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5

I agree with you and Nate Eldredge that the limit doesn't exist. However, if we replace $0$ with $+\infty$, then the limit exists and equals $e^{-a}$. Let $F$ denote the cdf of a standard normal, and let $\;f=F'$ be the pdf. Then $$\lim_{x \to \infty} \frac{P(X>x+\frac{a}{x})}{P(X>x)} = \lim_{x \to \infty} \frac{1-F(x+\frac{a}{x})}{1-F(x)}$$Now as $x ...


4

Let $U$ be standard normal. Let $V=RU$, where $R$ is a Rademacher random variable that takes values $-1$ and $1$, each with probability $1/2$. Suppose $U$ and $R$ are independent. Let $X=\frac{U+V}{2}$ and $Y=\frac{U-V}{2}$. Then $X+Y$ and $X-Y$ are each normal. But $X$ is not normal, for it takes on value $0$ with probability $\frac{1}{2}$. Remark: ...


3

Edit: This answer shows the identity $$\mathbb{E} \left( \left| \sum_{i=0}^{n-1} \int_{t_i^n}^{t_{i+1}^n} (W(t_i^n)-W(t)) \, dt \right|^2 \right) = \mathbb{E} \left( \sum_{i=0}^{n-1} \left[\int_{t_i^n}^{t_{i+1}^n} (W(t_i^n)-W(t)) \, dt\right]^2 \right). \tag{1}$$ Fix $i<j$. Then, by the tower property, $$\begin{align*} ...


3

$$ \int_0^{\infty} \left(-\frac{1}{2}\dfrac{d}{dx}\mathrm{e}^{-x^2}\right)\phi(ax+b)dx $$ use integration by parts $$ \left[-\frac{1}{2}\mathrm{e}^{-x^2}\phi(ax+b)\right]_0^{\infty} +\int_0^{\infty} \frac{1}{2}\mathrm{e}^{-x^2}\phi'(ax+b) dx $$ where $$ \phi'(ax+b) = \dfrac{d}{dx}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{ax+b}\mathrm{e}^{-s^2}ds = ...


2

This is an extremely important result known as the central limit theorem. The proof isn't terribly difficult either (at least considering how fundamental the theorem is).


2

When $b=0$ the integral is equal to $$\frac1{4 \pi \sigma \sqrt{a}} \int_0^{\infty} du \, e^{-\left (\frac{u}{\sigma^2} + \frac{\lambda}{a u} \right )} $$ Consider then $$ \int_0^{\infty} du \, e^{-p \left (u + \frac{q}{u} \right )} $$ Let $v = u+q/u$; then $u^2-v u+q=0$, or $$u = \frac{v}{2} \pm \frac12 \sqrt{v^2-4 q}$$ $$du = \frac12\left ( 1 \pm ...


2

I go through the solution step by step: The quantity to be calculated is $$E[|X_1+X_2||X_1,X_2\text{ are reals}]=2E[|B|||B|>1]$$ where $B$ is a standard Gaussian r.v. In order to calculate the $E[|B|||B|>1]$ we need the corresponding conditional distribution: $$P(|B|<x||B|>1)=\frac{P(|B|<x,|B|>1)}{P(|B|>1)}.$$ (1) ...


2

I agree with your analysis: the limit does not exist, since the right- and left-sided limits differ. I suspect that either the original problem was erroneous, or perhaps you made a mistake when copying it.


2

Let $W$ (for example) be a random variable which takes on values $0$ and $1$, each with probability $\frac{1}{2}$. let $Z$ be standard normal, and suppose $W$ and $Z$ are independent. Let $X=W+Z$ and let $Y=W$. Then $X-Y$ is normally distributed, but neither $X$ nor $Y$ is normal. Remark: Note that if $X$ and $Y$ are normal and not independent, then $X-Y$ ...


2

Consider the more general probability $$\Pr\left[ \mu - z \frac{\sigma}{\sqrt{n}} < \bar X < \mu + z \frac{\sigma}{\sqrt{n}} \right],$$ for some $z > 0$, where $$\bar X = \frac{1}{n} \sum_{i=1}^n X_i$$ is the sample mean of $n$ independent and identically distributed observations from a normal distribution with mean $\mu$ and standard deviation ...


2

It's not a lucky guess. It's at the heart of Stein's method: you need a characterizing equation for your distribution. There isn't a unique equation, but in fact many and depending on the situation, one might use other characterizing equations. In the case of the normal distribution, this is actually simple integration by parts. Below, all expectations are ...


2

I think the three methods you listed pretty much cover it. However I'd like to stress that Stein's method is, to use your terms, a real workhorse in probability theory, especially in stochastic geometry. There are two advantages to Stein's method: It gives rates of convergence. For example the Berry-Essen theorem that states that if $(X_{i})$ are iid ...


1

The normal distribution is symmetric about its mean. Therefore you can flip the CDF calculation as follows $1 - \Phi(x) = \Phi(-x) $ where $\Phi$ represents the CDF of a $N(\mu,\sigma)$ distribution. Edit: whoops forgot to flip the sign, thanks


1

The convolution between the pdfs of $X$ and $Y$, given by: $$ f_X(x) = \lambda e^{-\lambda x}\cdot\mathbb{1}_{x\geq 0},\qquad f_Y(x)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}} $$ leads to: $$\begin{eqnarray*} f_{X+Y}(x) &=& \frac{\lambda}{\sqrt{2\pi\sigma^2}}\int_{0}^{+\infty}\exp\left(-\lambda ...


1

If X and Y are independent, then the result follows from Cramer's Theorem. Here is a link. Note the requirement of independence here. In the non-independence case, if $Y=Z-X$, where $Z$ is normal, and $X$ has any distribution also works.


