# Tag Info

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.

0

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.

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 $X-Y$ ...

0

if $A$ and $B$ are independent, then $A^2$ and $A+B$ are not independent because $A^2$ and $A$ are correlated.

0

Every hypothesis test gives rise to a decision rule which instructs you when to reject the null hypothesis. In your example the decision rule is "Reject $H_0$ when $\bar X\ge402.3398$". The power of a hypothesis test is the probability that the decision rule leads to the right conclusion (i.e., the prob that you observe $\bar X\ge402.3398$) when the null is ...

0

Its saying that there is at least a 93% chance you will detect an error of at least +5 from the target of 400, assuming your process deviation is 10. Thus, the processor will only ship flawed materials at most 7% of the time. All they are doing is determining the area to the right of the cutoff, given that the mean is 405 and a deviation of 10 for each $x$. ...

0

You likely mean the product of $n$ independent normal random variables. In this case, the answer is no.

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 $\mathbb{P}(s^2=95)<0.05$.

0

A notice: There are better ways to approximate the binomial distrution. See On Normal Approximations to Discrete Distributions.

0

If you know that X is a bounded random variable, maybe you should use a different distribution to model it. I can't say which distribution without further information, but if it needs to be somewhat gaussian, you could use a truncated normal distribution (just an exemple). You could also use a normal distribution with low variance so that the probability to ...

0

Normal distribution two parameters Mean and Standard Deviation do effect Normal distribution. Mean Mean will shift the Normal distribution forward or backward on it's axis. Mean will not distort shape of Normal Distribution Standard Deviation Standard Deviation does effect shape A large Standard Deviation mean that normal distribution will approach to less ...

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 $$\mathbb{P}[\beta\leq X\leq \alpha]=\int_{\beta}^\alpha f(x)\,dx=\int_\beta^\alpha ... 0 There is no closed form solution for this. You can think of it as a Normal random variable with mean \mu distributed as a lognormal distribution. 1 Let X_i be the random variable which equals 1 when box i is filled, and 0 otherwise. Then the variable you want to know about is X = \sum_{i = 1}^{n}X_i. We have:$$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}} = ...

0

...or, consider your integral being the squared integral of (exp(-r^2)dr, from 0 to Infinity) in polar coordinates which can be solved easely. The result is sqrt{1/2*sqrt(pi)*erf (Infinity)}. erf(infinity) is a fancy way of saying "1".

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After investigation I've realized that I've made things unnecessarily complicated. If $X$ is the sum of enough n s-sided dice rolls to that the central limit theorem can be used then $X\in N(\mu_d,\sigma_d²)$ which means that $X*k+m\in N(\mu_d*k+m,\sigma_d²*k²)$. Index $d$ indicates that the value is of the actual distribution of the dice and not the one it ...

0

Let $Y = \lVert X \rVert^2$. Then $Y/\sigma^2 \sim \chi^2_p$. And you wish to find $EY^{-1/2}$. This is $$\frac{1}{\sigma}\cdot\dfrac{1}{2^{p/2}\Gamma(p/2)}\int_0^\infty y^{-1/2} y^{p/2 - 1} e^{-y/2}dy$$ $$= \frac{1}{\sigma} \dfrac{2^{(p-1)/2}\Gamma((p-1)/2)}{2^{p/2}\Gamma(p/2)} = \frac{1}{\sigma\sqrt{2}}\dfrac{\Gamma((p-1)/2)}{\Gamma(p/2)}$$

0

In cases with a small number of trials, like flipping a coin $10$ times, there will be some difference between the binomial probability and the normal probability. For these cases you should use the binomial one since, for instance, $P(\text{getting at most }3 \text{ heads})$ can be calculated by hand. In cases with a bigger number of trials, like flipping ...

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 ...

