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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 …

Your formula yields $$f(n_0)=\prod_{j=0}^{m-1}\frac{n-n_0-j}{n-j}\leqslant \left(\frac{n-n_0}{n}\right)^m= \left(1-\frac{n_0}{n}\right)^m\leqslant e^{-mn_0/n},$$ that is about $e^{-1/\gamma}$.

Another try, now I claim that the inequality holds. For a random variable $X$, denote $\|X\|=(\mathbb E |X|^\sigma)^{1/\sigma}$. It is indeed a norm. Then $$ \|X_1+\ldots+X_n\|\leqslant \|X_1+\ldots+X …

Denote $K_0=[-a,a]\times \mathbb{R}^{k-1}$, $K_{-}=[-\infty,-a]\times \mathbb{R}^{k-1}$, $K_{+}=[a,\infty]\times \mathbb{R}^{k-1}=-K_-$. We have $$\mu(L)(\mu(K_0)+2\mu(K_+))=\mu(L)=\mu(L\cap K_0)+\mu( …

The following argument is less or more the same as that of Iosif Pinelis, but with less computations and the symmetry is rather explicit. It may be explained without complex numbers, but the explanati …

I have no idea why Lucia removed her nice answer, but here goes a short elementary argument. We want to bound from below the expectation of $f(\pi)$, where $\pi$ is a random permutation and $f(\pi)=\p …

Here goes a combinatorial proof of the answer (formula (1)) by Christian Remling. If Alice wins, then after the last draw (or after the beginning if there were no draws) we have a sequence $B^jTA^N$, …

Here is a proof that the variance of $d_T(v)$ does not exceed $\frac14(\deg v-1)$. For every edge $e\in E$ take a variable $x_e$ and consider the polynomial $$P:=\sum_T \prod_{e\in T} x_e,$$ where the …

Imagine that for all $k=1,2,\ldots$ all events $A_n$ for $n\in [k!,(k+1)!-1)$ are the same event $C_k$. Then if $x$ belongs to all $A_i$ along a subsequence of positive density yields that it belongs …

The probability that all samples are less than $N^{19/10}$ is $(1-N^{-19/20})^{N}$ that tends to 0. The expected number of samples greater than $N^{1/2}$ is $N^{3/4}$, thus, the probability that we ha …

Denote $\alpha=\mathbb{E} u_1^2v_1^2$, $\beta=\mathbb{E} u_1v_1u_2v_2$. Then by the symmetry and linearity of expectation we have $$f(m):=\mathbb{E} (u_1v_1+\ldots+u_mv_m)^2=m\alpha+m(m-1)\beta.$$ We …

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 …

For arbitrary $C>0$ and positive i. i. d. $Y_1,\ldots,Y_k$ for $Z:=\sum_{i=1}^k Y_i$ we have $$\sum_{i=1}^k {\mathbf 1}(Z/Y_i\geqslant C)\geqslant k-C$$ (since $Z/Y_i<C$ means that $Y_i>Z/C$, which ma …

Every $\gamma>0$ seems ok. We have $$ \mathbb{P}(|x_k|\geqslant e^{\gamma k})\leqslant \mathbb{P}(1+|x_k|\geqslant e^{\gamma k})= \mathbb{P}(\gamma^{-1}\log(1+|x_k|)\geqslant k), $$ and the sum of th …

Denote $p_k=P(X=k)$. Then $E(R_n)=\sum_k P(k\in \{X_1,\ldots,X_n\})\leqslant \sum_k \min(1,np_k)$. We are given that $\sum kp_k<\infty$. Fix $\varepsilon>0$. The sum of $\min(1,np_k)$ over $k<\varepsi …