# Tag Info

## New answers tagged normal-distribution

0

This is a proof for $\mu = 0$. When $\mu \neq 0$, the proof can be done with some slight modification. You want to prove $\sqrt{n}(1-X_n^{-1}) - G_n = \sqrt{n}(2-X_n^{-1}-X_n)$ converges to $0$ in probability, which is equivalently that it converges to $0$ in distribution. To do this, we can apply the second order delta method(for reference see this) with ...

0

Let $X_i$ have mean $\mu_i$ and standard deviation $\sigma_i$, and write $X_i = \mu_i + \sigma_i Z_i$ where $Z_i$ are independent standard normal random variables. The joint density of $Z_1, \ldots, Z_n$ is $$f(z_1, \ldots, z_n) = (2 \pi)^{-n/2} e^{-(z_1^2 + \ldots + z_n^2)/2} = (2 \pi)^{-n/2} e^{-\|{\bf z}\|^2/2}$$ which is rotationally invariant, i.e. ...

0

Assuming you have a standard normal distribution table handy... (a) You can use the table once you have converted your random variable (let's call it $X$) to the standard normal distribution $Z$. The conversion is $$Z = \frac{X - \mu}{\sigma}$$ where in this case $\mu = 250$ and $\sigma = 15$. So $$P(X > 270) = P\left(Z = \frac{X - \mu}{\sigma} > ... 0 The important thing to remember is that you are dealing with multiple convolutions gaussian filter. If you have any probability background this is no different to multiple draws from a gaussian distribution and thus the same laws apply. Theorem: The normal distribution is stable. Specifically, suppose that X has the normal distribution with mean μ∈ℝ and ... 0 You did this right. The short way to look at it is that B+C-A is normally distributed with mean being \mu = \mu_B+\mu_C-\mu_A and \sigma^2 = \sigma^2_B + \sigma^2_C + \sigma^2_A. The key point you need to know is that a variate made of the sum of two independent normal variates is itself normally distributed, even if the means of those two variates ... 1 More generally, if X_1, \ldots, X_n are independent normal random variables with means \mu_i and variances \sigma^2_i, and c_1, \ldots, c_n are constants, Y = c_1 X_1 + \ldots + c_n X_n is normal with mean \mu_Y = c_1 \mu_1 + \ldots + c_n \mu_n and variance c_1^2 \sigma_1^2 + \ldots + c_n^2 \sigma_n^2. Here you're doing the case n=3 with ... 1 Let  X_1 \sim N(23.95,7.44), X_2 \sim N(16.29,7.79) , then, as mentioned by Dilip Sarwate in the comments,  X_1 - X_2 \sim N(7.66,{\sqrt{7.44^2 + 7.79^2}})  Thus the problem becomes$$\mathbb{P}(X_1<X_2) = \mathbb{P}(X<0),\ \ where \ \ X = X_1 - X_2 $$This can be calculated directly or reverting to the standard  N(0,1)  distribution as ... 0 \left(\dfrac{37}{38}\right)^{34} is the probability of losing 34 bets in a row. If this does not happen then you have gained 35 at least once and so are ahead after 34 bets. So this is not a Z score but, after subtraction from 1, an exact probability answer to (a). You do not need to consider expectations or approximations. 2 Let us evaluate the CDF of a Gaussian distribution:$$f(x) = \frac{1}{\sqrt{2\pi\sigma^2}} \int_\mathbb{R} \exp \left( - \frac{(x-\mu)^2}{2\sigma^2}\right)\ dx \label{orig}\tag{1}$$Let us make the transformation in \ref{orig} by \mu \mapsto N \mu and \sigma \mapsto \sigma \sqrt{N}.$$\begin{align*} f(x) &= ...

2

The variance of this random vector is $\begin{bmatrix} 1 & \rho \\ \rho & 1 \end{bmatrix}$. Level sets of the density are ellipses, which are circles only when the correlation $\rho$ is $0$. With a positive-definite symmetric matrix, one can always rotate the two coordinate axes to diagonalize the matrix; then one has two linear combinations of the ...

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I would have thought that unless you want to get tangled in the language of one-tailed hypothesis tests, a Bayesian approach might be best. But even then you would need a prior distribution for $\mu_x-\mu_y$; let's suppose there is an improper prior where this difference (call it $D$) is uniformly distributed. Then given $D$, the likelihood of observing ...

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In part (c) you need a continuity correction, just as in the earlier parts: $$\Pr\left( X\le 39.5 \right) = \Pr\left( \frac{X-50}{\sqrt{50}} \le \frac{39.5-50}{\sqrt{50}} \right) \approx\Pr(Z\le -1.485) \approx 0.069.$$

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The probability distribution function (pdf) of a normal distribution with mean $\mu$ and standard deviation $\sigma$ is defined to be: $$f(x) = \dfrac{1}{\sigma \sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$$ For $x= \mu\pm\sigma$ this simplifies to $f(\mu\pm\sigma) = \dfrac{1}{\sigma\sqrt{2\pi e}}$ For $x= \mu\pm 2\sigma$ this simplifies to $f(\mu\pm ... 1 If the total number of hours worked$N=13500$, and the number of employees is$n=450$, then the mean number of hours worked per employee is given by: $$\mu = \frac{N}{n}=30\:\text{hours}$$ We are given that the standard deviation$\sigma = 7\:\text{hours}$, and therefore we have that the number of hours worked per employee is normally distributed$T \sim ...