0

Central Limit Theorem applied to a mean of discrete random variables. (a) Saying that $X$ is normal is simply wrong; must be an error. We are told that the PDF (PMF) of $X$ is $f(x) = x/55,$ for $i = 1,\dots,10.$ As a check, we verify that $1/55 + \cdots + 10/55 = 1,$ as must be the case for a point mass function. Then, as you have already verified using ...

0

eltigrechino's answer is correct after the correction by Ian. That is, if your distribution is normally distributed ~N($\mu, \sigma$) then it is a property of normal distributions that 1 - N(x) = N(-x). In particular, using your expression, we have: x = $\frac{lnK - lnS_0 - (\mu - \frac{1}{2}\sigma^2)T}{\sigma\sqrt{T}}$ Note that $lnK - lnS_0 = ... 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: $$\begin{eqnarray*} f_{X+Y}(x) &=& \frac{\lambda}{\sqrt{2\pi\sigma^2}}\int_{0}^{+\infty}\exp\left(-\lambda ... 2 This is an extremely important result known as the central limit theorem. The proof isn't terribly difficult either (at least considering how fundamental the theorem is). 2 I think the three methods you listed pretty much cover it. However I'd like to stress that Stein's method is, to use your terms, a real workhorse in probability theory, especially in stochastic geometry. There are two advantages to Stein's method: It gives rates of convergence. For example the Berry-Essen theorem that states that if (X_{i}) are iid ... 0 For a given multivariate Gaussian distribution, the iso-density locus is an ellipsoid. Let \mathbf{x}\in\Bbb{R}^n be an n-dimensional random vector that is distributed normally around {\mu}\in\Bbb{R}^n with covariance matrix \Sigma\in\Bbb{S}_{++}^n, where \Bbb{S}_{++}^n denotes the set of all symmetric positive definite matrices with entries in ... 0 It means that X \sim N(0,t), i.e. X is a random variable with normal distribution with mean 0 and variance t, if and only if \sqrt{c} X \sim N(0,ct). Multiplication by \sqrt{c} always multiplies the variance of a random variable by c and the mean by \sqrt{c}. The normal distributions are rather special in that multiplication by a constant ... 3 Edit: This answer shows the identity$$\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*} ... 0 I may have made some sloppy mistakes, but I think this is the gist of it:\{(x,y) | x^2+y^2\leq d\} = \{(r,\theta) | r\leq d\}$, So$\int \int \frac{1}{2\pi} \exp\{-\frac{1}{2}(x^2+y^2)\} dy dx=\intop\limits_0^{2\pi}\intop\limits_0^d \frac{1}{2\pi}\exp\{-\frac{1}{2}r^2\}r dr d\theta=1-e^{-\frac{1}{2}d^2}$2 When$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 $$u = \frac{v}{2} \pm \frac12 \sqrt{v^2-4 q}$$ $$du = \frac12\left ( 1 \pm ... 1 If X_i are iid normal with mean \mu and variance \sigma^2, then \sigma^{-2} \sum_{i=1}^n X_i^2 has noncentral chi-squared distribution with n degrees of freedom and noncentrality parameter \lambda = n (\mu/\sigma)^2 (as defined e.g. in Wikipedia; conventions may sometimes differ). So what you have is a multiple of a noncentral chi-squared random ... 0 The function g(x) = e^x/(1+e^x) is monotone, so if Y = g(X) where X \operatorname{Normal}(\mu,\sigma^2), then$$f_Y(y) = f_X(g^{-1}(y)) \left| \frac{dg^{-1}}{dy} \right|.$$Note the support of Y is (0,1);$$g^{-1}(y) = \log \frac{y}{1-y},$$and$$\frac{dg^{-1}}{dy} = \frac{1}{y(1-y)}.$$Therefore,$$f_Y(y) = \frac{1}{\sqrt{2\pi} \sigma y(1-y)} ... 0 Here is a standard very-well-known example. Let$X \sim N(0,1)$,$Z$a discrete random variable taking on values$+1$and$-1$with equal probability$\frac 12$, and define$Y = XZ. Then, \begin{align} P\{Y \leq y\} &= P\{XZ \leq y\}\\ &= P\{X \leq y, Z = +1\} + P\{X \geq -y, Z = -1\}\\ &= P\{X \leq y\}P\{Z = +1\} + P\{X \geq -y\}P\{Z = -1\}, ... 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 = ... 0 For the density: Note that X_i \sim N(\mu_i, \Sigma_{ii}). Hence, for t \in \mathbf R^+, t \le 1, we have \begin{align*} \def\P{\mathbf P}\P(Y_i \le t) &= \P(e^{X_i} \le t)\\ &= \P(X_i \le \log t)\\ &= \frac 1{\sqrt{2\pi\Sigma_{ii}}} \int_{-\infty}^{\log t} \exp\bigl(-(x-\mu_i)^2/2\Sigma_{ii}\bigr)\, dx \end{align*} For t ... 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 The Normal distribution is a probability density function. You are correct in your intuition. This Wikipedia article explains it well - in particular the section entitled "Absolutely continuous univariate distributions" If the i-th bin is x_i \leq x\leq x_{i+1} then the probability, hence the proportion of points, in that bin isPr[x_i\leq X\leq ... 5 I agree with you and Nate Eldredge that the limit doesn't exist. However, if we replace0$with$+\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 ...

