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Let $X \sim Uniform(0,1)$ with pdf $f(x)$: Let $X_n$ and $X_{n-1}$ denote the largest and second largest order statistics in a sample of size $n$. Then, the joint pdf of $(X_{n-1},X_n)$, say $g(x_{n-1},x_n)$, is: where I am using the OrderStat function from the mathStatica package for Mathematica to automate, or do it manually following: Wiki -- ...

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The probability that $n-1$ of them are less than $x$ is $nx^{n-1}(1-x)+x^n$, so the PDF of the second-highest is the derivative of that. If the second-highest is $x$, then the expected value of the sum of the top one is $(1+x)/2$, and the sum of the top two is $(1+3x)/2$. Integrate the PDF of the second-highest, multiplied by $(1+3x)/2$.

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Let $\left(\Omega,\mathcal{A},P\right)$ be a probability space. Then a random variable on it is a function $X:\Omega\rightarrow\mathbb{R}$ such that $X^{-1}\left(B\right)=\left\{ \omega\in\Omega\mid X\left(\omega\right)\in B\right\} \in\mathcal{A}$ for each Borelset $B$. Denoting the collection of Borelsets on $\mathbb{R}$ by $\mathcal{B}$ we state that ...

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To help you prove that $Y$ is a random variable, we need to know what kind of definition of random variables you are familiar with. Regarding the distribution function: $F(t) = P(X \leq t)$. Let $F_Y$ be the distribution function of $Y$. Then $$F_Y(t) = P(Y \leq t) = P(aX + b \leq t) = P\left(X \leq \frac{t-b}{a}\right) = F\left(\frac{t-b}{a}\right)$$ if $a ... 0 Let$G$be the c.d.f of$Y$and$f$be the p.d.f of$U. Then, we have: \begin{align*} \displaystyle G(y)&=Pr(Y\leq y)=Pr\left(\frac{1}{1+U}\leq y\right)&\\ &=Pr\left(\frac{1-y}{y}\leq U\right)=\int_{(1-y)/y}^{1}f(x)\,dx=\frac{2y-1}{y}&\\ \end{align*} Thus,g(y)=G^{\prime}(y)=\displaystyle\frac{1}{y^{2}}$with$1/2\leq y\leq 1$PS: My way ... 2 It is known that for gaussian measures independent=uncorrelated. But, if I properly understand, you are worried about some expected relation. It is an analogy from the linear algebra, but I hope, that it may resolve your doubts. If$X$,$Y$are lineary indepent, such are$X+2Y$and$X-2Y$. 1 We can prove that$X$and$X^2$are independent if and only if$X^2$is constant. One possible proof is to observe that, whenever$0 \leq a \leq b$and$\Bbb{P}(a \leq |X| \leq b) > 0$, we have $$a^2 \Bbb{P}(a \leq |X| \leq b) \leq \Bbb{E}(X^2 \mathbf{1}_{\{a \leq |X| \leq b\}}) \leq b^2 \Bbb{P}(a \leq |X| \leq b).$$ Now if$X$and$X^2$are ... 1 There are cases where$X$and$X^2$are independent! Consider$X$with$P(X=-1)=P(X=+1)=\frac12$, for example. Let$X$be a random variable and assume that$X^2$is not a.s. constant (which is the case especially for$X$with uniform distribution, in fact for anything that is not of the form$P[X=a]+P[X=-a]=1$for some$a\in\mathbb R$). Then there exists an ... 0 How 'bout$P(0<X^2<1/2) \ne P(0<X^2<1/2\,\mid\,3/4<X)$? 0 If$a$and$b$are independent gaussian random variables and if$a$and$b$are centered with the same variance$\sigma^2$then, for every$t$,$x(t)=a\sin(2πf_0t)+b\cos(2πf_0t)$is gaussian centered with variance$\sigma^2\sin^2(2πf_0t)+\sigma^2\cos^2(2πf_0t)=\sigma^2$. If$a$and$b$are centered with different variances, then$x(t)$is gaussian centered ... 0 In the general case, the magnitude will be a Rice distribution. See: http://en.wikipedia.org/wiki/Rice_distribution If it is zero mean for x and y, then the phase will be uniform. 