# Tag Info

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This is possible. Note first that the constants in front of the exponential functions in $f_A$ and $f_B$ don't need to bother us, because we have $r$. This means we can concentrate on what happens in the exponents. The exponent of $f_A$ can be written as: $$-\frac{1}{2} x^T \Sigma_A^{-1}x + \mu_A \Sigma_A^{-1} x - \frac{1}{2}\mu_A \Sigma_A^{-1} \mu_A$$ The ...

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The vector $\vec X = [X_1,\dots,X_N]^T$ has a rotationally invariant distribution. That is, if $A$ is any orthogonal matrix, then the distribution of $\vec X$ and $A \vec X$ are the same. Hence by letting $A$ be an orthogonal matrix that takes $[1,\dots,1]^T$ to $[\sqrt N,0,\dots,0]^T$, your problem is the same as computing $$\Pr(\sqrt N X_1 \mid \sum ... 1 Hint: P(|Z|>2.4)= 1-P(|Z|\leq2.4)=1-P(-2.4\leq Z \leq 2.4)=1-2\times P(0\leq Z \leq2.4)=1-2 \times \big(\frac{1}{2}-P(Z>2.4)\big)=2\times P(Z>2.4)$$P(Z>x)= \int_{x}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}} dz$$0 HINT: If C(x)=x^2, then we can find a closed form for its expected value. Recall that$$I(a)=\int_{-\infty}^{\infty}e^{-ax^2}dx=\sqrt{\frac{\pi}{a}}$$and$$-I'(a)=\int_{-\infty}^{\infty}x^2e^{-ax^2}dx=\frac{\pi^{1/2}}{2a^{3/2}}$$Then, change variables from (x-\mu)/\sqrt{2\sigma} to x and expand the quadratic inside the integral. Can you ... 1 For the case of elliptically contoured distributions (of which the Gaussian is a special case), the distribution of the norm of Mv is available in the literature (See for example 1 and 2) M. Rangaswamy, D.D. Weiner, and A. Ozturk, "Non Gaussian Random Vector Identification Using Spherically Invariant Random Processes," Aerospace and Electronic Systems, ... 0 HINT... The 50th percentile is the mean \mu. You need to find \sigma, so get hold of a set of inverse \Phi tables and use \frac{712.92-\mu}{\sigma}=\Phi^{-1}(0.75). Can you take it from there? 1 We exploit the formula$$\min \{ X,Y \}=X-\min \{ X-Y,0 \}.$$Hence E[\min \{ X,Y \}]=E[X]-E[\min \{ X-Y,0 \}]=\mu-E[\min \{ X-Y,0 \}]. It remains to compute the second term. Since X,Y are iid N(\mu,\sigma^2), X-Y is N(0,2 \sigma^2). So the second term is$$\int_{-\infty}^0 0 dz + \int_0^\infty z n_{0,2\sigma^2}(z) dz$$where n_{m,s^2} ... -1 If I understand the question correctly, then X and Y each follow a normal distribution with the same mean \mu and the same variance \sigma^2. If so, then \mathrm{E}(X) = \mathrm{E}(Y) = \mu and \mathrm{E}(X - Y) = 0. But when \mathrm{E}(X - Y) = 0, then also \mathrm{E}(|X - Y|) = 0. Therefore,$$ \begin{eqnarray} \mathrm{E}(\min\{ X,Y \}) ...

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you can view the whole as an option pricing problem. $$\min(X,Y) = X - (X-Y)_+$$ So what is $$E((X-Y)_+)$$ ? $X-Y$ is distributed $N(0,2\sigma^2).$ Write it as $X-Y = \sigma \sqrt{2} Z$ with $Z$ standard normal, then $$E((X-Y)_+) = \sqrt{2} \sigma E(Z_+).$$ Now $$E(Z_+) = \frac{1}{\sqrt{2\pi}} \int \limits_{0}^{\infty} z e^{-z^2/2}dz = ... 0 N_4(\mu,\Sigma) is the pdf of the multivariate normal distribution. The marginal distributions are normal distributions whose parameters are given in the covariance matrix \Sigma and in the expectation vector \mu. If you recall the definition the entries of the covariance matrix, \Sigma,$$\sigma_{i,j}=E[(Y_i-\mu_i)(Y_j-\mu_j)].$$and ... 1 The formula doesn't actually have anything specifically to do with the normal distribution (as opposed to any other distribution). It simply yields the standard deviation, a measure of how spread out the values are from their average value (their mean). If the values are distributed according to the normal distribution, then about 68 percent of the ... 1 Suppose you have a Normal distribution X \sim N(\mu ,\sigma^2) we can transform the random variable into a standard Normal distribution using Z=\frac{X-\mu}{\sigma}. This is done so that we have Z \sim N(0,1) which is easier to work with. The cumulative distribution function of the standard Normal distribution is given by:$$\Phi ...

