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Take two positive iid Cauchy variates with density $$f(x)=\frac{2}{\pi}\frac{\mathbb I_{x>0}}{1+x^2}$$ and infinite expectation. The minimum variate $\min(Y_1,Y_2)$ then has density $$g(x)=\frac{8}{\pi^2}\frac{\pi/2-\arctan(x)}{1+x^2}\mathbb I_{x>0}$$ Since (by L'Hospital's rule) $$\frac{\pi/2-\arctan(x)}{1+x^2} \equiv \frac{1}{x^3}$$ at infinity, the ...

20

Let's find a general solution for independent variables $X$ and $Y$ having CDFs $F_X$ and $F_Y,$ respectively. This will give us useful clues into what's going on, without the distraction of computing specific integrals. Let $Z=\min(X,Y).$ Then, from basic axioms and definitions, we can work out that for any number $z,$ \eqalign{ F_Z(z) &= \Pr(Z\le ... 10 Per whuber's answer to Standardized Student's-t distribution, the density of the standardized t distribution on \nu degrees of freedom isf(x) = \frac{\Gamma(\frac{\nu + 1}{2})}{\Gamma(\frac{\nu}{2})} \frac{1}{\sqrt{\pi(\nu-2)}} \left[1+\frac{x^2}{\nu-2}\right]^{-\frac{\nu+1}{2}}.$$(Note that we need \nu>2 so we have two moments to standardize.)... 8 If X and Y are positive random variables both with finite means, then \frac{X}{X+Y} \in (0,1) and so E\left[\frac{X}{X+Y}\right] \in (0,1) also. Similarly, \frac{Y}{X+Y} \in (0,1) and so E\left[\frac{Y}{X+Y}\right] \in (0,1) also. Then, as noted in my comment on the question cited by the OP, we can write$$E\left[\frac{X}{X+Y}\right] + E\left[\...

6

Well, if you don't impose independance, yes. Consider $Z \sim Cauchy$ and $B \sim Bernouilli(\frac{1}{2})$. Define $X$ and $Y$ by: $$X = \left\{ \begin{array}[ccc] 0 0 & \text{if} & B = 0\\|Z| & \text{if} & B = 1\end{array}\right.$$ $$Y = \left\{ \begin{array}[ccc] . |Z| & \text{if} & B = 0 \\0 & \text{if} & B = 1\end{... 4 This answer is not as general as Whuber's answer, and relates to identical distributed X and Y, but I believe that it is a good addition because it gives some different intuition. The advantage of this approach is that it easily generalizes to different order statistics and to different moments or other functions T(X). Also when the quantile function is ... 3 If you also know the means of X,Y, you can use the definition of covariance:$$\begin{align}\operatorname{cov}(RX,RY)&=E[R^2XY]-E[RX]E[RY]\\&=E[R^2]E[XY]-E[R]^2E[X]E[Y]\\&=(\sigma_r^2+\mu_r^2)(c_{xy}+\mu_x\mu_y)-\mu_r^2\mu_x\mu_y\end{align}$$where c_{xy}=\operatorname{cov}(X,Y). 3 Provided e^2+c^2\ne 0, the event cX + d \lt eY + g defines a half plane. (When e^2+c^2=0, this event is either empty or universal, giving trivial answers.) This suggests we can simplify the problem with a change of coordinates. To this end, define a new random variable$$U = cX + (-e)Y + (d-g)$$so that the conditioning event is U \lt 0. The ... 2 It is a typo, but you should get zero for the integral anyway. You should have:$$\begin{aligned} \mathbb{E}(\tfrac{1}{2} \cos(t_1+t_2 + 2U)) &= \int \limits_0^{2\pi} \frac{1}{4 \pi} \cdot \cos(t_1+t_2 + 2u) \ du \\[6pt] &= \Bigg[ \frac{1}{8 \pi} \cdot \sin(t_1+t_2 + 2u) \Bigg]_{u=0}^{u=2\pi} \\[6pt] &= \frac{1}{8 \pi} \Bigg[ \sin(t_1+t_2 + 4 \...

2

I don't know what $t$ is doing in the OP's notation and so will ignore it here. The key is what is the meaning of "unrelated" in "random unrelated variables". If "unrelated" merely means different, then, as whuber points out in a comment on the OP's question, there is not enough information to answer the question. If "unrelated" means mutually independent ...

