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If $(Y_n)_{n\geqslant 1}$ is a sequence of real valued random variables which converge in distribution to $Y$ and $g\colon\mathbf R\to\mathbf R$ is a continuous function, then $(g(Y_n))_{n\geqslant 1}$ converges to $g(Y)$ in distribution.

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Hint: In full generality (if $U$ and $V$ are integrable), $$E(\color{red}{5}U+\color{blue}{3}V)=\color{red}{5}\cdot E(U)+\color{blue}{3}\cdot E(V).$$ If $U$ and $V$ are independent (and square integrable), $$\mathrm{var}(\color{red}{5}U+\color{blue}{3}V)=\color{red}{5}^2\cdot \mathrm{var}(U)+\color{blue}{3}^2\cdot \mathrm{var}(V).$$

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Each $Y_1, Y_2, \ldots$ follows a logarithmic distribution: $$\Pr(Y=k) = -\frac{(1-p)^k}{k\log(p)}, \quad \text{where } k\geqslant1, 0<p<1.$$ If $N \sim \mathsf{Poi}(\mu)$ independently of $Y_1, Y_2, \ldots$, then we find that $X$ has a negative binomial distribution with parameter $r = -\mu/\log(p)$ and success probability $p$. Hence, the parameter ...

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$X,Y$ are independent if and only if $$P\left\{ X\in A\wedge Y\in B\right\} =P\left\{ X\in A\right\} P\left\{ Y\in B\right\}$$ is true for measurable sets $A,B$. If this is the case and $f,g:\mathbb{R}\rightarrow\mathbb{R}$ are measurable functions then: $$P\left\{ f\left(X\right)\in A\wedge g\left(Y\right)\in B\right\} =P\left\{ X\in ... 0 Evidently so. See this link Are functions of independent variables also independent? which addresses independence. It's clear they are identically distributed, isn't it? 0 Yes. In fact one can show that X and Y are independent if and only if f(X) and g(Y) are independent, for every measurable function f, g I don't know how you introduced independence, but X and Y independent means that the \sigma-algebrae generated by them are independent; it is easy to see that the \sigma-algebra generated by f(X) is ... 0 No, it is not correct. Please note that A = (-\infty,x] is a subset of \mathbb{R} whereas \mathbb{P} is a (probability) measure on the probability space. This means that \mathbb{P}(A) is not even well-defined. For the first one, note that$$\mathbb{E} \big( 1_{(-\infty,x]}(V_i) \big) = \int 1_{(-\infty,x]}(V_i(\omega)) \, d\mathbb{P}(\omega) = ...

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Prescribe $g:\mathbb{R}^{2}\rightarrow\left\{ 0,1\right\}$ by $\left(x,y\right)\mapsto1$ if $x\in\left(0,1\right)\wedge y\in\left(0,\sqrt{x}\right)$ and $(x,y)\mapsto 0$ otherwise. Then the PDF for $X$ and $Y$ are respectively: $$f_{X}\left(x\right)=\int g\left(x,y\right)8x^{2}ydy$$ and: $$f_{Y}\left(y\right)=\int g\left(x,y\right)8x^{2}ydx$$ For a fixed ...

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Draw a picture. The joint density function "lives" over the part $D$ of the unit square that is below the half-parabola $y=\sqrt{x}$. To find the density function of $y$, we have to "integrate out" $x$. The function $8x^2y$ is the joint density only on $D$, so we have to confine attention to $D$. Note that at the beginning of $D$ we have $y=\sqrt{x}$, ...

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In general, no. A standard counterexample is to take your probability space to be $\Omega = [0,1]$ with Lebesgue measure, and consider the complex-valued random variables $X_n(\omega) = e^{2 \pi i n \omega}$, which are all bounded in absolute value by 1. They are orthonormal in $L^2(\Omega)$, so by Bessel's inequality, they converge weakly to 0 in ...

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For $y>0$: $$P\left\{ X=n\mid\lambda=y\right\} =e^{-y}\frac{y^{n}}{n!}$$ and PDF of $\lambda$ is $f_{\lambda}\left(y\right)=e^{-y}$. This leads to: $$P\left\{ X=n\right\} =\int P\left\{ X=n\mid\lambda=y\right\} ... 1 It's hard to tell from a small sample like this. I suggest using the same method in Excel to generate about 1000 points. Then plot the results on a histogram to see what general shape the distribution takes. If the histogram is relatively flat, you're probably getting the uniform distribution. If you see peaks or slopes, you aren't. 0 Let T denote the time between two successive X_2 arrivals, then T is exponential with parameter \lambda_2 and, conditionally on T=t, the distribution of N is Poisson with parameter \lambda_1t. Thus, E(s^N\mid T)=\mathrm e^{-\lambda_1T(1-s)} and E(s^N)=E(\mathrm e^{-rT}) with r=\lambda_1(1-s). Recall that, for every u\gt-\lambda_2, ... 1 We use the following two facts: for a (deterministic) sequence of real numbers (a_n)_{n\geqslant 1}, we have \limsup_na_n=\limsup_na_{n+k} for any integer k; if (Y_n)_{n\geqslant 1} is a sequence of random variables, then \limsup_n Y_n is \sigma(Y_n,n\geqslant 1) measurable. 2 The proof of the convergence in distribution of (X_n,Y) to (X,Y) is done here (the use of characteristic functions could make it shorter). Assume that X=h(Y) for some Borel function h. Using the assumption with g:=f\circ h, a bounded Borel-measurable function, we obtain that for each continuous and bounded function f,$$\lim_{n\to ...