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I assume that the "random numbers" you used come from the uniform distribution. The variance of the uniform distribution is $$\frac1{12} (b - a)^2$$ where $a$ and $b$ are the lower and upper limit, so in your case $$\frac13 \mathtt{noiseMax}^2.$$ The variance of the sum of independent random variables is the sum of the variances, so for $n$ random ...

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In this particular case, it means that you draw $d$ times a $N(0,1)$ (real) random variable, and these random variables are independent. It means that $\varepsilon=(\varepsilon_1,\dots,\varepsilon_d)$, the $\varepsilon_i$ are independent, and are normally distributed, with variance $1$ and mean $0$. If you don't have the identity matrix but another ...

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Technically, the definition of the vector Gaussian likelihood density when you have mean 0 and covariance matrix $K$ is proportional to $f(x) = e^{-(1/2)x^T K^{-1} x}$, with constant of proportionality determined by $K$ so that it is truly a probability density. This is just how it's defined, and if you know numerical sampling methods like MCMC then this is ...

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due to symmetry, half the women are shorter that 165, so you need to calculate $$p(M<165)=p\left(z<\frac{165-178}{8}\right)$$ Can you finish it?

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I would break this question up into two pieces. First, what height are half of the women taller than? Since the mean of the women distribution is $165$ and the median of a normal distribution is its mean, the answer is 165. The second half of the question is: What proportion of men are less than the height we just calculated. i.e. What proportion of men ...

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The idea behind is OK but the following expression is wrong: $$P(g | X=x) = \frac{P(X=x | g)P(g)}{P(X=x)}.$$ Here is the right approach. Say, first, that the a black random point (denoted by $X$) lies in the interval $[x,x+\Delta x]$. Then we have $$P(g\mid x\le X \le x+\Delta x)=\frac{P(\{x\le X \le x+\Delta x\} \cap g)}{P(x\le X \le x+\Delta ... 0 You have the right idea but the wrong Bayes formula. You need to replace P(X=x|g), which is indeed 0, by f_g(x) where f_g is the pdf of N(-1,1). Similarly, P(X=x)=0 but you replace it by f_g(x)P(g)+f_b(x)P(b). This is called a gaussian mixture. 1 A random Gaussian process v = (v_k) with a covariance matrix U can be represented by v = U^{1/2} g, where g is a vector of i.i.d. \mathcal N(0,1) random variables. So it would seem reasonable that a (n \times p) matrix is called "distributed according to a matrix valued normal distribution" if it has some kind of representation like:$$ X = ...

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More generally, consider a smooth function $f(x)$ on $\mathbb R^n$ which has a local maximum at $x_0$ with $f(x_0) = 1$, and such that the Hessian matrix $H$ at $x_0$ is negative definite. Thus $$f(x) = 1 + \dfrac{1}{2} (x-x_0)^T H (x - x_0) + O(\|x - x_0\|^3)$$ Now take $x = x_0 + y/\sqrt{n}$. We have $f(x)^n = \exp(n \log f(x))$ where $$n \log ... 2 If we ae told that X and Y are jointly Normal, then we know that$$E[X|Y] = E[X] + \rho \frac{\sigma_X}{\sigma_Y}( Y-E[Y])$$in which in your case reduces to$$E[X|Y] = \rho Y$$In general, if we only know that the variables are marginally Normal, then I don't think there's much to say. Calling E[X|Y]=g(Y) we know that write$$E[XY] = E[E[XY|Y]] ...

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This is a special case of the more general phenomenon that two i.i.d. variables with normal distribution, considered as coordinates in the plane, yield a rotationally invariant distribution over the plane. This is due to the fact that the product of the exponentials is the exponential of the sum of the exponents, so the combined exponent is ...

