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E.g. let $\vec{Y}$ be a random vector with normal distribution on $\mathbb R^n$ with $\mathbb E\vec{Y}=\vec{0}$ and the identity matrix as covariance matrix. Then $\vec{X}:=\vec{Y}/||\vec{Y}||$ is a uniformly distributed unit-vector.

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No, there is no such $Y$. Suppose there were. Let $A = \{\omega : Y(\omega) \le \frac{1}{2}\}$ be the event that $Y$ is at most $\frac{1}{2}$. Then for any $0 \le a \le b \le 1$ we have \begin{align*}m(A \cap [a,b]) &= P(Y \le \frac{1}{2}, a \le X \le b) \\ &= P(Y \le \frac{1}{2}) P(a \le X \le b) && \text{by independence} \\ &= ... 2 A normal distribution does not actually have any minimum and maximum. So you ignore these two parameters. But yes, a normal distribution is determined by its mean and standard deviation, and the formula is given by: f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-(x - \mu)^2 / 2\sigma^2}, $$where \mu is mean and \sigma is standard deviation. So just plug in ... 2 By the law of total expectation we have$$E[Z[i]]=E[E[Z[i]|X[i],Y[i]]]=E[\frac{1}{3}X[i]+\frac{2}{3}Y[i]]=\frac{1}{3}E[X[i]]+\frac{2}{3}E[Y[i]]$$For the product,$$E[Z[1]Z[2]]=E[E[Z[1]Z[2]|X[1],Y[1],X[2],Y[2]]]=E[\frac{1}{3}X[1]X[2]+\frac{2}{3}Y[1]Y[2]]=\frac{1}{3}E[X[1]]E[X[2]]+\frac{2}{3}E[Y[1]]E[Y[2]]$$This should solve your problem. 1 You could call Y a generalized multi-variate indicator function that gives values -1 or 1 on the different parts of the support. Note you have to choose which sign to assign to 0 (however, a single point that is not a point mass doesn't change the values of any integrals). In general there is no closed form for P(X = v). However you can express it ... 1 Note that$$P(X^2 = a)=P(X=\pm \sqrt{a})$$So for each a, you just sum the probabilities of the associated underlying X taking the requisite values. In general, if you are applying a function g to a discrete RV X, then if g is a 1-to-1 function of X, you are indeed just "shuffling" the probability masses about on the x-axis. However, if g ... 1 If I were you, I would consider the negative binomial distribution, which illustrates, collectively, the number of failures before the third head. You already have your probability of success, p=\frac 14 and r=3 and so, knowing that your distribution is parameterized, you go on to solve for the expected number of failures. 1 As you said, this is the moment generating function for Y=e^{X} evaluated at 1. Y has a log-normal distribution. Unfortunately the MGF, M_{Y}(t), for the log-normal distribution is not defined for any positive t. See, for example, this proof. 1 Consider the function f(\lambda)=e^{-\lambda}\lambda^n. Its derivative is e^{-\lambda}(n\lambda^{n-1}-\lambda^n), or equivalently e^{-\lambda}\lambda^{n-1}(n-\lambda). Take for example n=1. Then f(\lambda) is increasing until \lambda=1, and then decreasing. So there are infinitely many pairs (\lambda_0,\lambda_1) with \lambda_0\lt 1 and ... 1 Let X_1,X_2,X_3 be three independent random variables, uniformly distributed over A=\{1,2,3,4,5,6\}, and Z=\max(X_1,X_2,X_3). For any n\in A, the probability that Z\leq n is the probability that any X_i is \leq n, hence:$$ \mathbb{P}[Z\leq n] = \left(\frac{n}{6}\right)^3 $$and:$$ \mathbb{P}[Z=n]=\mathbb{P}[Z\leq ...

1

The probability you want is $P(X>a+b|X>b)$ where $X$ has an exponential distribution. Using the CDF of the exponential distribution... $$P(X\leq a+b|X>b) = \frac{P(X\leq a+b\cap X > b) }{P(X>b)} = \frac{P(b< X \leq a+b)}{P(X>b)} = \frac{ e^{-b\lambda} - e^{(a+b)\lambda} }{e^{-b\lambda}}$$ $$=1 - e^{-a\lambda}$$ This means ...