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.

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: $$\mathsf P(Z\leq ... 1 Q-function is defined as Q(x) = \frac{1}{\sqrt{2\pi}} \int_x^\infty \exp\left(-\frac{u^2}{2}\right) \, du P(−\frac{20}{σ}≤z≤\frac{20}{σ})=0.99 It implies \frac{1}{\sqrt{2\pi}} \int_{−\frac{20}{σ}}^{\frac{20}{σ}} \exp\left(-\frac{u^2}{2}\right) \, du = 0.99 As normal distribution is symmetric, \frac{1}{\sqrt{2\pi}} \int_{0}^{\frac{20}{σ}} ... 1 (a) If you are using at a normal table, first determine whether it gives areas P(Z \le z) [or P(0 < Z \le z)]. In the first instance look in the body of the table to find the closest probability to .9500; in this case, probably a tie between .9496 and .0505. Then find the corresponding value of z by looking at the margins; in this case, 1.64 or ... 1 At the same time, my notes state this formula (with no explanation..) that z = \frac{x - \mu}{\sigma}, where z is a z-score. But what is this and what does it have to even do with these problems? Suppose you have a random variable X, which is  \mathcal N(\mu,\sigma^2 ) distributed-with \mu\neq 0 and/or \sigma^2\neq 1. Now you want to ... 1 P(Z\le z)=\int_{-\infty}^z \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}\,dz=0.95 Or if probability density function has not been covered, just consider the picture of a normal distribution. We know that Pr(Z\le z)=0.5 when z=0 since 0 is in the middle. Now, move z to the right, the probability will become larger. When should we stop? 2 I go through the solution step by step: The quantity to be calculated is$$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)}.$$(1) ... 1 If all you want is the expected value of D, then write the expectation as an integral and switch to polar coordinates:$$ \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\\ ...

0

As there is a law of total expectation, so there exists a law of total variance. We will call the random variable representing the total number of widgets ordered in a month $W$. Note that $$W = Y_1 + Y_2 + \ldots, Y_X,$$ because if we observe $X$ orders in a month, that means we observed orders $Y_1, Y_2, \ldots, Y_X$ of widgets in each order. So, the ...

2

One rejects the null hypothesis when the P-value is small. A common criterion is to reject if the P-values is less than 0.05. In a Kolmogorov-Smirnov test, the D-statistic measures the maximum diagonal distance between the empirical cumulative distribution functions (ECDFs) of the two samples. (Everything is re-scaled so the ECDF fits inside the unit ...

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