4 Let$W$be the maximum of the$X_i$, where$n$is fixed. Then$W\le t$if and only if all the$X_i$are$\le t$. This has probability$\sim(1-t^{-\alpha})^n$. Thus $$\Pr(W\gt t)\sim 1-(1-t^{-\alpha})^n.$$ But for fixed$n$and large$t$we have, taking the first two terms of the binomial expansion,$1-(1-t^{-\alpha})^n\sim nt^{-\alpha}$. 1 You have$E [ \sum_n |X_n| ] = \sum_n E[|X_n|] < \infty$, hence$\sum_n |X_n| < \infty$ae. It follows that$\sum_n X_n$converges absolutely ae. 1 Your integral calculates$P(Y \geq X \geq 0.5)$, not$P(Y \geq X| X \geq 0.5)$. 5 Hint: for each$k$$$\mathbb{E}[X_1|X_1 + ... + X_n = x] = \mathbb{E}[X_k|X_1 + ... + X_n = x]$$ Now what happens if you make the sum? 1 No, in general we cannot expect this; at least if we do not want to extend the given probability space. Counterexample: Let$(\Omega,\mathcal{A},\mathbb{P})$be a probability space where$\mathcal{A} = \sigma(Z)$for some standard Gaussian random variable$Z$. Then we cannot decompose$Z$as the sum of two independent Gaussian random variables, i.e. write ... 0 That table is a joint probability distribution table for$X$and$Y$. For example, the first empty box has$P(X=1 \cap Y=1)$. I'll give two methods to calculate all the joint probabilities. The first a more obvious method and the second perhaps a more intuitive method. Solution$1$We mostly use the formula definition for conditional probability in ... 0 This is the same as$P \left(X_1-X_2>0\right)=\int_0^{\infty } \left(f_1\left( x\right)-f_2\left( x\right)\right) \, dx$$$=\frac{1}{2} \left(\frac{\text{erf} \text{\mu 2}}{\sqrt{2} \sigma }-\frac{\text{erf} \text{\mu 1}}{\sqrt{2} \sigma }\right)$$ 3 The random variables$X_{n+1}$and$M$have the same mean. Let$Y=X_{+1}-M$. Then$Y$has mean$0$. The random variable$Y$is a linear combination of independent normals, so it is normal. Note that$X_{n+1}$has variance$1$and$M$has variance$\frac{1}{40}$. Thus$Y$has variance$1+(-1)^2\frac{1}{40}=\frac{41}{40}$. Now that we know the mean and the ... 2 Let$X$be the number of bull's eyes from the first gun, and$Y$the number from the second gun. We want$E(4X+6Y)$, which is$4E(X)+6E(Y)$. Note that$E(X)=(10)(0.3)=3$and$E(Y)=4$. Remark: The number$X$of bull's eyes from the first gun has binomial distribution$n=0$,$p=0.3$. So$E(X)=np$. Your answer had the right structure, but did not take into ... 3 The starting point here is the definition of conditional probability: $$P(X|Y) = {P(X,Y) \over P(Y)}$$ Then just repeatedly applying that gets the result. Just to make it easier to view, write$Z\equiv Z_1, Z_2, ..., Z_n$. Then $$P(X|Z,Y)={P(X,Y,Z)\over P(Y,Z)}={P(Y|Z,X)\times P(Z,X) \over P(Y|Z)\times P(Z)} = {P(Y|Z,X)\times P(X|Z) \over P(Y|Z)}$$ ... 0 Suppose$X$is m.s. differentiable. Then for any$t$fixed, $$\frac{X(s)-X(t)}{s-t}\to_{m.s.}X'(t)\quad \text{as }s\to t$$ It thus follows that $$\frac{\mu_X(s)-\mu_X(t)}{s-t}\to_{m.s.}\mu_{X'}(t)\quad \text{as }s\to t \tag 1$$ because the limit of the means is the mean of the limit, for a m.s. convergent sequence (in fact if$X_n\to_{m.s.}X$then ... 0 Let$Y_{1} = X_{1}^{2}$and$Y_{2} = X_{1}X_{2}$. The inverse of this transformation is$x_1=\sqrt{y_1}$and$x_2=\frac{y_2}{\sqrt{y_1}}. The Jacobian is $$J=\frac{\partial(x_1,x_2)}{\partial(y_1,y_2)}= \left|\matrix{\frac{\partial x_1}{\partial y_1} & \frac{\partial x_1}{\partial y_2}\\ \frac{\partial x_2}{\partial y_1} & \frac{\partial ... 1$$X = X_1 + X_2 + \cdots + X_n.$$Then$$\begin{align*} X^2 &= (X_1 + X_2 + \cdots + X_n)^2 \\ &= X_1^2 + X_2^2 + \cdots + X_n^2 + 2(X_1 X_2 + X_1 X_3 + \cdots + X_1 X_n) \\ &\quad + 2(X_2 X_3 + X_2 X_4 + \cdots + X_2 X_n) \\ &\quad + \cdots + 2X_{n-1} X_n \\ &= \sum_{i=1}^n X_i^2 + 2 \sum_{i < j} X_i X_j. \end{align*}$$Then since ... 