Jan 29, 2019 · When two random variables are statistically independent, the expectation of their product is the product of their expectations. I found this on wikipedia : https://en ...

May 20, 2019 · What you need is the joint distribution of $ (X,Y)$. Then e.g. in discrete case $\mathbb EXY=\sum_ {x,y}xyP (X=x,Y=y)$. If there is a PDF then $\mathbb EXY=\int xyf_X (x,y)dxdy$ (both …

@modocache: Both Andre and I have told you what $E [XY]$ means. It means: think of $XY$ as a new random variable, compute its distribution, and then find the weighted ...

Dec 14, 2013 · The answer to the question in the title is that we say that $X$ and $Y$ are uncorrelated exactly when $E [XY]$ happens to equal $E [X]E [Y]$, and if $E [XY] \neq E [X ...

Mar 15, 2015 · The upper left cell of the table ($X=1, Y=3$) has a probability of $0.3$ of occurring. So this cell contributes $0.3\cdot (1)\cdot (3)$ to the expectation of $XY ...

Nov 19, 2017 · Thanks for your reply. Yeah, I see. That's the general intuition behind independence of events which as I see, can be easily focused on expectation. However, when I wrote my question, …

Feb 1, 2012 · First, handle the finite case using the theorem. Then handle the infinite case, but be careful to properly distinguish infinite from non-existent expectations (remember that for Lebesgue …

Let $X$ and $Y$ be independent random variables having the same distribution and the finite mathematical expectation. How to prove the inequality $$ E(|X-Y|) \\le E ...

Jul 26, 2015 · @Math-fun I understand it yields the average.But how do you interpret the first level integral $ \int h f (x,y)dx$ . I know that for $E (X)$ , the first level ...

note that this result holds under more general assumption. (you are actually assuming here that the vector $ (X, Y)$ is continuos)