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1

You are taking $X=\sigma_xZ+\mu_x$ and $Y=\sigma_yZ+\mu_y$. They are not independent. You should take $X=\sigma_xZ_1+\mu_x$ and $Y=\sigma_yZ_2+\mu_y$ where $Z_1$ and $Z_2$ are iid with standard normal distribution.


0

This question was sitting among a host of normal-distribution questions, so I automatically tried to use the normal distribution Variance equation. The Variance equation for a binomial random variable is $Var(X) = n*p*(1-p)$ which is the correct equation to use. Sorry about that, I'm still new to this area of mathematics and didn't even realize there were ...


0

Hint: use the fact that $$ P_V(I - P_V) = 0 $$ and the Cochran theorem.


2

\begin{align} \mathbb E(e^{tX}) & = \int_{-\infty}^\infty e^{tx} \frac 1 \sigma e^{-\left|(x-\mu)/\sigma\right|} \, dx = \int_{-\infty}^\infty e^{t(\mu+\sigma w)} e^{-|w|}\, dw \\[10pt] & = e^{t\mu} \int_{-\infty}^\infty e^{(t\sigma)w} e^{-|w|} \, dw \\[10pt] & = e^{t\mu} \left( \int_{-\infty}^0 e^{(t\sigma+1)w} \,dw + \int_0^\infty ...


1

$P_D$ is a symmetric idempotent matrix, i.e. $P_D^2 = P_D^T=P_D$. So if $X\sim N(\mu,\sigma^2 E_d)$, then $$ P_D X\sim N(P_D\mu,\, P_D(\sigma^2 E_D)P_X^T) = N(P_D\mu,\,\sigma^2 P_D). $$ (Generally if $X\sim N(\mu,\,V)$ then we would have $AX\sim N(A\mu,\,AVA^T)$.) (Since it's the projection onto $\{(x,\ldots,x)\mid x\in\mathbb R\}$, I suppose I should ...


2

ad a) The random variable is binomial distributed: $X\sim Bin(8,0.6)$ Thus the calculation is: $P(X \geq 6)=\sum_{k=6}^8 {8 \choose k} \cdot 0.6^k\cdot 0.4^{8-k}$ ad b) The formula for the approximation is: $P(X \leq x)\approx\Phi\left( \frac{x+0.5-\mu}{\sigma} \right)$ $\Phi(z)$ is the standard normal function. The value for $\Phi(z)$ can be looked up ...


0

What you mean to say is that the mean of the distribution should be $\mu_1$ - $\mu_2$, and the standard deviation of the distribution should be $\sqrt{\sigma^2_1 + \sigma^2_2}$. I do not know where you have $x$ and $y$. Now, given that the mean is $0$, and the standard deviation is $\sqrt{2}$, how many standard deviations of this distribution is $2$ oz?


1

Substitute $\frac{z^{2}}{2}=t$, then $zdz=dt$ and $z^{3}=(2t)^{3/2}$. \begin{eqnarray} \mathbb{E}[Z^{4}]&=&\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}z^{4}e^{-z^{2}/2}\;dz\\ &=&\frac{2}{\sqrt{2\pi}}\int_{0}^{\infty}z^{4}e^{-z^{2}/2}\;dz\\ &=&\frac{2}{\sqrt{2\pi}}\int_{0}^{\infty}(2t)^{3/2}e^{-t}\;dt\\ ...


0

HINT: I would say $\mu_{33}=33\mu=12.87,\,\,\,\sigma^2_{33}=33\sigma^2\Rightarrow \sigma_{33}=\sqrt{33}\sigma=1.49$ Consumption of ice cream = X (random variable). $P(X \le x) = F(x) = 0.95 \Rightarrow x = F^{-1}(0.95)$. F (x) is the distribution function of the normal probability distribution $N(\mu_{33},\sigma^2_{33})$.


2

Let $\mu_n$ be the gaussian distribution with mean $0$ and standard deviation $1/n$. Then $\displaystyle\int_\mathbb R f(x) \mu_n(\mathrm dx) \to f(0)$ for every bounded measurable function $f$ continuous at $0$.


