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First, a little intuition. When $\tau = 1$, $X = S$. So $P(X = S) = p$. That means that once we're given $S = s$, $X = s$ with probability $p$. When $\tau = 0$, then knowing that $S = s$ doesn't tell us anything about $X$, and $X \sim U[0,1]$ with probability $1-p$. This mean that $X|S = s$ is a mixture of a uniform distribution and a degenerate distribution....
If $Z_i\sim N(,\Sigma)$ then $Z_i=\Sigma^{1/2}Y_i$ are $N(0,1)$, where $\Sigma^{1/2}$ is any square root of $\Sigma$. So $f(Z_1,\dots,Z_n)=(f\circ g)(Y_1,\dots,Y_n)$, where $g$ is multiplication by $\Sigma^{1/2}$. Clearly $f\circ g$ is still Lipschitz. If $\lambda$ is the largest eigenvalue of $\Sigma$, the Lipschitz constant is bounded above by $L\sqrt{\... 0 Let$X_1,\cdots ,X_{100}\overset{iid}{\sim} F_X(x)$for some distribution$F_X(x)$that has an expected value$\mu$. Since$X_1$is distributed as$F_X(x)$,$\mathbb E[X_1]=\mu$. Since$X_2$is distributed as$F_X(x)$,… In the absence of other information about the subject, the value that you expect is the population value. 0 Yes, assuming independence and equal probabilities, this is just the binomial distribution. So the number of alleles of value 1 is$X\sim \mathcal{Binom}(20,p)$. 0 A parametric distribution is defined by its family and its parameters. Consider the Poisson family of distributions. This distribution has a single parameter$\lambda$. In GLMs, we wish to model the mean of the parametric distribution as a function of a linear model of covariates. The mean should be called, as you say, or is treated as "the ... 4 Your idea is right but you need to "weight" each conditional expectation by the associated probability. For a variable$X$and an event$E$, the conditional expectation$\mathbb E(X\mid E)$is defined as ($I$is the indicator function), $$\mathbb E(X \mid E) = \frac{\mathbb E(XI_E)}{\mathbb P(E)}$$ which gives $$\mathbb E(XI_E) = \mathbb E(X \... 3 There's no error. Start with X^\top X = (X^\top X+\lambda I) - \lambda I; premultiply both sides by (X^\top X+\lambda I)^{-1}. Simplify. Postmultiply by \beta 0 Two ways to approach the problem: Geometric order statistics For each member of the population, define (independent) random variables T_i, which is the day that member exits to the pool. T_i has geometric distributions,$$ \DeclareMathOperator{\P}{\mathbb{P}} \P(T_i=k)=(1-p)^k \cdot p, \quad k=1,2,3,\dotsc $$Then define the order statistics$$ T_{(... 2 To supplement the answer given by @Ben, consider the following plot which is produced with the following R code. # Moment function m <- function(k, n, p, a){ res <- rep(NA, length(n)) for(i in seq_along(n)){ xx <- 0:n[i] ff <- exp(-1/(xx+a)) xf <- k*log(ff) + dbinom(xx, n[i], p, log=TRUE) #For computational stability res[i]... 2 Since$X$is a discrete random variable with finite support, it is simple to compute the moments of any function of$X$using the law of the unconscious statistician. Applying this law gives:$\begin{align} m_k(n,p,a) &\equiv \mathbb{E} \Bigg( \exp \bigg( - \frac{k}{X+a} \bigg) \Bigg) \\[6pt] &= \sum_{x=0}^n \exp \bigg( - \frac{k}{x+a} \bigg) \... 0 IfX_1,...,X_n$are$n$independent draws from a population then$\text{Var}[\sum X_i]=n\text{Var}[X]$so that the standard error of$\sum X_i$is$\text{se}=\sqrt{\text{Var}[\sum X_i]}=\sqrt{n\text{Var}[X]}$. If you have an estimate for the population variance you can use this to estimate the standard error,$\hat{\text{se}}=\sqrt{n\hat{\text{Var}}[X]}$. ... 4 Break the problem into two parts. Work out the distribution of$n\bar y.$Assume$(y_1,\ldots, y_n)$has an$n$-variate Normal distribution. (Without this assumption, or some specific assumption about the joint distribution, the problem is insoluble.) Under this assumption$n\bar y,$being a linear combination of the$y_i,\$ has a Normal distribution. Let ...