0 When X_1, X_2, \ldots, X_n are independent random variables, we have$$\operatorname{Var}\left[\sum_{i=1}^n c_i X_i\right] = \sum_{i=1}^n c_i^2 \operatorname{Var}[X_i]$$for any fixed scalar constants c_1, c_2, \ldots, c_n. So for example, with n = 2, the above formula reduces to$$\operatorname{Var}[c_1 X_1 + c_2 X_2] = c_1^2 \operatorname{Var}[X_1] ... 0 Hint: use the variance of a Poisson distribution three times use the variance of a scaled distribution two times use the variance of a sum of independent random variables 1 The distribution of the sum of the two independent normally distributed variablesX \sim N(\mu_X, \sigma_X^2)$,$Y\sim N(\mu_Y, \sigma_Y^2)$is normally distributed, more specifically if$Z = X + Y$then$Z \sim N(\mu_X + \mu_Y, \sigma_X^2 + \sigma_Y^2)$, see e.g. the wiki for a proof. In your case that would result in$Z = Y_1 - Y_2 \sim N(-4, 10)$. 0 yes:$f(y_{1}|y_{2})*f(y_{2})=f(y_{1},y_{2})=f(y_{2}|y_{1})*f(y_{1})$0 Use fact that$P(X=1, Y=2, Z=3)=P(Z=3|X=1, Y=2)P(Y=2|X=1)P(X=1)$0 Think about the probability of getting a particular value for each variable. For$x=1$we need to roll a 1 with a single roll. So that's$\frac{1}{6}$. For$y=2$we need to roll a 2 with$x$dice rolls. So we need to get a 2 with 1 roll. That's$\frac{1}{6}$. For$z=3$we need to roll a 3 with one of the rolls, and then something smaller than or equal to ... 1 An indirect approach follows from the total expectation formula, since you know$\mathbb{E}(X)$,$\mathbb{P}(X > x)$, and$\mathbb{E}(X > x)$already. A direct approach works too: $$\mathbb{E}(X|X \leq x) = \frac{\int_0^x y e^{-y} dy}{\int_0^x e^{-y} dy}.$$ In general we have $$\mathbb{E}(X | A) = \frac{\mathbb{E}(X 1_A)}{\mathbb{P}(A)}$$ for any ... 1 To check if your intuition is correct, you can use that the density of$X$given that$X\in A$for a set$A$with$P(X\in A)>0is equal to $$f_{X\mid X\in A}(u\mid X\in A)=\begin{cases}\frac{f_x(u)}{P(X\in A)}, & u\in A\\ 0, & u \notin A\end{cases}$$ So in your case $$f_{X\mid X\le x}(u\mid X\le x)=\frac{f_X(u)}{P(X\le ... 1 We have$$E(X)=E(X|X\le k)\Pr(X\le k)+E(X|X\gt k)\Pr(X\gt k).\tag{1}$$You know E(X), E(X|X\gt k) and the probabilities mentioned in (1). Now E(X|X\le k) is determined. There are other ways to solve the problem, for example by finding the conditional distribution of X given X\le k. 1$$\operatorname{E}[\max(X,c)] = \operatorname{E}[X \mid X > c]\Pr[X > c] + \operatorname{E}[c \mid X \le c]\Pr[X \le c] = \frac{c+1}{2} \cdot (1-c) + c \cdot c.$$The first equality is applicable for any distribution on X. The second applies specifically to the case where X \sim \operatorname{Uniform}(0,1) with c \in (0,1). 1 Hint: Let Y=\max(X,c). If X\lt c then Y=c. Note that \Pr(Y=c)=c. If X\ge c then, by symmetry, Y has expectation \frac{1+c}{2}. 1 If we define \mathcal{F} := \sigma(\mathcal{G},\mathcal{H}), then \mathcal{F} contains by definition both \mathcal{G} and \mathcal{H} and therefore$$\mathbb{E}(X \mid \mathcal{F}) = X \qquad \text{and} \qquad \mathbb{E}(Y \mid \mathcal{F}) = Y.$$So, obviously the conditional expectations are again independent - there is no need for further ... 2 By the Jacobian formula,$$f_Y(y)=\frac{\mathbf 1_{|y|\lt1}}{\pi\sqrt{1-y^2}}.This is the so-called Arcsine distribution, which famously appears in probability theory and in number theory, as shown by two mathematical giants from the 20th century: \qquad\qquad\qquad\qquad 2 The support of the distribution is of course [-1,1]. For y\in[0,1], we have \begin{align} \Pr(Y\le y) & = \Pr(0\le X\le\arcsin y\text{ or }2\pi\ge X\ge \pi-\arcsin y) \\[10pt] & = \frac{\arcsin y + (2\pi-(\pi-\arcsin y))}{2\pi}. \end{align} The density is the derivative of that, and that is not a constant function, so it's not uniformly ... 