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esg
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Bounded convergence for expectation of random variables
No, uniform integrability of $X_n$ is not needed, only the uniform integrability of $Y_n$. In the situation above this follows from the uniform boundedness of $f_n$ and is not affected by the properties of $X_n$
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Total progeny of a Galton-Watson branching process - standard textbook question
$r(s):=sg_Y(s)$ is the generating function of the total progeny in a GW-process with reproduction function $a_d(s)=q+ps^d$. (3) thus shows that the no. of successors of one individual in the GW-process with reproduction $b_d$ is distributed like the sum of the successors of $d$ independent GW-processes with reproduction $a_d$. In particular, the extinction probablity $r(1)$ of a GW-process with reproduction $b_d$ is the $d-$th power of the extinction probability of a GW-process with reproduction $a_d$.
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Generalized Shared Birthday
Similar considerations show that for $t>2$ one needs roughly $n\approx (t!\cdot k \cdot d^{t-1})^{1/t}$ to get the probability of $k$ $t$-fold birthdays close to $1/2$
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sequencial shift on families =flipped powers. How?
Typo: it should be $W(z)=-zf_1(z)$
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Eigenvalues of a matrix with entries involving combinatorics
In my eyes it suggests that the Adams operation should have a nice combinatorial interpretation, at least for the case $K^*(U(n))$. (But I am no expert).
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Hankel determinants of binomial coefficients
Note that ${np \choose n}$ is a moment sequence. It follows from the "sin-omials" solution that ${np \choose n}=\mathbb{E}(Y_{1/p})^{np}=\mathbb{E}(Y_{1/p}^p)^n$ where $Y_\alpha=\frac{\sin(X)}{(\sin(\alpha X))^\alpha (\sin((1-\alpha)X))^{1-\alpha}}$ and $X$ is uniform on $[0,\pi]$. Thus $\det(H_n)>0$ (the distribution of $Y_{1/p}$ is not finitely supported).
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$\binom{x}{2}+\binom{x}{4}+\cdots+\binom{x}{2u}$ is a convex function on $[0,+\infty)$?
The case $x<1$ can also be settled without computation, by noting that $\ell(t):= \frac{1}{(1+t)^{1-x}}\left(\frac{\log(1+t)}{t}\right)^2$ is the Laplace transform of a nonnegative random variable $X$ possessing all moments (the first factor is the LT of $\Gamma(1,1-x)$, and $\log(1+t)/t$ is the LT of $g(x)=\int_{x}^\infty \frac{e^{-y}}{y}\,dy$), and $[t^{2u}] \frac{1}{1-t} \ell(t)$ is an even degree MacLaurin sum of $\ell$, evaluated at 1, and therefore exceeds $\ell(1)>0$. Can this observation be extended to give a more conceptual proof for the other cases?
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