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There are some issues here: The density is only positive when $0<x<y<1$ so you could use indicator variables, for example something like $f(x,y)=2 \, 1_{x<y}\, 1_{0<x<1}\, 1_{0<y<1}$ There can be confusion between the limits of integration and the variables being integrated over, so perhaps use something like $F(x,y)=\int\limits_{x'=-... 5 Sample from a Pareto distribution. If$Y\sim\mathsf{Exp}(\mathrm{rate}=\lambda),$then$X = x_m\exp(Y)$has a Pareto distribution with density function$f_X(x) = \frac{\lambda x_m^\lambda}{x^{\lambda+1}}$and CDF$F_X(x) = 1-\left(\frac{x_m}{x}\right)^\lambda,$for$x\ge x_m > 0.$The minimum value$x_m > 0$is necessary for the integral of the density ... 13 First, note that the range of$\DeclareMathOperator{\P}{\mathbb{P}} Y$is$(1, \infty)$. First find the cumulative distribution function of$Yin the usual way: \begin{align} F_Y(t) & = \P(Y \leq t) = \P(e^X \le t) \\ & = \P( X \leq \ln(t) ) \\ & = F_X( \ln(t) ) = 1-e^{-\lambda \ln(t)} \\ ... 4 How can we find the minimum and maximum of random variable/sample as it's a function, it's not a single number. As a number The maximum of a sample is the highest number of a sample. This is a single number. As a function But, when we consider the sample as a random variable that can take different values with different probabilities, then the maximum ... 1 You are right that in this case, for the prior predictive we get that \sum_{x=0}^\infty p(x)=\infty ,$$so if you need the prior predictive distribution, you will need to do the work to model a reasonable proper prior distribution. You did not tell us why you did choose the prior \frac1{\theta^{1/2}}, probably you need a better choice of prior. ... 1 The two statements :$$\left(\sqrt{n}(X_n - c) \overset{D}{\to}\mathcal{N}(0, 1)\right) \Longrightarrow \left(\sqrt{n}(g(X_n) - g(c)) \overset{D}{\to}\mathcal{N}(0, g'(c)^2)\right) $$and$$\left(\sqrt{n}(X_n - c) \overset{D}{\to}\mathcal{N}(0, \sigma^2)\right) \Longrightarrow \left(\sqrt{n}(g(X_n) - g(c)) \overset{D}{\to}\mathcal{N}(0, g'(c)^2\sigma^2)\... 0 You can writeY$as $$Y = WX$$ where $$W = \begin{pmatrix}1 &0&0&1\\0&1&0&-1\end{pmatrix}.$$ Then, $$\text{cov}(Y)=W\text{cov}(X)W'.$$ 0 I also struggled with the topic and the most intuitive reflection I've found is truncation. Indicator function effectively truncates the density. An example from (Train, 2009) can illustrate the point (e.g. some$v = 2): $$\text{P}(\varepsilon>-v) = \int_{-\infty}^{\infty}1_{\{\varepsilon \:>\: -v\}} \cdot f(\varepsilon)d\varepsilon \\ = \int_{-v}^{\... 0 In the definition of the conditional density$$f_{X|Y}(x|y) = \dfrac{f_{X,Y}(x,y)}{f_Y(y)}$$both f_{X,Y}(\cdot,\cdot) and f_Y(\cdot) are densities wrt some appropriate dominating measures. You need to find the proper dominating measure for (X,Y) when Y=\mathbb I_{X>5}. 2 To supplement the answer given by @Ben, consider the following plot which is produced with the following R code. # Moment function m <- function(k, n, p, a){ res <- rep(NA, length(n)) for(i in seq_along(n)){ xx <- 0:n[i] ff <- exp(-1/(xx+a)) xf <- k*log(ff) + dbinom(xx, n[i], p, log=TRUE) #For computational stability res[i]... 2 Since X is a discrete random variable with finite support, it is simple to compute the moments of any function of X using the law of the unconscious statistician. Applying this law gives:$$\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) \... 1 As you pointed out (and assuming that your random sample comes from a normal distribution), an estimator of the standard errorSE$is$\sqrt{\frac{S^2}{n}}$. In the same way you see the$\bar{X}$as a random variable that represents the sampling distribution of the mean, you can see$S^2$as a random variable representing the the sampling distribution of ... 2 Let's see how far we can towards characterizing functions$f$where such a formula will work. We know it works for linear functions, but are there any others? How about when the random variables$X$have restricted values? The setting of the question is one in which$f$is given but the distribution of the random variable$X$is unknown and arbitrary--... 1 Think about adding the powers of matrix. For example the following identity holds:$\mathbf{A}^{a}\mathbf{A}^{b} = \mathbf{A}^{a + b}$Applied to your problem:$\mathbf{A}^{0.5}\mathbf{A}^{0.5}\mathbf{A}^{-1} = \mathbf{A}^{0} = \mathbf{I}$1 (1) -(3) each have one unrestricted dimension. As suggested in comments under the question, place constructed points on the plane$W - H - Q = 0$:$ \\ (W,H,Q) = \begin{cases}X\ge 0,~Z\ge 0:&W = X,&Q=Z,&H=W-Q\\ \\ Y\ge 0,~Z<0:&H=Y,&Q=Z,&W=H+Q\\ \\ X<0,~Y<0:&W=X,&H=Y,&Q=W-H\\ \end{cases} $This ... 3 The exact formula for the variance of$Y$requires use of the function$f$and the full distribution of$X$(not just its variance). Nevertheless, while there is no exact formula of the kind you want, you can get approximate formulae using Taylor approximation (also called the "delta method"). To facilitate analysis using the delta method, ... 7 Let$X\sim \mathcal N(0,\sigma^2)$denote a normal random variable and let$f$be the function $$f(x) = \begin{cases}+1, & x > 0,\\-1, &x \leq 0.\end{cases}$$ Then,$f(X)$is a random variable taking on values$\pm 1$with equal probability and so$f(X)$has variance$1$. On the other hand, if$X\sim \mathcal N(1,\sigma^2)$, then$f(X)\$ takes on ...