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Theory and applications of probability and stochastic processes: e.g. central limit theorems, large deviations, stochastic differential equations, models from statistical mechanics, queuing theory.

70 votes
Accepted

integral of a "sin-omial" coefficients=binomial

Tonight I read here [the answer by esg to another your question] that $\frac1{2\pi}\int_{-\pi}^\pi e^{-ik t}(1+e^{it})^ndt=\binom{n}{k}$, which is, well, obvious at least when both $n$ and $k$ are pos …
Fedor Petrov's user avatar
22 votes
Accepted

Is measure preserving function almost surjective?

Yes, by Luzin's theorem. Fix $\varepsilon>0$ and take a compact subset $K$ of measure at least $1-\varepsilon$ such that $F$ is continuous on $K$. Then $F(K)$ is a compact set of at least the same me …
Fedor Petrov's user avatar
19 votes

If $X$ and $Y$ independent and identically distributed, then $E(|X-Y|)\leq E(|X+Y|)$. Are ot...

It suffices to prove that for arbitrary $X_1,\ldots,X_n$ we have $$\sum_{i, j} |X_i-X_j|\leqslant \sum_{i, j} |X_i+X_j|, \quad \quad \quad (\star)$$ then applying $(\star)$ for a random sample from y …
Fedor Petrov's user avatar
13 votes

integral of a "sin-omial" coefficients=binomial

(not an answer) Denote $\alpha=k/n$, $f(x)=(\frac{\sin x}{\sin \alpha x})^\alpha (\frac{\sin x}{\sin (1-\alpha) x})^{1-\alpha}$. Then your claim may be rewritten as $\pi^{-1}\int_0^\pi f^n(x)dx=\frac …
Fedor Petrov's user avatar
11 votes
Accepted

On the limit of partial sum of infinite doubly stochastic matrix

No. Enumerate all positive integers which are not powers of $2$: $3=n_1<n_2<n_3<\dots$ and partition positive integers into two-element sets $\{n_k,2^{k-1}\}$. Let $a_{i,j}=1$ if the set $\{i,j\}$ is …
Fedor Petrov's user avatar
10 votes
Accepted

How to prove the sum of n squared binomial probabilities does not increase as n increases

Denote $q=1-p$, write $[x^a]f(x)$ for a coefficient of $x^a$ in the polynomial (or Laurent polynomial) $f(x)$. We have $$F(n)=[1](px+q)^n(px^{-1}+q)^n=[1]((p^2+q^2)+qp(x+x^{-1}))^n=\\=(2\pi)^{-1}\int_ …
Fedor Petrov's user avatar
10 votes

Identities and inequalities in analysis and probability

Many inequalities are proved by identities representing the thing which must be proved to be non-negative as integral (or sum, or expectation) of squares. For example: CBS inequality $$ \int_X f^2 \cd …
10 votes

Explicit example of second Borel–Cantelli lemma

All irrational numbers in $(0,1)$ have unique representation as $\sum_{k=2}^\infty c_k/k!$, where $c_k\in \{0,1,\dots,k-1\}$. 'Digits' $c_k$ are independent, so you may choose the events $E_n$ as '$c_ …
Fedor Petrov's user avatar
8 votes
Accepted

Shannon entropy of $p(x)(1-p(x))$ is no less than entropy of $p(x)$

Denote $f(p)=p(1-p)$, $H(p)=-p\log p$, let $p_1,\dots,p_n$ denote all positive probabilities of our distribution, then $\sum p_i=1$, finally denote $s=\sum_i f(p_i)$. Then we need to prove the inequal …
Fedor Petrov's user avatar
8 votes
Accepted

Approximating binomial coefficient sum

Actually we may simply compute this sum (and I guess that your expectation may be computed differently to give the answer in the below simplified form). We start with $1/{nk\choose m}=(nk+1)\int_0^1 x …
Fedor Petrov's user avatar
8 votes
Accepted

Can I get away without using Arzela-Ascoli?

Of course you can, and this is how Arzela-Ascoli is often proved. You may fix a finite $\varepsilon/3$-net $D\subset X$ and partition $[0,1]$ onto disjoint subsets $A_1,\ldots,A_N$ of diameter less th …
Fedor Petrov's user avatar
7 votes

The average of reciprocal binomials

Let me elaborate on Fry's suggestion and your forthcoming comment. $$\frac1{2^{n+1}}\sum_{k=1}^{n+1}\frac{2^k}k=\frac1{2^{n+1}}\int_0^2(1+x+\dots+x^n)dx=\frac1{2^{n+1}}\int_0^2\frac{1-x^{n+1}}{1-x}dx …
Fedor Petrov's user avatar
7 votes
Accepted

If $g$ is differentiable, how can we show that $z\mapsto1\wedge e^{g(z)}$ is differentiable ...

Yes, it is true. Since $2\min(a,b)=a+b-|b-a|$ for positive numbers $a,b$, and $F(z)=e^{g(z)}-1$ is differentiable, it suffices to prove that $|F|$ is differentiable except at a countable set of points …
Fedor Petrov's user avatar
7 votes
Accepted

An expansion from Ramanujan related to birthday problem

We have $$Q(n)=\sum_{k=0}^{n-1} (1-1/n)(1-2/n)\ldots(1-k/n).$$ Write $1-x/n=e^{-x/n-x^2/(2n^2)-\ldots}$, then $$Q(n)=\sum_{k=0}^{n-1} \exp\left(-\frac{k(k+1)}{2n}-\frac{k(k+1)(2k+1)}{12n^2}-\ldots\rig …
Fedor Petrov's user avatar
7 votes

A limit obtained from a probability distribution on the positive integers

This result belongs to P. Erdös, W. Feller, and H. Pollard (1949). It worth mention that if $\sum a_n p_n<\infty$, this follows from Wiener division theorem in the algebra of absolutely summable Fouri …
Fedor Petrov's user avatar

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