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Your proposition is not correct. Here is a counter-example. Set $$\alpha(x):=e^{\frac{x^2}2}\int_0^{x}e^{-\frac{t^2}3+t} {\rm d} t.$$ Then \begin{align} \alpha'(x) & =x\:e^{\frac{x^2}2}\int_0^{x}e^{-\frac{t^2}3+t}{\rm d}x +e^{\frac{x^2}2}e^{-\frac{x^2}3+x}=x\alpha(x)+e^{\frac{x^2}6+x} \end{align} and $$... 5 Since you seem to be turning around this question and some of its variants again and again, let us try to answer it (almost) completely. First, as mentioned partially by the text you are reading, to know the characteristic function of every normal random vector, it is enough to know the characteristic function of a standard one-dimensional normal random ... 3 It is the difference between the z score for a datum from an entire population and a sampling. The z score for a datum x is z = (x - \mu)/\sigma where \mu is the population mean and \sigma is the population standard deviation. If the datum x is not from the entire population but rather from a sampling from that population then the standard ... 2 Hint: Complete the square in the exponent and do a change of variables. 2 Hints Denote by p the density of X, i.e.$$p(x) =\frac{1}{\sqrt{2\pi \sigma^2}} \exp \left(- \frac{x^2}{2\sigma^2} \right).$$Deduce from p(x) = p(-x) that$$\mathbb{E}(X^{2n+1}) = \int_{\mathbb{R}} x^{2n+1} p(x) \, dx =0.$$Check that \mathbb{E}(X^{2n}) = \frac{(2n)!}{2^n n!} \sigma^{2n} holds for n=1. Now perform an inductive step, i.e. assume ... 2 The statement$$\delta(x) = \lim_{v\to 0}\frac{e^{-x^2/2v}}{\sqrt{2\pi v}}$$means that for every C^\infty smooth function \varphi with compact support we have$$\varphi(0) = \lim_{v\to 0} \int_{-\infty}^\infty \frac{e^{-x^2/2v}}{\sqrt{2\pi v}} \varphi(x)\,dx \tag{1}$$(I would not expect computer algebra systems to correctly handle various modes of ... 2 If X and Y are independent and normal (\mu_X,\Sigma_X) and (\mu_Y,\Sigma_Y) respectively, then:$$E(\|X-Y\|^2)=\|\mu_X-\mu_Y\|^2+\mathrm{tr}(\Sigma_X+\Sigma_Y)To show this, note that, by independence, X-Y is normal (\mu_X-\mu_Y,\Sigma_X+\Sigma_Y) and that every random variable Z normal (\mu,\Sigma) can be written as Z=\mu+LU where ... 2 If x,y are independent, and thus, uncorrelated, then p(x,y) is their joint probability distribution, which is Gaussian again, with mean: [\mu_x,\mu_y]^T and covariance \text{diag}\{\Sigma_x,\Sigma_y\} (dimensions are 2N\times2N). Then, ... 2 If you're having trouble finding distributions (or densities), it's usually not a bad idea to start from the cdf, F_X(x) = P(X \leq x). Let Y = |X|, X \sim N(0,\sigma^2) and let F_X be the cdf of X. Then, for y \geq 0, \begin{align*} P(Y \leq y) &= P(|X| \leq y) \\ &= P(-y \leq X \leq y) \\ &= P(-y \leq X \leq 0) + P(0 < X \leq y) ... 2 Here's some intuition: Suppose you have 15 coins as follows: O O O O O O O O O O O O O O O O How many ways can you divide up these 15 coins into 4 different categories of coins? For instance: O | O O | O O O O O O O O O | O O O (Pennies | Nickels | Dimes | Quarters) Hope this is helpful, let me know if you need another hint :) 2 First you have to transform the random variable: Z=\frac{X-\mu}{\sigma} What you are looking for is P(2.99 \leq X \leq 3.01)=2\Phi(\frac{3.01-3}{\sigma})-1=0.9 \Phi(.) is the Standard normal distribution. What you have to do next is to solve this equation for \sigma. The value for \Phi(.)^{-1} (inverse function) can be looked up in a table. 1 Using independence, we obtain that\mu\{S_n>\varepsilon\sqrt{n\log\log n}\}=\mu\{N>\varepsilon\sqrt{\log \log n}/\sigma\},$$where the distribution of N is standard normal. So we are reduce to estimates on the tail function of a standard normal distribution. We can use the equivalent$$\mu\{|N|>x\sqrt 2\}\overset{x\to +\infty}{\sim}\frac ...

