@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 ...

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 …

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 ...

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, …

thank you, you must right. Just an irritating use of implication to mean premise, got really lost.

Use Lagrange multipliers to find the maximum and minimum values of the function :$$f(x,y)=e^{xy}$$ constraint $$x^3+y^3=16$$ This is my problem in my workbook. When I ...

Jan 23, 2013 · Prove $E[XY]=E[YE[X|Y]]$. I tried proving it using the definition of covariance, but I ended up going in a circle. Any hints on how to go about the proof?

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 …