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3

There are quantitative versions of the CLT which provide an estimate of the rate of convergence. One is given at http://en.wikipedia.org/wiki/Berry%E2%80%93Esseen_theorem. This says that the sequence of CDFs of the rescaled sample means converges uniformly to the appropriate normal CDF, provided the third moment of the variables exists. Assuming the ...


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The joint distribution of $x$ and $y$ is circularly symmetric around the origin of the $x,y$-plane. The set of points $A$ where $x > y$ consists of all points below the line $x = y$; in polar coordinates, it is all points $(r,\theta)$ such that $r > 0$ and $-\frac34\pi < \theta < \frac14\pi.$ The probability distribution integrated over this ...


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The density of $(x,y)$ depends on $x^2+y^2$ only hence its distribution is rotationally invariant and the argument $\theta$ of the point $(x,y)$ is uniformly distributed on $(-\pi,\pi)$. The events of interest are $[x\gt0]=[-\pi/2\lt\theta\lt\pi/2]$ and $[x\gt y,x\gt0]=[-\pi/2\lt\theta\lt\pi/4]$, with respective probabilities $1/2$ and $3/8$, hence $P(x\gt ...


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Have you taken calculus? It is because $$\lim_{t, x \to -\infty}\int\limits_{t}^{x}f(s)\text{ d}s = 0$$ where $f$ is the equation of the graph of the standard normal distribution.


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Note: The following is not an answer, but merely some thoughts which might or might not be helpful to you. First note that you confused(?) your inequality signs. I think you want $$ \gamma_{n}\left(\left\{ x\in\mathbb{R}^{n}\,\mid\,\left\Vert x\right\Vert ^{2}\geq\frac{n}{1-\varepsilon}\right\} \right){\color{red}\leq}e^{-\varepsilon n/4} $$ and $$ ...


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Let $Z$ be a standard normal variable. Since: $$\frac{1-\Phi(x)}{\phi(x)}=\frac{1}{\mathbb{E}[Z\,|\,Z>x]}=e^{x^2/2}\int_{x}^{+\infty}e^{-z^2/2}\,dz=\int_{0}^{+\infty}e^{-zx}e^{-z^2/2}\,dz$$ we have that: $$\mathbb{E}[Y]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}e^{-(x-\mu)^2/2}\int_{0}^{+\infty}e^{-zx}e^{-z^2/2}\,dz\,dx\tag{1} $$ so: ...


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I don't know where you could have seen that claimed, but it doesn't make sense unless $x$ is a fixed constant. If $x$ is a random variable, it's not clear what $y\sim N(0,x^2)$ would mean. If $x$ is constant, then one can say that if $x^{-1}y\sim N(0,1)$ then $y\sim N(0,x^2)$. It is also true that if the conditional distribution of one random variable ...


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$n \ge 30$ is Rule-of-Thumb. Basically to make distribution less skewed, uni-modal, and to make it look more like Normal Distribution. The number of variables can be less than 30, for example if, population distribution is normal, or etc. It really depends on population or sampling distribution,


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A good intuition would be to consider the shape of the level sets of the likelihood. That is: the level sets of the density of a Gaussian correspond exactly to the level sets of the $\ell_2$ norm. The $\ell_2$ norm is induced by an inner product, so generally all your linear algebra and geometry intuitions usually transfer pretty well into the Gaussian case. ...


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If you must do this analytically then you need a convolution. I would not. If your two normal distributions $X_1\sim \mathcal N(a,b^2)$ and $X_2\sim \mathcal N(c,d^2)$ are independent then an easier approach is to say $fX_1\sim \mathcal N\left(fa,f^2 b^2\right)$ and $(1-f)X_2\sim \mathcal N\left((1-f)c,(1-f)^2d^2\right)$ so $$X=fX_1+(1-f)X_2 \sim \mathcal ...


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You want a function $h$ such that for all $\theta\in\mathbb R$, the following integral is zero: $$\int_{-\infty}^\infty \exp\left(-\frac{n}{2}(t-\theta)^2\right)h(t) \, dt.$$ This is $$ \int_{-\infty}^\infty \exp\left(-\frac n 2 \theta^2\right) \exp(nt\theta)\exp\left(-\frac n 2 t^2\right)h(t) \, dt. $$ The first factor does not depend on $t$ so it can be ...


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Let $X \sim N(0,1)$ and $Y \sim N(0,1)$. Then, as noted, the pdf of $Z = X Y$ is $f(z)$: The mgf of $Z$ is $E[e^{t Z}]$: where I am using the Expect function from the mathStatica package for Mathematica to automate. Let $(Z_1, \dots, Z_n)$ denote a random sample of size $n$ drawn on $Z$, and let $Q = \sum_{i=1}^nZ_i$ denote the sample sum. Then, by ...


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$\newcommand{\var}{\operatorname{var}}\newcommand{\cov}{\operatorname{cov}}$You have not specified that the pair $\begin{bmatrix} X_1 \\ X_2 \end{bmatrix}$ is normally distributed, and that does not follow from the fact that the two components separately are normally distributed, nor have you said what the covariance between $X_1$ and $X_2$ is. You say $X_1 ...


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A number of observations: (i) As functions of independent random variables, $\dfrac{X_n}{\sqrt{n}}\,\,$ is independent of $\,\,\displaystyle \sum\limits_{m=1}^{n-1}{X_m^2}\,$. (ii) $\,\,\dfrac{X_n}{\sqrt{n}}\sim\mathcal{N}\left(0, 1/n\right)\,\,$ and $\,\,\displaystyle \sum\limits_{m=1}^{n-1}{X_m^2}\sim\chi^2(n-1)\,$, since they are linear combinations of ...



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