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If you do $n$ tests of size $\alpha/n$, then $\alpha$ is the Bonferroni bound on at least one of the tests succeeding. It is conservative because it is the worst possible bound without any further inf …

Even though there is no closed form for the CDF of the binomial distribution, there is one for the derivative with respect to the $p$ parameter. Namely, if $$ F = \sum_{i=0}^s \binom ni p^i(1-p)^{n-i …

This type of discretized normal distribution occurs in some practical problems. For example, the median of the vertex degrees of an Erdős-Renyi random graph with fixed edge probability has such a dis …

You need a weaker conjecture, since the cdf $(1+x)e^{-x}$ for $x\ge 0$ has moments of the desired form yet it isn't like $1-Ae^{-cx}$. More generally, the distribution with cdf $(1+x^k)e^{-x}$ has mom …

Suppose $F~$ is a probability distribution symmetrical about 0, for which all moments exist. Let $\mu_i~$be the $i$-th moment (of course $\mu_i=0$ if $i~$ is odd). We know there are some conditions …

The classical approach to this is called "smoothing" and is described in Chapter 16 of Feller's Introduction to Probability Theory, Vol 2. Basically you use the bounds on the moment differences to bou …

Steven's example is indeed the simplest. See chapter 3 of this book for counterexamples to lots of similar possibilities.

I suggest you try Gauss-Hermite integration. You can guess the precision by increasing the number of abscissae. Tables of abscissae and weights are here.

There is a method, provided the times can be scaled to be integers not too large. Consider that the times are $t_1,\ldots,t_{40}$. The problem can be described like this: for some given $N$, how many …

If $Y(x)=(1-\Phi(x))/\phi(x)$, it is easy to check that $Y'(x)=xY(x)-1$ and from this anything you like follows by standard methods.

If the edge weights are scaled by a sufficiently high factor, a minimum weight matching will have the least greatest weight. By also removing the edges with weight less than $w$, a minimum weight matc …

I think this is a hard unsolved problem, though I'm far from expert on it. It strikes me as a harder cousin of the orthant problem, which is already quite hard. See this paper for references.

In my paper "On Littlewood's estimate for the binomial distribution", Adv. Appl. Prob., 21 (1989) 475-478, copy at http://cs.anu.edu.au/~bdm/papers/littlewood2.pdf , I find sharp exact bounds on this …

The distribution you describe is called a Poisson-binomial distribution. If you do a search with this name you will find a substantial literature on it.

It has lots of other solutions, at least for some alphas. Try $x = -.225312055012178104725014013952$, $y = 0.813419847597618541690289359893$, $\alpha = 0.775232509215700110280368495370$. In general, …