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I take it that by "mean sum of squares" you mean sum of squares divided by degrees of freedom. Sometimes those are called mean squares; I don't think I've heard "mean sum of squares" before. The F-distribution is usually characterized as the distribution of $$\frac{\chi^2_n/n}{\chi^2_m/m}$$ where the two chi-square random variables have respective ...

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Since $X$ and $Y$ are independent, the joint density of $X$ and $Y$ is the product of the individual densities, so it is $$\frac{1}{2\pi}e^{-(x^2+y^2)/2}$$ on the entire $x$-$y$ plane. We want to find $$\int_D \frac{x^2+y^2}{2\pi}e^{-(x^2+y^2)/2}\,dy\,dx,$$ where $D$ is the unit disk. Switch to polar coordinates. We want to find ...

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The density is $$f(x) = \begin{cases} \frac2{\sqrt{2\pi}} e^{-x^2/2} & \text{if }x>0, \\[6pt] 0 & \text{if }x<0. \end{cases}$$ So the expected value is $$\int_0^\infty xf(x)\,dx = \frac2{\sqrt{2\pi}} \int_0^\infty e^{-x^2/2} \big(x\,dx\big) = \frac2{\sqrt{2\pi}} \int_0^\infty e^{-u}\,du = \frac2{\sqrt{2\pi}} \cdot 1 = \sqrt{\frac{2}{\pi}}. ... 1$$ \begin{align} P(x=x_0 \mid x>0) &= \frac{P(x=x_0)}{P(x>0)} \\ &= \frac{\frac{1}{\sqrt{2 \pi}} e^{-x_0^2/2}}{\int_0^\infty \frac{1}{\sqrt{2 \pi}} e^{-z^2/2} dz} \\ &= \sqrt{\frac{2}{\pi}} e^{-x_0^2/2} \end{align} $$So$$ E(x \mid x>0) = \sqrt{\frac{2}{\pi}} \int_0^\infty e^{-z^2/2} z\, dz = \sqrt{\frac{2}{\pi}} $$1 Hint: Let ~\alpha=\dfrac{R^2}{1+R^2},~ and ~J=\displaystyle\int_{\mathbb R^2}\exp\bigg[-\alpha~(x+iy)^2-y^2\bigg]~dx~dy.~ Now, evaluate this simpler expression, and see what happens as you differentiate repeatedly with regard to \alpha, a total number of k times. :-) 1 Try twiddling with the sliders on this page. 1 The answer is yes and no: You only really need to use it if the area under the normal curve will change substantially with a 0.5 unit change in the location on the x axis. In the normal approximation, your mean and variance are large, so it is not necessary. In the second case, the underlying distribution is not defined over the integers, but over a ... 1 I would say \Phi(\frac{c-\mu}\sigma)=0.08\Rightarrow \frac{c-\mu}\sigma =\Phi^{-1}(0.08)=-1.406 \Phi(u)=\,\, distribution function of the standard normal distribution N(0,1) You'll find in the tables N(0,1), in Ecxel (function "=NORMSINV(0,08)) etc. 1 The z-score calculation is designed to answer the question: "How far from typical is this result?" When dealing with a single datum, the single-value formula z=\frac{x-\mu}{\sigma} gives us the answer. But when dealing with a whole pile of individual measurements, we expect the Law of Large Numbers, and its more formal cousin the Central Limit Theorem, ... 1$$\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{z^2(t-\frac{1}{2})}dz=\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-z^2(\frac{1}{2}-t)}dz$$The general PDF for a normal distribution is given by:$$ f(x)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}  You should attempt to solve the integral by fitting a normal distribution and cancelling ...

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Let $x$ be the maximum height of an adult that admits a clearance of at least $17$ cm for the doorway. Let $h$ be the height of such a doorway, so $h = x + 17$. Now we wish to find the value of $x$ such that $99\%$ of adults have height less than or equal to $x$; i.e., this is $\Pr[X \le x] = 0.99$ = invnorm(0.99,187.5,9.5) = 209.600. Therefore, $x + 17$ ...

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