2

YES, $E[XAX^h]=E[XX^h]\circ A$, where $\circ$ is for Hadamard product. $E[XAX^h]$ mixes the columns, but since variables are i.i.d., it must that it will be function of variance $\sigma^2$ and a combination of weights in $A$. $$M=E[XAX^h]$$ $$M_{tk}=E[\sum_{ps} X_{tp} A_{ps} X^*_{ks}] =E[\sum_{p} X_{tp} A_{pp} X^*_{kp}]=\sigma^2\delta_{tk}A_{tk}$$ because $... 2 I am not sure if there is a simple way to frame this in combinatorial terms, other than to observe the equivalency that you have already observed. Nevertheless, a preliminary aspect of framing the result in combinatorial terms is to look at the moment generating function of the standard normal distribution. The MGF and its Maclaurin expansion are given by: ... 2 Summary Suppose the problem is described as follows:$X$and$Y$have a bivariate normal distribution with respective parameters$\mu_X$,$\mu_Y$,$\sigma_X$,$\sigma_Y$, and$\rho$and it is desired to find the mean of$R=X/(X+Y)$given that$X>0$and$Y>0$, then using Mathematica I was only able to find a symbolic result for$E(R|X>0, Y>0)$... 2 It is not. Take$X_n \sim Bernoulli(p)$and$Y_n=X_n+1$.$E(X_n/Y_n)=p/2E(X_n)=pE(Y_n)=p+1p/2\neq p/(p+1)$in general. 1 You can avoid conditioning on X in the proof only if you are assuming that X is fixed (i.e., non-random, a constant) or that X and the error term epsilon are independent. You need these assumptions to bring X outside the expectation operator. But these are very restrictive assumptions that can be replaced by the weaker assumption that the expected value of ... 1 For integer$n \geq 0$, suppose that$X$takes on value$n$with probability$p_n$where the$p_n$'s are nonnegative numbers such that$\sum_{n=0}^\infty p_n = 1$. Note that we don't assume that all the$p_n$are nonzero; in fact, in many cases of interest,$p_n = 0$for all$n > M$for some finite integer$M$. Be that as it may, notice that the graph of ... 1 Every value that$Y$takes on is the result of applying$g(\cdot)$to one or more values that$X$takes on. In your own example,$Y=3$is the value of$g(0)$and$g(42)$and no other values that$X$might take on. That is, we can partition the set of all possible values that$X$takes on into disjoint subsets (e.g.$\{0, 42\}$maps to$Y=3$,$\{1,99\}$maps ... 1 Let$X_1=\mathbf 1(Y\in A)$be a Bernoulli variable with some probability$p>0$. And, let$X_2$be a constant random variable that equals to$C$with probability$1$.$\mathbb E[XX^T]is surely full rank. Let's check the other one: \begin{align}M=\mathbb E[XX^T\mathbf 1(Y\in A)]&=\mathbb E[XX^TX_1]=\mathbb E \begin{bmatrix}X_1^3 & X_1^2X_2\\... 1 I've thought about this a lot (I am OP), and I've come up with the following conclusion through my readings: Consider RV X distributed wrt a density f(x\mid\theta_0) (some observed) Here we assume the data can be generated through a parameterization (justify the presence of conditioning), and that in particular samples wrt X are generated wrt a true ... 1 The issue is that you aren't considering the full support of cdf ofmax\{y_1,_2,_y\}. The full support is (0, \infty). Taking a look here: https://en.wikipedia.org/wiki/Uniform_distribution_(continuous) at the definition of F(x). Then consider that you'll have 1 minus this value, so for your problem you'd have: a=200, b=600 and then 1-F(y) = 1 if ... 1 Your notation is quite confusing to me, but about item number two: "random sample" means that the random variables X_1, \dots, X_n are independent and identically distributed. Particularly, they all have the same mean and variance. That is why \text{E}(X_i) = \mu and \text{Var}(X_i) = \sigma^2. I think you are confused about the difference between a ... 1 Use E(Z|A)=\frac{E(Z1_A)}{P(A)} #Conditional_expectation_with_respect_to_an_event So without any assumption about distribution of (X,Y),By defining A=\{ cX + d < eY + g\} we haveE(aX + b | cX + d < eY + g)=E(aX + b | A)=a E(X | A)+b=a\frac{E(X 1_A)}{P(A)} +b=a\frac{E(X 1_A)}{P(cX + d < eY + g)} +ba\frac{\int_{\{\omega \in \Omega\... 1 After an hour of testing I finally figured it out. It seems that the argument ub=7 means you're stopping the integration at 7, and that means that if you get more than 7 wins then it doesn't calculate it and basically the reward is 0. thus, right after you calculate ev you should add to it the probability of getting more than 7 wins times the reward for all ... 1 A typo: the\log$is missing from the rhs: $$\mathbb{E}_{p(\cdot|X)}[\log p(X,Z)|X]=\underset{k}{\sum}\sum_{z_k\in\mathcal Z}p(z_k|X_k) \log p(z_k|X_k)p(X_k)$$ Furthermore, given that the EM algorithm involves two types of parameters, the current one,$\theta^0$say, and the free one,$\theta$say, it would be safer to write the conditional expectation ... 1 The methods presented in this work (https://people.maths.ox.ac.uk/gilesm/files/SLOAN80-056.pdf) concern Multi-Level Monte Carlo (MLMC) methods for expectations of this form. MLMC is typically not designed to provide unbiased estimators per se, but can usually be modified to do so using the trick of McLeish. Broadly, if you are interested in$\$\mathfrak{I}=...

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