# Tag Info

## New answers tagged normal-distribution

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 $$\mu\{|N|>x\sqrt 2\}\overset{x\to +\infty}{\sim}\frac ... 0 Hint for the first question: It is P(X\leq x)=\Phi\left( \frac{x-\mu}{\sigma} \right) and P(X\leq -x)=1-\Phi\left( \frac{x-\mu}{\sigma} \right) And P(-x \leq X \leq x)=P(X\leq x)-P(X\leq -x) \Phi(\cdot ) is the cdf of the standard normal distribution. X\sim \mathcal N \left( 0.9, 0.003^2 \right)  2 The statement$$\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}$$(I would not expect computer algebra systems to correctly handle various modes of ... 1$$ \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}} $$1 The density is$$ 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}}. ...

0

$\phi(.)$ is the density function for $N(0,1)$ distribution. Now we know that if $X\sim N(\mu,\sigma^2)$, then $aX+b\sim N(a\mu+b,a^2\sigma^2)$

0

$$f_{X_1,X_2,\theta}(x_1,x_2,t)=f_{\varepsilon_1}(x_1-t)\cdot f_{\varepsilon_2}(x_2-t)\cdot f_\theta(t)$$ "integrate" over $t$ using $f_\theta(t)=1$ to get $$f_{X_1,X_2}(x_1,x_2)$$ proportional to $$\exp\left(-(x_1-x_2)^2/(4\sigma^2)\right)$$ do a similar reasoning to get $f_{X_2}$ and then use Bayes' rule.

0

The normal bivariate distribution is characterized by the parameters $$E(X),\qquad E(Y),\qquad\mathrm{var}(X),\qquad\mathrm{cov}(X,Y),\qquad\mathrm{var}(Y).$$ To compute these, note that $X=\mu_X+\sigma_XU$ and $Y=a+bX+\sigma V$ where $(U,V)$ is bivariate standard normal, that is, such that. ...

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. :-)

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) ...

0

$945$ mL is $5$ mL below the mean. $5$ mL is half of the standard deviation. So the question is, what is the probability that a standard normal random variable is less than $-1/2\ {}$? Usually you'll get that from a table or from software.

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

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 ...

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.

0

Compare $\Phi(z)$ with $z=\displaystyle\frac{(x-μ)}{σ}$ (your first computation), and $\displaystyle F(x)=\Phi \frac{(x-μ)}{σ}$ (your second computation).

6

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 $$... 0 If X is a random variable, then E(X + k) = E(X) + k, E(cX) = c E(X), Var(X + k) = Var(X), and Var(cX) = c^2 Var(X). (These are facts for any random variable, including normal variables.) If you already know these facts, you can use them to determine the mean and variance of Y - \mu, then use them again to find your desired result. 5 Since you seem to be turning around this question and some of its variants again and again, let us try to answer it (almost) completely. First, as mentioned partially by the text you are reading, to know the characteristic function of every normal random vector, it is enough to know the characteristic function of a standard one-dimensional normal random ... 1 I take it that by "mean sum of squares" you mean sum of squares divided by degrees of freedom. Sometimes those are called mean squares; I don't think I've heard "mean sum of squares" before. The F-distribution is usually characterized as the distribution of$$ \frac{\chi^2_n/n}{\chi^2_m/m} where the two chi-square random variables have respective ... 0 If a^2=1-2t, then t-\frac12= -\frac12\left(1-2t\right)=-\frac{a^2}{2}, so \begin{align} \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 a^2/2} \, dz \\[10pt] & = \frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-w^2/2} \left(\frac{dw}a\right) \\[10pt] & = ... 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}} $$You should attempt to solve the integral by fitting a normal distribution and cancelling ... 0 Partition your covariance matrix \Sigma and its inverse \Psi as follows, with the subscripts _A denoting known values and _B denoting missing values:$$ \Sigma = \begin{pmatrix}\Sigma_A & \Sigma_{AB} \\ \Sigma_{BA} & \Sigma_B\end{pmatrix}, \qquad\Psi = \begin{pmatrix}\Psi_A & \Psi_{AB} \\ \Psi_{BA} & \Psi_B\end{pmatrix}. $$If we ... 1 Let x be the maximum height of an adult that admits a clearance of at least 17 cm for the doorway. Let h be the height of such a doorway, so h = x + 17. Now we wish to find the value of x such that 99\% of adults have height less than or equal to x; i.e., this is \Pr[X \le x] = 0.99 = invnorm(0.99,187.5,9.5) = 209.600. Therefore, x + 17 ... 0 Here is one way to count the number of possibilities. Suppose you only had pennies to work with then there is only one way. Now if you have pennies and nickels, then there can be (1 penny, 14 nickels) or (2 pennies, 13 nickels) and so on upto (14 pennies, 1 nickel). That is there are 14 possibilities. In general, if you have x coins, there are x-1 ways ... 0 In the case of Brownian motion, we know from the stationarity of the increments that B_t-B_s \sim B_{t-s}. This implies$$\mathbb{E}((B_t-B_s)^2) = \mathbb{E}(B_{t-s}^2)=t-s.$$But in general, B_t \sim N(0,t), t \geq 0, does not imply that B_t-B_s has variance t-s. Counterexample: Let B_t \sim N(0,t) and B_s \sim N(0,s) be independent random ... 2 Here's some intuition: Suppose you have 15 coins as follows: O O O O O O O O O O O O O O O O How many ways can you divide up these 15 coins into 4 different categories of coins? For instance: O | O O | O O O O O O O O O | O O O (Pennies | Nickels | Dimes | Quarters) Hope this is helpful, let me know if you need another hint :) 2 Hints Denote by p the density of X, i.e.$$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.$$Check that \mathbb{E}(X^{2n}) = \frac{(2n)!}{2^n n!} \sigma^{2n} holds for n=1. Now perform an inductive step, i.e. assume ... 2 Hint: Complete the square in the exponent and do a change of variables. 2 If X and Y are independent and normal (\mu_X,\Sigma_X) and (\mu_Y,\Sigma_Y) respectively, then:$$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, ...

0

Suppose $X \sim \mathrm{Normal}(\mu,\sigma)$ (mean $\mu$ and standard deviation $\sigma$). Then $Z = \dfrac{X-\mu}{\sigma} \sim \mathrm{Normal}(0,1)$. This much should be clear. Now suppose we observe $n$ independent and identically distributed values drawn from $X$; i.e., we consider these $x_1, x_2, \ldots, x_n \sim \mathrm{Normal}(\mu,\sigma)$. The ...

0

First, Dilip's suggestion should be heeded - without context there is no clear answer to your question. However, in general, the standard deviation in the denominator should match the standard deviation of the variable being normalized. otherwise a Z-score will not be approximated by a standard normal. I think it is possible, that the two are identical, but ...

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, ...

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

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.

0

Since $X$ is normally (hence symmetric with respect of its mean) distributed, the equation $P(X>3.01)=P(X<2.99)$ allows the conclusion that $\mathbb E(X)=\frac{1}{2}(3.01+2.99)=3$. Then $X=3+\sigma U$ where $U\sim Norm(0,1)$, so that $P(U>\frac{0.01}{\sigma})=P(X>3.01)=0.05$. This equation can be written as ...

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