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It means that $X$ and $Y$ have finite second moments, and so $\mathbb E[|XY|]<\infty$ by Hölder's inequality. Therefore the covariance (and hence the correlation) of $X$ and $Y$ exists. Recall that $$\operatorname{Var}(X+Y) = \operatorname{Var}(X) + \operatorname{Var}(Y) +2\operatorname{Cov}(X,Y),$$ and so $$\operatorname{Cov}(X,Y) = ... 1 First off, an event is a subset of the sample space; it doesn't make sense to talk about a probability distribution for E. As for what "prior" means, T is an exponential random variable with unknown parameter \lambda. The problem is saying that \lambda itself is a random variable, with U(a,b) distribution. So the probability density function of is ... 1 On this website http://economictheoryblog.com/2015/02/19/ols_estimator/ you can find the proof that the beta estimator of OLS is indeed an unbiased estimator of the true beta in the population. HTH 1 The normal distibution, also called the Gaussian distribution is the probability distribution that assigns to every measurable set A of real numbers the probability$$ \int_A \frac 1 {\sqrt{2\pi}} \exp\left( \frac{-1} 2 \left( \frac{x-\mu} \sigma \right)^2 \right) \, \frac{dx} \sigma. $$In particular, there is the cumulative probability distribution ... 1 There are t=\binom{n}{3} ways to choose 3 distribution centres. For i=1 to t, define random variable X_i to be 1 if the set of three distribution centres indexed by i mutually send trucks to each other, and X_i to be 0 otherwise. Then the number W of "triangles" is X_1+\cdots+X_t. By the linearity of expectation and symmetry, we have ... 1 If H_6 is the number of heads in 6 tosses:$$E(H_6) = E(H_6 | \text{biased coin})P(\text{biased coin}) + E(H_6 | \text {unbiased coin})P(\text{unbiased coin})$$Both conditional expectations are binomial ones: 6p for the biased, 3 for the unbiased one. So we get \frac{66}{20}\frac{1}{3} + 3 \frac{2}{3} = \frac{11}{10} + 2 = 3.1 indeed. The other one ... 1 You could use Newton's Method to solve the equation numerically: Set f(n) = 1−0.955^n −0.005^n\cdot0.995^{n −1}\cdot n− 0.005^2\cdot0.995^{n −2}\cdot \frac{n(n−1)}{2} -0.5 Then n is a solution of your equation exactly when n is a root of f,that is f(n) = 0. Newton's method goes like this: x_{m+1} = x_m - \frac{f(x_m)}{f'(x_m)} where f'(n) ... 1 I would plot the function: > n=0.001:0.001:20; > plot(n,1−(0.995.^n) −(0.005.^n).*(0.995.^(n −1)).*n− (0.005.^2).*(0.995.^(n −2)).*n.*(n−1)/2)) Then see where the graph cuts 0.5 1 In Matlab and Octave "fsolve" is a nice function for finding roots to functions. First you need a "function handle" for your function, you can do this defining it as a lambda expression and assigning it a name, say f: f = @(n) 1−(0.955^n) −(0.005^n)*(0.995^(n −1))n− (0.005^2)(0.995^(n −2))n((n−1)/2) - 0.5; now typing n_solved = fsolve(@f, ... 1 As Clement pointed out, the the sigmoid function is not a valid pdf because its integral is not equal to 1. I am almost certain that you mean that it is the cdf, as it is a nondecreasing function that tends to 0 and 1 as x tends to -\infty and \infty respectively. Hints: In general, E[Y]=E\left[\sum_i X_i\right] = \sum_i E[X_i]. If the X_i ... 1 Hint: Let X\sim Bin(250,0.47)$$P(X\geq 125)=1-P(X\leq 124)\approx 1-\Phi\left( \frac{x+0.5-\mu}{\sigma} \right)=1-\Phi\left( \frac{x+0.5-n\cdot p}{\sqrt{n\cdot p\cdot (1-p)}} \right)= 1-\Phi\left( \frac{124+0.5-0.47\cdot 250}{\sqrt{250\cdot 0.47\cdot (1-0.47)}} \right)$$\Phi(z) is the cdf of the standard normal distribution. 0.5 is the ... 1 For y in the interval (1/e,1) we have$$\small F_Y(y)=\Pr(Y\le y)=\Pr(e^{-X}\le y)=\Pr(-X\le \ln y)=\Pr(X\ge -\ln y)=1-(-\ln y)=1+\ln y.$$For completeness, we add that F_Y(y)=0 if y\le 1/e, and F_Y(y)=1 for y\ge 1. Note, in the above calculation, the switch in the direction of the inequality, when \Pr(-X\le \ln y) was replaced by \Pr(X\ge ... 1 Hint: P(Y \le y \mid X=x)= \dfrac{y-x}{1-x} provided that x \le y So \displaystyle P(Y \le y) = \int_{x=0}^{x=y} \frac{y-x}{1-x} \,dx Then differentiate with respect to y to get the density 1 You want this:$$ f_{W,Z}(w,z) = \frac{\partial^2}{\partial w\,\partial z} \int_{-\sqrt{z}}^{\sqrt{z}}\int_{-\sqrt{w}}^{\sqrt{w}} f(x,y) \, dx \, dy.  Use the chain rule: Write $u= \sqrt z$ so that $u^2 = z$ and $2u\,du = dz$ and then you have \begin{align} \frac{df}{dz} & = \frac{df}{du}\cdot\frac{du}{dz} = \left( \frac d {du} \int_{-u}^u ...

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