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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.


6h
comment Expected Value of conditional expectation
@AsafKaragila Wow. Thanks for the information.
7h
comment Expected Value of conditional expectation
Duplicate of math.stackexchange.com/q/953014, closed by the OP 9 minutes ago. What is going on?
7h
comment Can I know the value of an infinite serie?
@Lucian Oh no, not again...
7h
comment Separation of points in a Poisson point process
Of course (but you know the answer, don't you?).
7h
answered Prove that the square root and exponent of a function in a $\limsup$ equals the the square root and exponent of a $\limsup$ of the function?
7h
comment What is the variance of this random variable: number of items
The identity E(I)=n/mu does not hold in general.
8h
revised Show the following definition does not give a $\sigma$-addtive measure pathwisely
added 101 characters in body
8h
comment Showing that $-\ln{X} \sim \exp{\alpha}$ for $X \sim Beta(\alpha, 1)$
Actually you are given the PDF for X, not its CDF, hence one can deduce directly the PDF of $Y=u(X)$ from the PDF of $X$ thanks to the identity $$f_Y(y)=f_X(u^{-1}(y))\mathrm{Jac}(y)$$ where $\mathrm{Jac}$ denotes the Jacobian of the transformation $y\mapsto u^{-1}(y)$, instead of using the long road PDF of $X\to$ CDF of $X\to$ CDF of $Y\to$ PDF of $Y$. This is explained in many posts on the site (as well as in every decent textbook).
8h
comment Is there a result that the density function for $\chi^{2}$ must be related to the standard normal density?
One cannot deduce the density of $X$ from the density of $Y=X^2$ alone (for example, $X$ and $|X|$ correspond to the same $Y$) but if one adds that $X$ must be symmetric then $X=S\sqrt{Y}$ with $(S,Y)$ independent, $S$ symmetric Bernoulli, hence the general identity $f_Y(y)=(f_X(\sqrt{y})+f_X(-\sqrt{y}))/(2\sqrt{y})$ yields the reverse relation in my post since $f_X(x)=f_X(-x)$.
9h
awarded  Refiner
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awarded  Explainer
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awarded  Revival
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awarded  Enlightened
12h
awarded  Nice Answer
13h
comment Can I know the value of an infinite serie?
Well, for example, here, you received two answers and two comments making explicit suggestions, and your reaction to these was... a complete lack of visible reaction. Does this mean that you are still lost, that one answer helped you solve the question, that one comment helped you solve the question, that all of them did, that you are not interested anymore, or still something else? No way to know.
13h
comment Can I know the value of an infinite serie?
What happens with your other questions?
13h
answered Thinking about a probability problem in terms of sets.
13h
comment Expected value of a normal distribution over a range
You don't know the quantity you want to compute? Well, nobody can decide for you.
13h
answered Is there a result that the density function for $\chi^{2}$ must be related to the standard normal density?
13h
comment Expected value of a normal distribution over a range
The trouble is that there (at least) two ways to that, either one considers $$\int_a^bxf(x)dx,$$ or one considers $$\frac1{F(b)-F(a)}\int_a^bxf(x)dx.$$ Please explain.