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2d
comment An upper bound for an average exponentiated weighted sum of a vector from $\{-1,1\}^n$
@sciona Well, if you are correct, then $A_{f(n)}(n)=\operatorname{cosh}^n f(n)$. In fact, I am certain that you are correct, but I still can't wrap my mind around your last equality. But that could be because I'm quite tired...
2d
comment An upper bound for an average exponentiated weighted sum of a vector from $\{-1,1\}^n$
I'm not sure how you obtain the second equality. In fact, $(e^{f(n)}+e^{-f(n)})^n\geq e^{nf(n)}\geq A_{f(n)}(n)$, with the second equality holding only when $f(n)=0$...
2d
asked An upper bound for an average exponentiated weighted sum of a vector from $\{-1,1\}^n$
Dec
15
accepted Trace distance between “weighted” Hermitian matrices
Dec
15
asked Trace distance between “weighted” Hermitian matrices
Nov
28
accepted How many ways can one “fit” $m$ non-overlapping sub-segments of length $k$ into a segment of length $n$?
Nov
27
revised Random variable related to binomial
corrected the renaming of the random variable
Nov
27
comment How many ways can one “fit” $m$ non-overlapping sub-segments of length $k$ into a segment of length $n$?
Thanks for a nice answer! One follow-up question: you say that "we know how to count solutions of [diophantine equation $E_0+E_1+\ldots+E_m=x$]". Now this indeed looks like a very simple diophantine equation: a sum of $m+1$ non-negative variables and same constant (unity) multiplying each variable. It looks like, in general, there are $\binom{x+t-1}{t-1}$ solutions for $\sum_{i=1}^ta_i=x$. But could you please point me to a source of this statement or explain how one derives it? There must be a textbook that contains this...
Nov
27
comment Random variable related to binomial
By @victorsouza's solution to my related question, shouldn't $\Pr(X=x)=\binom{n-x(k-1)}{x}p^x(1-p)^{n-xk}$?
Nov
27
comment Random variable related to binomial
Posted a related question.
Nov
27
asked How many ways can one “fit” $m$ non-overlapping sub-segments of length $k$ into a segment of length $n$?
Nov
27
revised Random variable related to binomial
corrected subquestion 2 ($k$ is constant) and renamed random variable to avoid confusion.
Nov
27
comment Random variable related to binomial
@GrahamKemp To obtain the p.m.f. I think the key is finding the expression for the number of ways one can "fit" $m$ non-overlapping sub-segments of length $k$ into a segment of length $n$. However, combinatorics isn't my strong suite and I can't figure that out. As for Gaussian approximation, for the binomial distribution it arises due to the Central Limit Theorem and the divisibility property of a binomial random variable (i.e. if $X\sim\text{Binomial}(n,p)$ and $Y\sim\text{Binomial}(m,p)$, then $X+Y\sim\text{Binomial}(n+m,p)$). The sums here won't be as nice, but maybe they aren't too bad.
Nov
27
asked Random variable related to binomial
Oct
20
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