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If $F$ is the standard normal CDF, $$(Q^{-1})'(p) = \dfrac{1}{F'(t)} = \sqrt{2\pi} \exp(t^2/2)$$ where $p = F(t)$. The maximum for $\epsilon \le p \le 1-\epsilon$ is at the endpoints. So $$|Q^{-1}( …

The characteristic function is $$\eqalign{ E\left[e^{itY}\right] &= E\left[ E\left[ e^{itY}|X \right]\right] \cr &= E \left[ \exp(it\mu X - t^2 X^2/2 \right] \cr &= \frac{\exp \left(\left(-(\alpha^2+ …

If $A$ is the matrix with rows $x_1$ and $x_2$, then $A \beta$ has a bivariate normal distribution with mean $A b$ and covariance matrix $A S A^T$. From that you can get the joint distribution of you …

I'm assuming that your "continuous" is actually "absolutely continuous", i.e. $X$ has a density. $$P(D=d) = E[P(D=d|X)] = \int dx \ P(D=d|X=x) f_X(x) = \int dx \sum_\ell \ P(D=d|X=x) f_{X|L}(x|\ell) …

Presumably your Gumbel random variables are independent. A Gumbel random variable $X_i$ with location parameter $\alpha_i$ and scale parameter $\beta_i$ has CDF $$ F_i(x) = \exp(-\exp((\alpha_i - x)/\ …

As Nate Eldridge noted, the answer is no. For a positive result, you need some extra condition. Suppose e.g. the variances $\text{Var}(X_n) < c (E X_n)^2$ for $n$ sufficiently large, with some const …

For $k > 0$, $P(Y=k)$ is the sum of coefficients of terms in the expansion of $(1 + f(k) (x_k-1) + \ldots + f(m) (x_m - 1))^n$ in which $x_k$ has degree $1$ and all $x_j$ for $j > k$ have degree $\n …

Using $\max(a,b) = \dfrac{a+b}{2} + \left| \dfrac{a-b}{2}\right|$, write $u = w_1 + |w_2| - |w_3|$ where $$ \eqalign{ w_1 &= C (y_1 + y_2) \cr w_2 &= \dfrac{1}{2} (x_2 - x_1 + C (y_2 - …

No. Take $n=3$. The CDF's of the order statistics are $F_1(u) = u^3 - 3 u^2 + 3 u$, $F_2(u) = 3 u^2 - 2 u^3$, $F_3(u) = u^3$. The function $F(u) = 4 (u - 1/2)^3 + 1/2$ is a CDF, and its unique repr …

The $p \in \mathbb R^{n(n-1)/2}$ corresponding to distributions on partitions form a convex polytope, the extreme points of which correspond to individual partitions. I don't know if there's a simple …

It's not true for sampling without replacement. Consider e.g. $N=9$, $p = 2/3$, $r = 6$. Note that the possible numbers of red balls in the sample are $3,4,5,6$, with $4$ corresponding to $\widehat{ …

According to Maple, the variance is $$ \dfrac{\sqrt{3}\; 27^{1/\alpha}}{6\pi} \Gamma\left(\dfrac{3+\alpha}{3\alpha}\right) \Gamma\left(\dfrac{3+2\alpha}{3\alpha}\right) \lambda^{-2/\alpha} $$

If $X$ and $Y$ have joint density $f(x,y)$, we have $p(q,\alpha) = \int_{-\infty}^\infty dy\ \int_{-\infty}^{q - \alpha y} dx\ f(x,y)$ and thus (assuming sufficient regularity) $\partial_\alpha p(q,\a …

Generalizing Efron, we can get probability $\ge 70\%$ with six 10-sided dice: 10 sides $=6$ 3 sides $=11$, 7 sides $=5$ 4 sides $=10$, 6 sides $=4$ 5 sides $=9$, 5 sides $=3$ 6 sides $=8$, 4 sides $ …

They are not equivalent. Suppose $X = [0,1]$, $\mu$ is a unit mass at 0, and $x_n = 1/n$. This sequence is $\mu$-uniformly-distributed-B, because for any continuous $f$, $\int f(x) \, d\mu = f(0) = …