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Yes: as soon as at least one of the random variables $X$ and $Y$ has a density and if $X$ and $Y$ are independent, the sum $X+Y$ has a density. To understand the magic, assume that $(X,Y)$ is independent, that $Y$ has density $f$ and that $X$ is purely discrete with $P(X=x_n)=p_n$ for every $n$, then $X+Y$ has density $g$ with $$g(x)=\sum_np_n\,f(x-x_n).$$ ...

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For a fixed $t$, the series defining $X(t)$ has only one term which can be different from $0$ (for the potential index $n$ for which $nT\lt t\leqslant nT+T/2$). Therefore, we can switch the sum and the expectation and we find that $\mathbb E[X(t)]=0$ for each $t$ because $p(t-nT)$ is not random and $\mathbb E[A_n]=0$. For the covariance, by the argument ...

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No. Assume $X$ be a binary (Bernully random variable) having values $-1,1$ with probabilities $(1/2,1/2)$. Then $Y+X$ will not have a density.

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From the graph, $Y$ cannot be less than $-b$ nor greater than $b$. $$\Pr(Y<-b)=0\\ \Pr(Y>b)=0$$ However, at the points of inflection, we know $Y$ has a probability mass: $$\Pr(Y=-b)=\Pr(X\leq -a) \\ \Pr(Y=b)=\Pr(X\geq a)$$ In the interval between there is a linear relation between $X$ and $Y$ so $$\Pr(Y\le y)=\Pr(X\le ay/b) , \forall y\in (-b, b)$$ ...

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If $b\leq y$ then $g\left(x\right)\leq y$ is true for every $x$ so that: $$F_{Y}\left(y\right)=P\left\{ g\left(X\right)\leq y\right\} =1$$ If $-b\leq y<b$ then $g\left(x\right)\leq y\iff x\leq\frac{a}{b}y$ so that: $$F_{Y}\left(y\right)=P\left\{ g\left(X\right)\leq y\right\} =P\left\{ X\leq\frac{a}{b}y\right\} =F_{X}\left(\frac{a}{b}y\right)$$ If ...

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First, the text below requires that you know definitions of a real random variable (or vector), measurable function, Borel $\sigma$-algebra and cumulative distribution function. Let $X$ be an $n$-dimensional random vector with the CDF ${F}$ (this is a fundamental assumption which begs the question does the appropriate probability space and the random ...

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The answer to your question is a matter of interpretation. If you formalize the sample spaces abstractly enough, then the sample spaces for $X_1$ and $X_2$ can be the same. After all, both weight and height are continuous quantities, and so can be modeled as real numbers. If the real numbers aren't a general enough setting, you can go into more abstract ...

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By the strong law of large numbers for i.i.d. integrable sequences, $$\frac1N\sum_{i=1}^N\log X_i\to\nu\quad\text{almost surely},$$ where $\nu=E(\log X_1)$. This is equivalent to the assertion that $$\lim_{N \to \infty} \left( \prod_{i = 1}^{N} X_{i} \right)^{1/N}=\mathrm e^\nu\quad\text{almost surely},$$ hence the expected value of the LHS exists and is ...

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This is my crack at the question, please if anybody sees any flaws, point them out I am assuming that $X_{i}$'s are independent. From this it is important to know that $$\left(\prod_{i=1}^{N}X_{i}\right)^{\frac{1}{N}}=e^{\frac{1}{N}\sum_{i=1}^{N}\ln(X_{i})}$$ Well notice that $\frac{1}{N}\sum_{i=1}^{N}\ln(X_{i})$ is just the average of the log of the ...

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The function $\omega \mapsto(X_1, X_2, \cdots X_n)$ is a measurable function from $\Omega$ to $\mathbb{R}^n$. It is, more strongly, also $\sigma(X_1, X_2, \cdots X_n)$-measurable. Since $f$ is measurable, the composition $f(X_1, X_2, \cdots X_n)$ is $\sigma(X_1, X_2, \cdots X_n)$-measurable.

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If your set consisted of one value only, then the probability to guess it correctly would be $p=\frac{1}{2^{8 \cdot 32}}$. Since your records are all different, the probability is $1,000,000 \cdot p$.

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$X_1,\ldots,X_n$ are the $n$ i.i.d. random variables. Let $X_{(1)}<\cdots< X_{(n)}$ be the order statistics, i.e. the same random variables sorted. (By continuity of the c.d.f., we need not write "$\le$".) Then $X_i=X_{(j)}$. Given $i$, what is the distribution of $j\text{ ?}$ Let $Y_1,\ldots,Y_n$ be $X_{\sigma(1)},\ldots,X_{\sigma(n)}$, where ...