3

For every $t\gt0$, $$P(X\gt t)=\int_0^\infty\frac1{\sqrt{2\pi}}\mathrm e^{-(x+t)^2/2}\mathrm dx,$$ and, for every $x\geqslant0$, $$\mathrm e^{-(x+t)^2/2}\leqslant\mathrm e^{-t^2/2}\mathrm e^{-x^2/2},$$ hence $$P(X\gt t)\leqslant\mathrm e^{-t^2/2}\int_0^\infty\frac1{\sqrt{2\pi}}\mathrm e^{-x^2/2}\mathrm dx=\mathrm e^{-t^2/2}\,P(X\gt0).$$


0

Idea for an approximate solution: divide the ellipsoid in small enough, flat enough quasi-rectangles $P_i$, choose an almost-area-preserving parametrization of each piece: $$f_i:R_i\longrightarrow P_i$$ with $R_i\subset\Bbb R^2$ a rectangle. Choose randomly an index $j$ with probability $\text{area}(R_j)/\sum_i\text{area}(R_i)$, choose a point in $R_j$ and ...


1

If $ 29 \leq X \leq 40$, then $X \geq 27$, this means $P(29 \leq X \leq 40 \cap X \geq 27) = P(29 \leq X \leq 40)$.


0

The basic idea is the following: Assume you have two independent random variables $X_1 \sim N(\mu_1, \sigma_1^2)$ and $X_2 \sim N(\mu_2, \sigma_2^2)$, then for their difference the following holds: $$ X_1 - X_2 \sim N(\mu_1 - \mu_2, \sigma_1^2 + \sigma_2^2 )$$ For a proof, see for example the wikipedia article. Now, for your case, notice that you are ...


2

The distribution of the sample mean $X_n$ is approximately normal with parameters $μ=x_{mean}$ and $σ/\sqrt{n}=x_{sd}/\sqrt{n}$, in symbols $$X_n \sim N\left(μ, \dfrac{σ}{\sqrt{n}}\right)$$ approximately (where the approximation is better as $n$ grows to $\infty$). This follows from the Central Limit Theorem and you do not need any further particular ...


1

Since $x=R\mathrm e^{\mathrm i\Theta}$ where $R$ and $\Theta$ are independent, $E(x^n)=E(R^n)E(\mathrm e^{n\mathrm i\Theta})$. Furthermore, $E(\mathrm e^{n\mathrm i\Theta})=0$ for every integer $n$ because $\Theta$ is uniform on $(0,2\pi)$, hence $E(x^n)=0$.


2

Complete the square of the exponent $x^2+y^2-2\rho xy = \left(x-\rho y\right)^2+y^2(1-\rho^2)=:u^2+y^2(1-\rho^2)$ $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{2 \pi \sqrt{1-\rho^2}} e^{-\frac{x^2+y^2-2 \rho x y}{2(1-\rho^2)}} dx dy=\\ =\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{2 \pi \sqrt{1-\rho^2}} e^{-\frac{u^2}{2(1-\rho^2)}} du ...


0

Try to use polar coordinates. Put $x=r\cos\theta$, $y=r\sin\theta$ and Jakobian will be $r$.


0

Let $X_i=\text{N}(220,25^2)$ be the weight of the $i^\text{th}$ wrestler. Then $X=\sum_iX_i$ be the total weight. Now read http://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables and http://en.wikipedia.org/wiki/Fubini's_theorem and note that Fubini's theorem applies to random variables.


0

Probability of a bulb that it will work after 44 months: normsdist(44,60,10,true) How many of 1400 bulbs 1400*normsdist(44,60,10,true)


5

One way to proceed is to generate a point uniformly on the sphere, apply the mapping $f : (x,y,z) \mapsto (x'=ax,y'=by,z'=cz)$ and then correct the distortion created by the map by discarding the point randomly with some probability $p(x,y,z)$ (after discarding you restart the whole thing). When we apply $f$, a small area $dS$ around some point $P(x,y,z)$ ...