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Suppose $X_1, X_2,..., X_k$ are any k random variables (not necessarily independent). Then variance of $X_1+X_2+...+X_n$ is calculated by: $Var(X_1+X_2+...+X_n)=\sum_{i=1}^kVar(X_i)+2*\sum_{1\le i\le k}Cov(X_i,X_j)$ When $X_1, X_2,...X_k$ are independent, then $Cov(X_j, X_j)=0$ $\therefore$ $Var(X_1+X_2+...+X_n)=\sum_{i=1}^kVar(X_i)$ When you have $X, ... 1 Hint: $$P(\ln(X + c) \le x) = P(X \le e^x - c) = \begin{cases} 0 & e^x \le c \\ P(\ln(X) \le \ln(e^x - c)) & \text{else} \end{cases}$$ 2 Outline: We do indeed want$\Pr(X\gt 5Y/4)$or equivalently$\Pr(W\gt 0)$where $$W=X-\frac{5}{4}Y.$$ If we assume that$X$and$Y$are independent, then$W$is normally distributed, with mean$E(X)-\frac{5}{4}E(Y)$and with variance$\text{Var}(X)+\frac{25}{16}\text{Var}(Y)$. Compute the mean and variance of$W$. The rest should be standard. 1 This will get you half way there. Suppose we have obtained an observation from Y. Call it$y$. I want to express these probability densities as functions. So let g(x) be the probability density of X and h(y) be the probability density of Y. You can formally write the probability that an observation from X is at least 25% greater than it as$P(x > ...

1

Sufficient statistics should be $M = \text{card}\{i: Y_i > 0\}$, $S = \sum_i Y_i$ and $T = \sum_i Y_i^2$, with likelihood function \eqalign{L(Y) &= {n \choose M} \Phi\left(\dfrac{50-\mu}\sigma\right)^{n-M} \prod_{i: Y_i > 0} \dfrac{\exp(-(Y_i-\mu)/(2\sigma^2)}{\sqrt{2\pi \sigma^2}}\cr &= {n \choose M} ... 2 \begin{align} \Pr(Y=0) = \Phi\left( \frac{50-\mu} \sigma \right) = {} & \int_{-\infty}^{50} \varphi\left(\frac{z-\mu}\sigma\right) \, \frac{dz} \sigma = \int_{-\infty}^{(50-\mu)/\sigma} \varphi(z)\,dz \\[12pt] & \text{ where } \varphi(z) = \frac 1 {\sqrt{2\pi}} e^{-z^2/2}. \end{align} For y>50, \Pr(50 \le Y \le y) = \int_{50}^y \varphi\left( ...

0

Deal all, I just noticed someone answered my question but soon deleted. She/he mentioned that performing integral calculation will tell the differernce. And actually it works and I realized and strongly believed that the equation on the book was wrong. I input the equations to an online integral calculator (well, I know it was an bad idea). And found the ...

1

Yes, you can go online for $t$-test $p$-value calculators as well as $z$-table calculators to verify that for around $40$ degrees of freedom or more, the two tests behave almost exactly the same. If you really want to use $t$-test though, you can use $t$-test $p$-value calculators online that can handle larger numbers of degrees of freedom. Your question ...

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No. It is a concave function which is equivalently to an increasing hazard rate. With $$h(x) = \frac{\varphi(x)}{1-\Phi(x)}$$ we have $$h'(x) = \varphi(x) \frac{\varphi(x)-x[1-\Phi(x)]}{[1-\Phi(x)]^2} \geq 0 \;.$$ The last inequality holds since it is sufficient to show $$g(x) = \varphi(x)-x[1-\Phi(x)] \geq 0 \;.$$ At $x=0$, $g(x)=\varphi(0)>0$ and ...

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With a bit of thought, checking your text or notes, clues below and in Comments, plus maybe some exploring online, you should be able to handle all parts--and learn some important relationships about variances, covariances, and the normal distribution in the process. Clues: (a) Independence as noted by @Saty (b) Linearity of Cov in both arguments as in ...

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a) Independents implies zero covariance. As $X\perp Y$ then $\mathsf{Cov}(X,Y)=0$. b) Covariance is linear: $\mathsf{Cov}(aX+bY+c,dX+eY+f) = a\,\mathsf{Cov}(X,dX+eY)+b\,\mathsf{Cov}(Y,dX+eY) \\ = ...$ c) Hint: The sum of independent normally distributed random variables, is a normally distributed random variable whose (a) mean is the sum of their means, ...

2

You have a problem with your geometric PMF: the sum of from $x = 0$ to $\infty$ is not equal to $1$. As such, you must write either $$\Pr[X = x] = (1/4)^x (3/4), \quad x = 0, 1, 2, \ldots,$$ or $$\Pr[X = x] = (1/4)^{x-1} (3/4), \quad x = 1, 2, 3, \ldots.$$ Which one you mean, I cannot tell, and because the supports are different, the resulting ...