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If $U_i$ is independent of $X_i$ then $E(Y_i\mid X_i)=E(X_i\mid X_i)+E(U_i\mid X_i)=X_i+E(U_i)$ hence $E(Y_i\mid X_i)=X_i$. In particular, $E(Y_i\mid X_i=1)=1$ and $E(Y_i\mid X_i=0)=0$. (Also, I calculated $P(Y > \frac{1}{2}) = 0.6728p + 0.1932$ from an earlier part of the question). Actually, if $\sigma^2=\frac13$ then $P\left(Y > ... 3 Let$Y$be uniform random variable on$(0,1)$and conditionally on$Y$,$X$is a centered normal random variable with variance equal to$Y$. Then$E(X|Y) = E(X) = 0$, but$X$is not independent of$Y$since$E(X^2 | Y) = YE(X^2) = E(Y) = \frac{1}{2}$implies$X \in L^1$1 Your result is correct. It can be done more efficiently by: $$P\left\{ z<X<Y\right\} =\int_{z}^{\infty}f_{X}\left(x\right)P\left\{ z<X<Y\mid X=x\right\} dx=\lambda\int_{z}^{\infty}e^{-\lambda x}e^{-\mu x}dx=\frac{\lambda}{\lambda+\mu}e^{-\left(\lambda+\mu\right)z}$$ 1 Consider independent Poisson processes$N_t$,$M_t$with rates$\lambda$,$\mu$, such that$X$is the time until the first occurrence of$N_t$and$Y$is the time until the first occurrence of$M_t$. Then$N_t + M_t$is a Poisson process with rate$\lambda + \mu$, and$Z$is the time until its first occurrence. One way to realize this is to start with a ... 0 By Lévy's continuity theorem it's enough to proof that$\left(\varphi_{1/X_1}(t/n)\right)^n \rightarrow e^{-a|t|}$, with$\varphi_X(t):=E^X\left[e^{itX}\right]$. It is a well-known fact that if$(c_n)_{n\in \mathbb{N}}$is a complex succession with limit$c$then$\lim_{n\rightarrow +\infty}\left(1+c_n/n\right)^n=e^c$. In this case ... 1 If$y=0$, then$Y_y=0$or$Y_y=1$, depending on the convention. If$y\gt0$, then:$P(Y_y\gt0)=pP_1(T_y\lt T_0)P(Y_y\gt n+1\mid Y_y\gt n)=P_y(T_y\lt T_0)=pP_{y+1}(T_y\lt\infty)+(1-p)P_{y-1}(T_y\lt T_0)$Similar recursions hold when$y\lt0$. If one is able to compute the various probabilities$P_a(T_b\lt T_c)$involved (and one should be), these ... 6 For$i=1$to$50$, define random variable$X_i$by$X_i=1$if box$i$has no balls, and$ X_i=0$otherwise. Then$X=X_1+\cdots +X_{50}$, so we want$E(X_1+\cdots+X_{50}$). By the linearity of expectation this is$E(X_1)+\cdots+E(X_{50})$. But$\Pr(X_i=1)$is the probability all the balls miss box$i$. This is$\left(\frac{49}{50}\right)^{100}$, and is ... 1 Prob$(X=i)$=Prob(All 100 balls go into$(50-i)$boxes)=${50\choose{50-i}}*\frac{(50-i)^{50+i}}{50^{100}}$The denominator is the total number of ways 50balls can be put in 100 bxes, each ball can be put on any of the 50 boxes, i.e in 50 ways, so 100 balls can be put in$50^{100}$ways. The numerator is the product of number of ways to choose the (50-i) ... 1 Define$\xi_i$to be a family of independent identically distributed Bernoulli variables with success probability$p$. Then$E[\xi_i]=p$for all$i$. Note that $$Y=\sum_{i=1}^X\xi_i$$ Wald's Identity states that if$N$is positive integer-valued random variable with finite expectation and$\eta_i$are independent identically distributed random ... 1 Formally, whenever you have a parametrized family of distributions$D(\theta)$for the parameter set$\theta\in \Theta$and some random variable$X$which takes values in$\Theta$and has the distribution$q$, if you define a new random variable as$Y \sim D(X)$, then the distribution of$Y$is given by $$\mathsf P(Y\in A) = \int_\Theta ... 1 E[Y] = E[E[Y|X]] The conditional expectation is pX. Then, E[pX] = pE[X] = \mu p 3 A simple example is X uniformly distributed on the usual Cantor set, in other words,$$X=\sum_{n\geqslant1}\frac{Y_n}{3^n},$$for some i.i.d. sequence (Y_n) with uniform distribution on \{0,2\}. Other examples are based on binary expansions, say,$$X=\sum_{n\geqslant1}\frac{Z_n}{2^n},$$for some i.i.d. sequence (Z_n) with any nondegenerate ... 0 The iff is not true:$$f(x)=\begin{cases} 2 \text{ if } 0\le x \le \frac {1}{4}\\ \frac {2}{3} \text{ if } \frac {1}{4}< x \le 1\\ 0 \text{ otherwise}\end{cases} f\not\in C^0F_X(x)=\int_{-\infty}^x f(x) dx$X\$ is continuous r.v.

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