2

If the $N$ points and the parameters $\mu, \sigma$ of the distribution are given, find the distance $r$ of the farthest point from $\mu$, and the probability is $2 \Phi(-r/\sigma)$. If the $N$ points are not given, any ordering of the distances of the $N+1$ points are equally likely, so $1/(N+1)$.


0

Hint: $\Phi (-z)=1-\Phi(z)\quad \Longrightarrow \quad \Phi (1)-\Phi(-2) \ = \ \Phi(1)-(1-\Phi (2)) \ = \\ \Phi(1)+\Phi (2)-1$


1

$P(Y^2 < 9) = P(-3 < Y < 3) = P(-3 -1 < Y-1 < 3 - 1) = P((-3-1)/2 < (Y-1)/2 < (3-1)/2) = P(-2 < Z < 1) = \phi(1) - \phi(-2)$ where $Z$ is standard normal, $\phi$ is the normal CDF.


0

If you are familiar with the Gamma function, then there is a shorter way to calculate the moments: First of all, by symmetry, $$\mathbb{E}(|X|^{2k+1}) = \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} |x|^{2k+1} \exp \left(- \frac{x^2}{2} \right) \, dx = \frac{2}{\sqrt{2\pi}} \int_0^{\infty} x^{2k+1} \exp \left(- \frac{x^2}{2} \right) \, dx.$$ Now substitute $y:= ...


1

Here's one way to look at it: \begin{align} f_Y(x) \, dx & =\frac{1}{\sqrt{(2\pi)^n}} \exp\left(-\frac{1}{2}\Big({\boldsymbol\Sigma}^{-1/2}(x-m)^T\Big)^T \Big( {\boldsymbol\Sigma}^{-1/2} (x-m) \right) \, \frac{dx}{|\boldsymbol{\Sigma}^{-1/2}|} \\[12pt] & = \frac{1}{\sqrt{(2\pi)^n}} \exp\left( -\frac 1 2 (u-\ell)^T(u-\ell) \right) \, du \end{align} ...


0

The number $X$ of dust particles in the filters is a discrete random variable that counts number of occurances (dust particles) in a specified interval (filter). Thus, it is a reasonable assumption that it can be probabilistically described by the Poisson distribution.


1

In the diagonal case, the determinant of $\Sigma$ is the product of its diagonal entries and each diagonal entry is the variance $\sigma_i^2$ of the $i$th coordinate hence, to divide by $\sqrt{|\Sigma|}$ is to divide by the product of the standard deviations $\sigma_i$, as desired. Said otherwise, $\sqrt{(2\pi)^n|\Sigma|}$ is the product of the ...


1

Because we want to make the integral of the density over $R^n$ equal to 1. More precisely, if we compute $\int_{R^n}\exp\left(-\frac{1}{2}({x}-{m})^T{\boldsymbol\Sigma}^{-1}({x}-{m}) \right)dx$, firstly we will make a variable change $y = \Sigma^{-1/2}(x-m)$, then we have $$\int_{R^n}\exp\left(-\frac{1}{2}({x}-{m})^T{\boldsymbol\Sigma}^{-1}({x}-{m}) ...


1

The sample covariance maxtrix estimator $$\Theta(X) = \frac{1}{N}\sum_{i=1}^N (\mathbf{X}_i-\bar{\mathbf{X}})(\mathbf{X}_i-\bar{\mathbf{X}})^\top$$ is a maximum likelihood estimator of the covariance matrix of a multivariate Gaussian distribution, see. But it is biased and we have $$E[\Theta(X)]=\frac{N-1}{N}\Sigma$$ Correcting for Bias we get an unbiased ...


