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29m
revised Use convolution theorem to evaluate $\int_0^\infty e^{-((|a-su|)/c)^b}e^{-(u/k)^p}du$
added 3 characters in body
30m
comment If $y'+y=|x|$ and $y(-1)=0$, what is $y(1)$?
@Ron True. Note that the Great Man used other substances as well, but one cannot recommend these publicly I am afraid.
32m
comment Product Spaces and Integration
Can you formalize it?
33m
comment If X and Z are independent and Y and Z are independent random variables, is cov(XY, Z) = 0?
The revised suggestion is wrong as well, try $(X,Y)$ uniform on $\{-1,1\}^2$ and $Z=XY$. Then any two random variables from $X$, $Y$ and $Z$ are independent and $\mathrm{cov}(XY,Z)=1$.
38m
comment Product Spaces and Integration
Why do you think what you say you think?
40m
comment If $y'+y=|x|$ and $y(-1)=0$, what is $y(1)$?
@Ron Coffee before mathematics. Always. :-)
3h
comment Finding the expected value in the given problem.
... If you do reject these, please be aware that I will not repeat the exercise and will not feel concerned anymore. All the best.
3h
comment Finding the expected value in the given problem.
... as is the case in the present situation. All in all, it seems your lack of training in these matters makes you err widely (which is not a crime in itself), but also that you do not see fit to listen to the explanations experts are providing (and this, according to me, is a serious problem). Finally, please consider the following: I took upon me to write these detailed explanations, which you will most probably reject, about the underlying mathematics and about what I can perceive of your behaviour. ...
3h
comment Finding the expected value in the given problem.
... that $$\int_\Omega(X(\omega)+Y(\omega))dP(\omega)=\int_\Omega X(\omega)dP(\omega)+\int_\Omega Y(\omega)dP(\omega).$$ What is making you obsessed with this triviality is probably that you are thinking about the equivalent version $$\int_\mathbb RzdP_{X+Y}(z)=\int_\mathbb R xdP_X(x)+\int_\mathbb R ydP_Y(y),$$ which indeed requires a two-lines proof to show it is equivalent to the first version, but is again valid in full generality. The "mathematical" problem (numbered 2.) you try to introduce again and again is moot as well as long as one sums a finite number of random variables, ...
3h
comment Finding the expected value in the given problem.
Jumping directly to what seems to be at the heart of your misconceptions, note first that the "conceptual" problem you raise (numbered 1.) is moot and actually points at a conceptual mistake you repeat endlessly: usually, one is exposed during the very first lectures of any theoretical probability course to the fact that $E(X+Y)=E(X)+E(Y)$ for every integrable random variables $X$ and $Y$ defined on the same probability space $(\Omega,\mathcal F,P)$, that is, to be quite explicit, ...
4h
comment Finding the expected value in the given problem.
@NikosM. This comment to me should have been appended to your answer, not here, and I am answering it (briefly) there.
4h
revised If $y'+y=|x|$ and $y(-1)=0$, what is $y(1)$?
added 41 characters in body; edited title
4h
comment If $y'+y=|x|$ and $y(-1)=0$, what is $y(1)$?
Usually, the notation $\int g(x)dx$, leaving the bounds of the integral unspecified, already includes an additive constant.
6h
revised SLLN when the expectation in infinite
deleted 14 characters in body; edited tags
6h
revised If $V$ is a vector subspace of $R^n$, prove that $V^ \bot$ is a vector subspace of $R^n$
edited title
6h
revised How to compute $\lim\limits_{x \rightarrow 0} \frac{1}{x^2}\int_0^{G(x)} \arctan(s+2s^2) ds$
added 78 characters in body; edited title
6h
comment Solve the equation $\Bbb|x-\sqrt{x}|=\lfloor x\rfloor-\sqrt{x}$
"With a great desire for you" This is frightening.
6h
revised Solve the equation $\Bbb|x-\sqrt{x}|=\lfloor x\rfloor-\sqrt{x}$
edited title
6h
comment SLLN when the expectation in infinite
Note that $S_n\geqslant S_n^x$ for every $x$, where $S_n^x=X_1^x+\cdots+X_n^x$ and $X_n^x=\inf(X_n,x)$ for every $n$, thus $\liminf S_n/n\geqslant\lim S_n^x/n=E(X_1^x)$ almost surely, then the limit $E(X_1^x)\to+\infty$ when $x\to\infty$ yields the conclusion.
6h
comment If $y'+y=|x|$ and $y(-1)=0$, what is $y(1)$?
No, "integrating both sides" (and using your somewhat faulty notations), one gets $e^xy=\int e^x |x|dx$, hence $$ey(1)=e^{-1}y(-1)+\int_{-1}^1e^x|x|dx=\ldots$$