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The parameters $p$ if that is construed in the usual way, would remain the same, and $r$ would be $36$. But that's not the best way to proceed. Find the expected value and variance of your geometric distribution. Multiply them by $36$. Those would be the expected value and variance of the distribution of the sum of independent random variables. Use the ...

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You can write a "closed form" expression for the bivariate normal integral using the Owen's T function (see references therein). Even in the univariate case, the normal integral has no closed form solution but its value can be written in terms of other standard functions.

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The distribution of individual observations: $$X_i \sim \operatorname{Normal}(\mu, \sigma^2 = 4).$$ The sample size: $n = 10$. The hypothesized mean: $\mu_0 = 0$. The hypothesis: $$H_0 : \mu = \mu_0 \quad \text{vs.} \quad H_1 : \mu \ne \mu_0.$$ The sampling distribution of the sample mean: $$\bar X = \frac{1}{n} \sum_{i=1}^n X_i \sim ... 1 Hint: P(Z<-1.04)=0.14\color{red}{92}. The last two digits has to be the other way round. Therefore the calculation is$$105(.1492)^2 \cdot(.8508)^{13}+15\cdot(.1492)(.8508)^{14}+(.8508)^{15}=0.61$$1 \dbinom{15}2 = \dfrac{15\times14}{2\times1} = 105, so you need 105 where you have 150. Other than that your answer looks OK. (I doubt anyone would really tolerate a standard deviation that big, but I don't think this was intended to be realistic.) 1 (1) Your answer is correct. It appears you might be using software instead of normal tables to get so many decimal places of accuracy. [To use normal tables, you would have to 'standardize' (convert to standard normal distributions), then get something like four digits of accuracy.] In R software, this computation is as follows, without standardizing. ... 1 For t\ge 0$$P\left\{\frac{a}{\sqrt{b+cX}}\le t\right\}=P\left\{\frac{a^2}{b+cX}\le t^2\right\}=P\left\{X\ge\frac{(a/t)^2-b}{c}\right\}=\int_{\frac{(a/t)^2-b}{c}\vee0}^\infty \frac{1}{x\sigma\sqrt{2\pi}}\exp\left\{-\frac{(\ln(x)-\mu)^2}{\sigma^2}\right\}dx$$Then for t\in\left[0,\frac{a}{\sqrt{b}}\right] ... 3 Obviously not: the support of a log-normal random variable is necessarily on (0, \infty); that is to say, if Y = \log X \sim \operatorname{Normal}(\mu,\sigma^2), then 0 < X < \infty. But the transformation$$X' = \frac{a}{\sqrt{b+cX}}$$is such that the support of X' is bounded above by$$\frac{a}{\sqrt{b}},$$which occurs as X \to 0^+, ... 0 We have at least an implicit solution to problem (1). In the n=2 case (which is the one I am mostly interested), possible solutions are parametrized by a single parameter. Therefore, the final solution can be obtained easily through numerical minimization. This problem illustrates fundamental differences in the geometry of the space of matrices endowed ... 2 Start by transforming X_i-50 to N(0,1). Then what distribution is the sum of the squares of standard normal distributions? 1 Following up on @Bombyx comment the derivation would go as follows (posted here just for completeness and future reference, and amplifying the steps mentioned in Casella and Berger Statistical Inference so that everyone can follow): With minimal rearrangement the t statistic formula becomes \large t= ... 0 See https://en.wikipedia.org/wiki/Multivariate_normal_distribution where it is stated that a multi-variate distribution is multi-variate normal if and only if every linear combination of the variables is normally distributed. If I understand correctly, your "projection" defines a linear combination that you are interested in of the variables, so that is ... 1 You have$$Y=Diag(a_1,\dots,a_n)(X_1,\dots,X_n)^T$$So Y is normally distributed as a linear transform of a normally distributed vector. Now you have the following properties for random vector X and a (non-random) matrix A:$$E(AX)=AE(X)$$This implies that E(Y)=0.$$V(AX)=AV(X)A^T$$From this you can find the covariance matrix of Y. You can also ... 1 Writing out the Gaussian densities explicitly leads to$$p(x+\delta t) \mid x(t), t)\\= \int \frac{1}{2\pi}\sqrt{\frac{t+\sigma^{-2}}{(\delta t)}}\exp\left(-\frac{1}{2(\delta t)}\left[x(t+\delta t)-x(t))-\mu (\delta t)\right]^2\right)\\\times\exp\left(-\frac{(t+\sigma^{-2})}{2}\left(\mu-\frac{x(t)}{t+\sigma^{-2}}\right)^2\right) d\mu$$Thus$$p(x+\delta t) ...

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