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1h
comment Find the sum of a series
"because of your opinion?" Again? You seem radically unable to discuss anything without personalizing the debate and transforming it into a boring he-said-she-said dead-end. Sorry but I am not interested.
2h
comment How to show the series of expectations for truncated symmetric random variables is convergent
Yes I thought you were talking about symmetry about zero.
2h
comment Find the sum of a series
The message got through the first time: you do not see fit to discuss the two precise questions I asked. No need to invent rudeness where there is none, nor to put words in my mouth that I never used. (For example, I did not advocate that nobody on math.se should answer low hanging fruits, only that I found somewhat surprising to see you, an experienced user, do so repeatedly... And yes, I find difficult to believe that you "did not understand the question" from the start.)
3h
awarded  Enlightened
5h
awarded  Nice Answer
6h
comment proof that some expected value equal to $\theta (\log n - \log k)$
Actually, both the lower and upper bounds are absurd on n > k.
6h
revised Find the sum of a series
deleted 11 characters in body
6h
comment Find the sum of a series
1. This is really a pity if you fail to understand it. (Let me mention that the question is routinely mentioned on the meta site. Since you do haunt the meta site, your reaction here is slightly surprising, to stay polite.) 2. Isn't it obvious? To let them a chance to think and to learn. Compare your approach with Alonso's and Essam's.
6h
comment Find the sum of a series
1. Why jump on such low hanging fruits? 2. Why a complete solution?
6h
comment proof that some expected value equal to $\theta (\log n - \log k)$
If n>k this asks to deduce a positive lower bound from some lower bound which may happen to be negative. Sure about the text? (Not to mention the zero-input feature.)
6h
comment Find the sum of a series
@dimitrij Do you know the sum of the series $\sum\limits_{n\geqslant1}\frac{z^n}n$, when it exists?
6h
comment How find $a \in (0, \pi)$ such that $(\cos(2^n a))_{n \ge 1}$ is convergent?
Sorry but this is not the purpose of the site.
6h
revised How to show the series of expectations for truncated symmetric random variables is convergent
added 37 characters in body; edited title
6h
comment How to show the series of expectations for truncated symmetric random variables is convergent
If the distribution of every $X_i$ is symmetric then $E(X_i:|X_i|<i)=0$ for every $i$, no? (And the i.i.d. and integrability conditions are not required.)
7h
comment How find $a \in (0, \pi)$ such that $(\cos(2^n a))_{n \ge 1}$ is convergent?
Four questions asked today -- none has a milligram of personal input. How comes?
7h
revised Mean return time in Markov chain
deleted 27 characters in body
7h
comment Consider 2 Stocks. If Stock 1 sells \$10(0.8) or sells \$20(0.9). If Stock 2 sells \$10(0.9) or \$25(0.8). Which stock sells for higher price?
Again? Let me repeat: this is a problem about one single stock, you should compute the average price A(x,y,p,q) that one single given stock of parameters (x,y,p,q) sells for, and then compare A(x1,y1,p1,q1) and A(x2,y2,p2,q2) for the sets of parameters (xi,yi,pi,qi) in the exercise. So, what is A(x,y,p,q)?
7h
comment Does the law of large numbers pin down the distribution of an infinite sample?
@M.Wind Nothing wrong with being cautious, if you ask me... :-) I mainly wanted to stress that you had already explained the content of my comment.
7h
comment How do I find $\frac{d}{dz}\left(\frac{2z-i}{z+2i}\right)\text{?}$
Simply to note that this is a rational function $Q(z)$ of $z$ hence one can differentiate $Q$ directly without decomposing it into its real and imaginary parts. And $Q(X)=(2X-i)/(X+2i)=2-5i/(X+2i)$ hence $Q'(X)=-(-5i)/(X+2i)^2=5i/(X+2i)^2$ hence the desired derivative is $5i/(z+2i)^2$ for every $z\ne-2i$.
11h
comment $\lim_{z\to 0} \frac{z}{\overline{z}}\text { does not exist }$
?? Two points: 1. My first comment indicates precisely the way to a fully rigorous proof, just folllow it completely instead of at 50%. 2. The "quotient group" question in your last comment is squarely impossible to understand.