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As somebody used to say:

Does research. Smokes. Battles administration. Smokes. Wishes he could stop battling administration so that he could have more time to do research. Smokes some more.

The same. Except I do not smoke.


How to ask a good question?

This paragraph is for my personal use but freely available:

Welcome to Math.SE! Please, consider updating your question to include what you have tried and where you are getting stuck. That way, people on this site will know exactly what help you need.


1d
comment Check for independence of variables when the density (or distribution) is known.
I answered 1. in a comment on the other page. If something is unclear to you about this, please explain what is.
1d
answered Solving system of differential equations $\dot{x}=3x - 2y$, $\dot y = 2x - y + 15 e^t \sqrt{t}$
1d
comment Solving system of differential equations $\dot{x}=3x - 2y$, $\dot y = 2x - y + 15 e^t \sqrt{t}$
Please, no \begin{cases}...\end{cases}in titles.
1d
revised Solving system of differential equations $\dot{x}=3x - 2y$, $\dot y = 2x - y + 15 e^t \sqrt{t}$
edited title
1d
revised Density function and Integration to $1$
added 1 character in body
1d
comment Determine correlation and independence when only the joint density is given?
You seem to be willing to make the comments drift to different, albeit related, questions. Please don't. If you have other problems, post other questions.
1d
answered Showing convergence of recursive sequence $A_{n+1}=\frac 1 {1+A_n}$
1d
comment Determine correlation and independence when only the joint density is given?
By "the support of the density is not a product", I mean that the support of the density is not a product. Recall that the support of some measure $\mu$ is the set of points $x$ such that $\mu(B(x,t))\gt0$ for every $t\gt0$.
1d
comment is conditional expectation less than unconditional one?
The only case when this happens is when $E(X_k\mid F_{k-1})=E(X_k)$ almost surely, that is, when $E(X_k\mid F_{k-1})$ is deterministic.
1d
comment Determine correlation and independence when only the joint density is given?
No, symmetry of density cannot disprove independence. The fastest way to disprove independence would be to note that the support of the density is not a product.
2d
comment Sequence $0\leq a_{n}-l\leq \dfrac{\pi^{2} }{2^{2n+1}}$
But now you spoiled it. Why?
2d
comment Show that $E(X|Y, Z) = E(X|Y)$ almost surely with condition Z is independent of $(X, Y)$
There are too many parentheses in your post, please correct them as I did in the title.
2d
revised Show that $E(X|Y, Z) = E(X|Y)$ almost surely with condition Z is independent of $(X, Y)$
edited title
2d
comment Determine correlation and independence when only the joint density is given?
?? I did not go from this to what you wrote, but to $E(X_i)=0$ and $E(X_iX_j)=0$.
2d
revised Determine correlation and independence when only the joint density is given?
added 412 characters in body
2d
answered Determine correlation and independence when only the joint density is given?
2d
comment Show that $E(X|Y, Z) = E(X|Y)$ almost surely with condition Z is independent of $(X, Y)$
No, the characterization you ADDED to the question is allright.
2d
answered Show that $E(X|Y, Z) = E(X|Y)$ almost surely with condition Z is independent of $(X, Y)$
2d
comment Is this two-dimensional version of the Intermediate Value Theorem correct?
What is not true?
2d
answered Bounding an expected hitting time