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1

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 $$ ...


0

Question: Let $X_i \sim N(0,\sigma_i^2)$. I would like to compute $E[|X_1-X_2|^k|X_2|^k]$ Bounds shmounds¬°? For given values of integer $k$, it is possible to obtain exact closed-form solutions. In particular, if $X_1$ and $X_2$ are independent (which appears to be your intention), the joint pdf of $(X_1,X_2)$, say $f(x_1,x_2)$, is: Then, ...


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The hard way to answer this question is to compute all terms of the distribution of $N$ up to and including $P(N=25)$, add those probabilities and subtract the sum from $1$. An easier way is to consider what would happen if you simply committed to rolling that same die $25$ times in succession, then compare the outcomes to the outcomes of the original ...


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Yes there is something similar to the summation, for the product. If all your random variables are independent, then you just apply the log function $$E[\log\prod_i X_i]=E[\sum_i \log(X_i)]=\sum_i E[\log X_i]$$


0

This is not a complete answer to your question, but it might help solving it. The minimum number of throws required in order to get over $100000$ is $11$: Either you get $3^{11}=177147$ Or you get $3^{10}\cdot2=118098$ So you basically need to calculate $1-\sum\limits_{k=11}^{25}P(N=k)$. You can calculate ...


1

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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I've repeated Higuchi calculations and got precisely the same answer as he promised D=-1.5143. Pay close attention to the number of times he divides the series length Lmk by k. That was my mistake for the first time, when I've lost the final averaging between the set of k series and we moved from Lkm to <Lkm>. Here the pointy brackets stands for ...


1

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 ...


0

If $(x_i)_{1\leqslant i\leqslant n}$ is a sequence of real numbers and $X$ is a random variable with $P(X=x_i)=\pi_i$, then Jensen's inequality with the convex function $x\mapsto x^2$ shows that $$ \sum_{i=1}^n \pi_ix_i^2=\mathrm{E}[X^2]\geqslant \mathrm{E}[X]^2=\Big(\sum_{i=1}^n \pi_ix_i\Big)^2. $$ More generally, Jensen's inequality shows that $$ ...


0

Hint, use Cauchy-Schwarz inequality accordingly: $$\sum \pi_{i}(f(x,\phi_{i})^2\sum \pi_{i}\ge \left(\sum \pi_{i}f(x;\phi_{i})\right)^2$$


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Hint $g_i:=point(i)$ $$int(((x-c(i))^2*norm(diff(f(x, -0.04, sqrt(0.11))),p),x,g_i,g_{i+1})$$


0

The joint density given $\sigma_1^2, \sigma_2^2$, of observing the sample $\boldsymbol x, \boldsymbol y$ is simply $$f(\boldsymbol x, \boldsymbol y \mid \sigma_1^2, \sigma_2^2) = \prod_{i=1}^n \frac{e^{-x_i^2/(2\sigma_1^2)}}{\sqrt{2\pi}\sigma_1} \prod_{i=1}^m \frac{e^{-y_i^2/(2\sigma_2^2)}}{\sqrt{2\pi}\sigma_2}.$$ Thus a likelihood function for $\sigma_1, ...


1

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 ...


0

i think the answer is $$\frac{(-1)^{n+1} 2^{\frac{1}{2}-\frac{3 n}{2}} \left((-1)^n+1\right) \sigma ^{3/2} \left(-\frac{1}{\sigma ^2}\right)^{n/2} \Gamma (n+1) \, _2F_1\left(\frac{n}{2}+\frac{1}{2},\frac{n}{2}+1;\frac{n}{2};-\frac{1}{2 \sigma ^2}\right)}{\sqrt{\pi } n! \Gamma \left(\frac{n}{2}\right)}$$


0

The z-score is the standardisation that you should plot. Full-stop. (And you have the correct formula for the z-score.) The z-score might usually range from -3 to +3 and you can then plot both z-score distributions on the same graph. The z-score distributions plot with their centres at z=0. You mention you want to plot on a 0-10 scale. What do you mean ...


0

Robustness is sort of a subjective matter. In a nutshell, if you produce an estimate with a robust estimator, and then you add a very extreme data point and re-estimate, you shouldn't produce an estimate that is too different from your first estimate. What does "extreme" mean? What does "too different" mean? This is precisely where the ambiguity comes in. ...


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I recommend first reading the Wikipedia article for robust statistics. After doing so, what do you intuitively conclude about the estimators you described above? Would you say they are robust or not? Could you furnish an example of an estimator that would be more robust than the MLEs?


1

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: ...


1

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. ...


0

$$\int_{0.30}^{\infty}P(x)\,dx=\int_{-\infty}^{\infty}P(x)\,dx-\int_{-\infty}^{0.30}P(x)\,dx$$ The first integral is equal to $1$ since $P(x)$ is a probability density function. The second one is not possible to evaluate with elementary functions. However using the function $$\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^x\exp(-t^2)\,dt,$$ the ...


-1

You sort of got it right, but unintentionally, a distribution is usually refers to cumulative distribution function (cdf) or the probability density function (pdf) or the probability mass function (pmf). You use pmfs for discrete cases and pdfs for continuous cases. Then say X is a r.v. If it's discrete then the cdf of X at x is the sum of the pmf from ...


1

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 ...


0

Abramowitz and Stegun give a number of approximations. One with only five constants is accurate to $1.5 \cdot 10^{-7}$ Numerical Recipes page 221 has an expansion that quotes slightly better errors. Neither explains where they come from, but you could look at the references.


2

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 ...


3

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 ...


1

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.


0

Assume without loss of generality that $(\mu_n)$ is decreasing. The simplest approach could be to divide any null linear combination $\sum\limits_na_nf_n=0$ by the gaussian density with parameters $(0,\sigma^2)$, yielding $$\sum_na_n\exp(x\mu_n-\mu_n^2/2)=0.$$ When $x\to\infty$, every exponential term except the first one is negligible with respect to the ...



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