Search Results

A generating function proof. As $\arcsin(z)=\sum_{k\geq 0} \frac{1}{2k+1} {2k \choose k} \frac{z^{2k+1}}{4^k}$ and $\frac{1}{\sqrt{1-z^2}}=\sum_{k\geq 0}{2k \choose k}\frac{z^{2k}}{4^k}$ we have that …

One can also use the binomial transform. (If $A(z)=\sum_{i\geq 0} a_i z^i$ is a (formal) power series, the (formal) power series $B(z):=\frac{1}{1-z} A(\frac{z}{1-z})$ has coefficients $[z^n] B(z)=\su …

(You are considering the uniform distribution on a discrete simplex. I'm not aware of specific results in the literature, and a brief internet search didn't reveal anything.) A simple way is to use …

(0) Preliminaries: (a) factors: above you should change the factors $\frac{1}{4}$ on the rhs of your equations (4) and (5) to $\frac{1}{2}$ (Kündgens distance is half the distance of Ahlswede/Katona …

One can also use Jensen's inequality. Let (for $\sigma>0$) $G_\sigma$ denote a random variable with $\Gamma(1,\sigma)$-distribution, i.e. having Lebesgue density $$f_\sigma(t)=\frac{t^{\sigma-1}}{\G …

I don't know a reference. One way to show the eigenvalues starts from the observation (which can be proved using generating functions) that $\sum_{i=0}^n {i\choose k} A_{i,j}={2n+1 \choose n-k} {j+k …

(I use the notation from my comments). By symmetry, $$\mathbb{P}(E^c)={n \choose m}\,\mathbb{P}(E^c, B=\{1,\ldots,m\})\;\;.$$ If $D_1=M_m\;\;$ we have to compute $$I_1:=\mathbb{P}(D_1^\prime>t+M_m,D_ …

Call the left resp. right hand sum $R_m$ resp. $R_{m+1}$. As $L$ is $\exp(\lambda)$ and independent of $(R_m,R_{m+1})$ , taking expection with resp. to $L$ first gives $$\mathbb{P}(R_m\leq L < R_{m+1 …