0 First generate a vector \mathbf{x} with independent N(0,1) normal random components:x_i \sim N(0,1).$$Find the Choleski decomposition of the covariance matrix \Sigma:$$\Sigma = L L^T .$$Then \mathbf{y} = L^T\mathbf{x} is jointly normal with covariance matrix$$E(\mathbf{y}^T \mathbf{y})=E(\mathbf{x}^TLL^T \mathbf{x})=\Sigma E(\mathbf{x}^T ... 1 The skew normal distribution has finite variance for all\alpha$. All stable laws with finite variance are Gaussian (without skew). So, if the skewness parameter$\alpha\neq0$, the skew normal is not stable. To put it differently, the skew normal is either skewed or stable, but not both. 1 Define$E_n$to be the event that$|X_1 + \dots + X_n| \ge \max_{1 \le i \le n} |X_i|$. With$P(E_2) \ge \frac{1}{2}$established, we proceed by induction. Assuming that$P(E_n) \ge \frac{1}{2}$, we will argue that$P(E_{n+1}) \ge \frac{1}{2}$. The proof begins by conditioning$P(E_{n+1})$on whether the maximum is achieved at$X_{n+1}$. Letting$A$be the ... 1 Temporarily imagine working with (positive) numbers$a$and$b$. We want to maximize$ax_1+bx_2$in the part of the first quadrant that is on or below the lin$x_1+x_2=1$. It is obvious that the maximum is reached on the line$x_1+x_2=1$, and indeed reached (perhaps among other places) at a corner, that is, at$(1,0)$or$(0,1)$. So the maximum value is ... 1 The event$Z \ge z$occurs if and only if$X \ge z$and$Y \ge z.$So $$P[Z \ge z]=P[X \ge z]P[Y \ge z]$$ $$P[Z \ge z]=e^{- \lambda_1z} e^{- \lambda_2 z}=e^{-(\lambda_1 + \lambda_2)z}$$ Now it can be seen that$Z$is itself an exponential random variable with parameter$\lambda_1 + \lambda_2.$So it has mean $$E[Z]={1 \over {\lambda_1 + \lambda_2}}= {1 ... 1 You are correct that there are only two possible results. He gets 900 if all 21 show up and 1050 otherwise. So his expected gain is 900 \cdot P(21) + 1050 \cdot (1-P(21)). If less than 21 show up, you don't care how many it is, so you don't need to calculate the chance of 19, 18, \dots (though you could). 0 Hint: The probability that W\ge 2, or equivalently the probability that W\gt 2, is the probability that 0 cars go through the intersection in 2 minutes. 0 a)$$P(X\le x)=\int_{-1}^{x} f(t) dt$$b)$$\int_{-1}^{1} xf(x) dx=m$$c)$$Var(X) = E(X^2)- E(X)^2 = 2-\frac{1}{4}=\frac{7}{4}$$2 Let us suppose she meets people in sequence. Write L for lower, M for middle, and U for upper. Her classifications can be thought of as a word of length$6$over the alphabet L, M, U. We find the probability of LLMMUU. This word means that the first person fell in the lower category, as did the next one. Then came two middles, and then two uppers. The ... 1 Here is an answer for$n=2$. Let$A=\{x | |x_1+\cdots x_n| \ge \|x\|_\infty \}$. Also note that if$x\in A$, then$-x \in A$. Let$B=\{ x | x_1 = 0 \text{ or } x_2 = 0 \}$, we see that$B \subset A$. Let$Q_k^\circ$be the interior of the four quadrants, we see that$Q_1^\circ \subset A$and hence$Q_3^\circ \subset A$. Note that$B,Q_k^\circ$form a ... 0 Hint: for both directions, use the Borel-Cantelli lemma with the independent events$A_{n,\varepsilon}:=\{|X_n|\gt \varepsilon\}$. 0 Hint: Use:$\mathsf {Var}(a\,Y) = a^2\,\mathsf {Var}(Y)$Also:$\mathsf {Cov}(a\,X,b\,Y) = a\,b\,\mathsf {Cov}(X,Y)$PS: remember that:$\mathsf{Cor}(X,Y) =\frac{\mathsf{Cov}(X,Y)}{\sqrt{\mathsf{Var}(X)\mathsf{Var}(Y)}}\$

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