\mathsf E(XY)$ is a constant.
What we can say is that: $\mathsf E(X\,Y) = \mathsf E\big(X\,\mathsf E(Y\mid X)\big)$
But we really do need to know what the joint distribution is to say more (or at least a marginal and conditional distribution).
$$\begin{align} \mathsf E(XY) & = \iint_{\Bbb R^2} x\,y\;f_{X,Y}(x,y)\operatorname d x\operatorname d y & = \sum_x\sum_y x\, y\; \mathsf P(X=x,Y=y) \\[2ex] \mathsf E\big((X\,\mathsf E(Y\mid X)\big) & = \underbrace{\int_{\Bbb R} x\,f_X(x) \int_{\Bbb R} y\; f_{Y\mid X=x}(y)\operatorname d y \operatorname d x }_\textsf{continuous valued random variables} & \underbrace{= \sum_x x\;\mathsf P(X=x)\,\sum_y y\;\mathsf P(Y=y\mid X=x) }_\textsf{discrete valued random variables} \end{align}$$
And so forth.
$\endgroup$ 2