0

Yes. You are wrong: $E[|X^n|]=\int_{-\infty}^{\infty}|x|^np(x)\space dx=\int_{-\infty}^{0}(-1)^nx^np(x)dx+\int_{0}^{\infty}x^np(x)\space dx=\int_{0}^{\infty}x^np(x)\space dx-\int_{-\infty}^{0}x^np(x)dx$ Now if you let $x=-u$, you will have: $\int_{0}^{\infty}x^np(x)\space dx-\int_{-\infty}^{0}x^np(x)dx=\int_{0}^{\infty}x^np(x)\space ...


1

The expected value is $$ E_n = E[|X|^{2n+1}] = 2 \int_0^\infty \frac1{\sqrt{2\pi}} \exp\left(-\frac{x^2}2\right)x^{2n+1} dx $$using parity to get rid of the $\{x<0\}$ part. As you know that $$ \frac d{dx} \exp\left(-\frac{x^2}2\right) = -x \exp\left(-\frac{x^2}2\right) $$ you can integrate by parts: $$ E_n = \left[-2 \frac1{\sqrt{2\pi}} ...


1

It does look that way empirically, and would do even more if you looked at $$\lim_{k\to \infty}\dfrac{\displaystyle \sum_{i=1}^{k} \phi\left[\Phi^{-1}\left(\frac{i-\frac12}{k}\right)\right]}{k}$$ which is essentially a discrete approximation to $$E[\phi(X)] = \int_{-\infty}^\infty \phi(x)^2\, dx = \int_{-\infty}^\infty \frac{1}{2\pi}\exp\left(-{x^2}\right) ...


0

In mathematics, $0$ is often used to mean the zero vector in any vector space. In this case, the covariance matrix is a $2\times 2$ matrix, then the mean is a vector in $\mathbb{R}^2$, so the zero vector is $\begin{bmatrix}0\\0\end{bmatrix}$.


2

The multivariate normal distribution in $\mathbb{R}^{d}$ is denoted by $\mathcal{N}(\mu,\Sigma)$ where $\mu \in \mathbb{R}^{d}$ and $\Sigma \in \mathcal{M}_{d}(\mathbb{R})$ is as symmetric positive matrix. The associated probability density function is : $$ x \in \mathbb{R}^{d} \, \longmapsto \, \frac{1}{(\sqrt{2\pi})^d} \frac{1}{\sqrt{\det(\Sigma)}} \exp ...


1

If this is supposed to represent a bivariate normal distribution then the mean should be a $1\times 2$ vector and the covariance matrix should be symmetric, such as $\mathcal{N}\left(\begin{bmatrix} 0 \\ 0 \end{bmatrix},\begin{bmatrix} 0.1 & 0.02 \\ 0.02 & 0.3 \end{bmatrix}\right)$


0

Hint: use the fact that the sum of normally distributed variables is a normal variable, and then compute the expected value and the variance of X_bar.


2

Here are some results : the first come from the definition of the $erf$ function, the other are obtained by integration by parts. $$\int e^{-ax^2}dx=\frac{\sqrt{\pi } \text{erf}\left(\sqrt{a} x\right)}{2 \sqrt{a}}$$ $$\int x e^{-ax^2}dx=-\frac{e^{-a x^2}}{2 a}$$ $$\int x^2 e^{-ax^2}dx=\frac{\sqrt{\pi } \text{erf}\left(\sqrt{a} x\right)}{4 a^{3/2}}-\frac{x ...


1

The first formula is correct. This is a direct consequence of the definition of the convolution and the definition of the normal and uniform laws. For the second formula, there are 2 missing factors: $$ \frac 1{\prod_{k=1}^N (b_k - a_k)} $$ which is the inverse of the total weight of the uniform distribution, and $$ \frac 1{(2\pi)^{N/2}} $$ to make sure ...


1

The only way I can see how to answer this question with pencil and paper only--no tables, no calculators--is to have the quantiles of the standard normal distribution memorized: $$\begin{align*} \Pr[Z \le 0.8] &\approx 0.842 \\ \Pr[Z \le 0.9] &\approx 1.282 \\ \Pr[Z \le 0.95] &\approx 1.645 \\ \Pr[Z \le 0.975] &\approx 1.96 \\ \Pr[Z \le ...