1

If all you want is the expected value of $D$, then write the expectation as an integral and switch to polar coordinates: $$ \begin{align} ED=E\sqrt{X^2+Y^2}&=\iint\sqrt{x^2+y^2}f(x)f(y)\,dy\,dx\\ &=\int_{x=0}^\infty\int_{y=0}^\infty\sqrt{x^2+y^2} \frac1{2\pi (a\sigma)^2}\exp[-(x^2+y^2)/2(a\sigma)^2]\,dy\,dx\\ ...


1

Note if $X\sim N[\mu,\sigma]$, then $\mathbb{P}[\beta\leq X\leq \alpha]=\mathbb{P}[z_\beta\leq z\leq z_\alpha]$ where $z\sim N[0,1]$ and $$z_{\alpha}=\frac{\alpha-\mu}{\sigma}$$ and similarly for $z_\beta$. Well by definition, we have $$\mathbb{P}[\beta\leq X\leq \alpha]=\int_{\beta}^\alpha f(x)\,dx=\int_\beta^\alpha ...


1

I've attached a picture of various beta densities from the wiki on the Beta Distribution. It really matters what your density looks like beyond the first two moments you specified, but in general, this will ensure the values are restricted to a bounded interval. If you aren't concerned with some density living outside your interval you might consider ...


1

To be a lot simpler than the other answers, just use the fact that the sum of two Gaussian random variables is gaussian and the fact that the scalar multiple of a Gaussian random variable is also Gaussian. The first of these facts can be worked out by computing the convolution of two arbitrary Gaussian distributions. The second requires some thinking about ...


1

Yes. Write this in matrix form, so you have $\mathbf {AX} = \mathbf Z$ where $\mathbf Z$ is bivariate normal. Then $\mathbf X = \mathbf {A^{-1}Z}$. The RHS is a linear combination of Gaussians, therefore the LHS is Gaussian.


1

One can frequently read that the product of two multivariate Gaussian pdfs, $f_1(x)$*$f_2(x)$, is itself a Gaussian, This is an incorrect statement no matter how many times or in how many different places you have read it; indeed it does not hold even for univariate Gaussian random variables. If the link you have provided does indeed say exactly what ...


1

I understand everything except for steps 3, 4, and 5. Mainly why do we decide to subtract 1−0.99 and then why do we randomly decide to divide by 2? Finally, why would we want to subtract it by 1 again? Standard Normal Tables (or Z-Tables) look up the the cumulative probability ($\alpha$) and report the value $z_\alpha$, such that: $$\mathsf P(Z\leq ...


1

Q-function is defined as $Q(x) = \frac{1}{\sqrt{2\pi}} \int_x^\infty \exp\left(-\frac{u^2}{2}\right) \, du$ $P(−\frac{20}{σ}≤z≤\frac{20}{σ})=0.99$ It implies $\frac{1}{\sqrt{2\pi}} \int_{−\frac{20}{σ}}^{\frac{20}{σ}} \exp\left(-\frac{u^2}{2}\right) \, du = 0.99$ As normal distribution is symmetric, $\frac{1}{\sqrt{2\pi}} \int_{0}^{\frac{20}{σ}} ...


1

Let $X_i$ be the random variable which equals $1$ when box $i$ is filled, and $0$ otherwise. Then the variable you want to know about is $X = \sum_{i = 1}^{n}X_i$. We have: $$E(X) = E(\sum_{i = 1}^{n}X_i) = \sum_{i = 1}^{n}E(X_i)$$ And, since the $X_i$ are independent: $$\sigma^{2} = Var(X) = Var(\sum_{i = 1}^{n}X_i) = \sum_{i = 1}^{n}Var(X_i)$$ Since ...


1

The distribution is not a normal distribution, it is a binomial distribution. But for large $N$ the normal distribution is an excellent approximation to the binomial distribution. At any rate, the exact answer for the binomial distribution is that the mean is $\frac{n}{r}$ and the standard deviation is $$ \sqrt{npq} = \sqrt{n \frac{1}{r} \frac{r-1}{r}} = ...


1

Your intuition is exactly correct. If you make the bins small enough then there is effectively a zero probability of seeing a sample in that exact bin. Hence for bins approaching zero width you get either a count of one or zero for your histogram, which is why it looks nothing like your continuous density function. Mathematically we have the following ...


1

Take some $Z$ following a normal distribution, and $X$ any other. Then set $Y=X-Z$. $X-Y$ follows a normal distribution, while $X$ doesn't.


1

The random variable $(n-1)\frac{s^2}{\alpha^2}$ is $\chi^2(n-1)$ distributed and $\mathbb{P}(\chi^2(n-1)>55.758)=0.05$ $(n-1)\frac{s^2}{\alpha^2}=59.375>55.758$. So $\mathbb{P}(s^2=95)<0.05$.


1

(a) If you are using at a normal table, first determine whether it gives areas $P(Z \le z)$ [or $P(0 < Z \le z)].$ In the first instance look in the body of the table to find the closest probability to .9500; in this case, probably a tie between .9496 and .0505. Then find the corresponding value of $z$ by looking at the margins; in this case, 1.64 or ...


1

At the same time, my notes state this formula (with no explanation..) that $z = \frac{x - \mu}{\sigma}$, where $z$ is a $z-score$. But what is this and what does it have to even do with these problems? Suppose you have a random variable X, which is $ \mathcal N(\mu,\sigma^2 )$ distributed-with $\mu\neq 0$ and/or $\sigma^2\neq 1$. Now you want to ...



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