2

Here is way to check what the distribution is like and leads to a conclusion. Let us call $\frac{(X_1+X_2)^2}{(X_1-X_2)^2} = W$ Then in the following images Series 1 is $W$, Series 2 is $\chi^2_2$, Series 3 is $F_{2,2}$ and Series 4 is $F_{2,1}$ and Series 5 is $F_{1,1}$. What I have done is I have generated $X_1$ and $X_2$ from $N(0,1)$ and computed W ...


0

To win the game you must find the two 'good' squares before the one 'bad' square, out of seven uncovered one at a time. Call the other four squares 'okay'. Suppose it takes $k$ turns to until you win, where $k$ is some number between $2$ and $6$.   Then to have won you must have selected $k-2$ 'okay' squares and $1$ 'good' square out of all the ways ...


0

The mean and variance of the marginal density can be directly evaluated from the expression $\mathbf{x}=\mathbf{Wz}+\boldsymbol{\mu}+\boldsymbol{\epsilon}$. Nevertheless, the more lengthy approach of evaluating the integral is a useful exercise in how to deal with integrals of exponential functions whose argument is a quadratic function of vectors, as is ...


0

My mistake, I have assumed that m>a. If the assumption is m<0 the answer from Maxiama is ${{-{{\sqrt{2}\,{\it \gamma}\left({{3}\over{2}} , {{m^2+2 \,a\,m+a^2}\over{2\,s^2}}\right)\,m\,s^2\,\left| s\right| }\over{ \left| m+a\right| }}-{{\sqrt{2}\,{\it \gamma}\left({{3 }\over{2}} , {{m^2+2\,a\,m+a^2}\over{2\,s^2}}\right)\,a\,s^2\, \left| s\right| ...


0

Parts (a) and (c) ask to compute probabilities of the marginal distributions. Parts (b) and (d) ask to compute probabilities of conditional distributions. If $(X,Y)$ is bivariate normal with mean vector $(\mu_x, \mu_y) = (3,1)$ and covariance matrix $$\begin{bmatrix} \sigma_x^2 & \rho \sigma_x \sigma_y \\ \rho \sigma_x \sigma_y & \sigma_y^2 ...


0

The integration of $e^{-y^2/2}$ is: $\displaystyle\int e^{-y^2/2} \operatorname d y = \sqrt{\pi/2}\; \operatorname{erf}(y/\sqrt{2})+\text{constant}$ Where $\operatorname{erf}(\cdot)$ is the Gauss error function, which is a special function (it cannot be represented in terms of elementary functions). That's why we have tables to look up values for Gaussian ...


2

If $a\,q\ne0$, the integral does not converge. Let $s^*=-\ln(a\,q)/2$. The denominator of the integrand vanishes when $s=s^*$. Close to $s^*$ we have $$ (e^{-2s}-a\,q)^2\sim (s-s^*)^2, $$ meaning $$ \lim_{s\to s^*}\frac{(e^{-2s}-a\,q)^2}{(s-s^*)^2}=c\ne0. $$ Finally, $(s-s^*)^{-2}$ is not integrable on any neighborhood of $s^*$.


0

One of the most important result you will learn in an elementary probability/statistics course is the following: Central Limit Theorem. Let $X_1,X_2,X_3,\ldots$ be a sequence of i.i.d. random variables with mean $\mu$ and variance $\sigma^2$. Then, as $n\to\infty$, \begin{align} \frac1{\sqrt n}\sum_{i=1}^n(X_i-\mu)\tag{1} \end{align} converges in ...


2

The table below shows the values given by R for $\Phi(x)=\Pr(X \le x)$ for some $x$. In addition $\Phi(-x)=-\Phi(x)$. For $x$ large and negative, a reasonable approximation is $\dfrac{-x}{x^2+1}\dfrac{\exp\left(-x^2/2\right)}{\sqrt{2\pi}}$. For example, with $x=-15$, this gives 3.670825e-51 So for $x$ large and positive, a reasonable approximation is ...



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