1

Consider orthogonal matrix $$A = \left( \begin{array}{cc} \frac{x}{\sqrt{x^2+y^2}}& \frac{y}{\sqrt{x^2+y^2}} \\ -\frac{y}{\sqrt{x^2+y^2}} & \frac{x}{\sqrt{x^2+y^2}} \end{array} \right).$$

2

No. The lognormal distribution doesn't have a moment generating function, so you can't use that approach. Instead, suppose $Y = \log X$. Then $Y \sim \operatorname{Normal}(\mu, \sigma^2)$ by definition, and $X = e^Y$. Therefore for a positive integer $k$, $$\operatorname{E}[X^k] = \operatorname{E}[e^{kY}] = M_Y(k),$$ where $M_Y(k)$ is the moment ...

1

Let $Q(x)$ be the Q-function, i.e., $Q(x)=P(N\geq x)$, where $N$ is a Gaussian random variable with mean $0$ and variance $1$. Then \begin{align*} P(Y \geq y) & =P(XZ \geq y) \\ &= P(Z=1)P(XZ \geq y|Z=1) + P(Z=-1)P(XZ \geq y|Z=-1) \\ &= \frac{1}{2}P(X \geq y|Z=1) + \frac{1}{2}P(-X \geq y|Z=-1) \\ &= \frac{1}{2}P(X \geq y) + \frac{1}{2}P(X ... 0 We have Y\le y if and only if X\le y and Z=-1 or if X \gt -y and Z=-1. If f_X is the density function of the standard normal, thenF_Y(y)=\frac{1}{2}\int_{-\infty}^y f_X(x)\,dx+\frac{1}{2}\int_{-y}^\infty f_X(x)\,dx.$$Differentiate, using the Fundamental Theorem of Calculus. We get f_X(y). 2 One hint that might help is to note that the function you are integrating is an odd function, so ... 2 Notice how the function is odd. Therefore, \displaystyle \int_{-\infty}^{0} f(x)=-\int_{0}^{\infty} f(x)\implies \int_{-\infty}^{\infty}f(x)=0 If we can show that \displaystyle \lim\limits_{x\to0,\pm\infty}f(x) exists, which are where it can diverge, then the integral exists. As we can plainly see \lim_{x\to0 \text{or} \pm\infty} is 0. Therefore the ... 1 'The general procedure looks correct. After calculating the probability that a can is over 8 ounces, you might want to use the normal to approximate the binomial. Your "p" for the binomial is not right. We want \Pr\left(Z\gt \frac{0.04}{0.22}\right). This is much larger than 0.145. 1 Let N be a normally distributed variable, with mean 0 and variance 1 (substract the mean and divide by the standard deviation in order be in this case). You look for a certain (smooth, increasing) f:\Bbb R\to[0,1] such as  f(N)  is uniform, that is:$$ P(f(N) \le q) = q $$for every q\in(0,1). Under regularity assumptions, this is$$ q = P(N \le ...

1

Your mistake: you can't divide by $Y$ without knowing its sign (if it's negative, the inequality would reverse). \begin{align*} P(XY<z) &= P(XY<z \mid Y=1) P(Y=1) + P(XY<z \mid Y=-1) P(Y=-1)\\ &=\frac{1}{2} P(X<z) + \frac{1}{2} P(-X<z)\\ &= \frac{1}{2} P(X<z) + \frac{1}{2} P(X>-z)\\ &= P(X<z). \end{align*} The last ...

1

You need to use the symmetry of the normal distribution about the Y axis. If $Y=-1$, $Z=-X\sim -N(0,1)$. Due to the symmetry of $N(0,1)$, that means $X\sim N(0,1)$. If $Y=1$, $Z=X\sim N(0,1)$. Since the probability of either one is evenly split, it means $Z\sim N(0,1)$. If you want to do this with the conditional probability equations in your answer, $$... 1 This cannot be true, as the example X_n=Y_n=X=-Y standard normal shows. If X and Y must be independent, try X_n=Y_n=X with Xand Y i.i.d. standard normal. If X_n and Y_n must be independent, try X_n=X=Y and Y_n=Z with X and Z i.i.d. standard normal. 1$$ f_{Z|Y}(z|y) = \frac{f_Z(z)f_{Y|Z}(y|z)}{\int_{-\infty}^{+\infty}{f_Z(a)f_{Y|Z}(y|a)da}} = \frac{\frac{1}{\sqrt{2\pi}} e^{-\frac{z^2}{2}} \frac{1}{\sqrt{2\pi}} e^{-\frac{(y-z)^2}{2}}} {\int_{-\infty}^{+\infty}{\frac{1}{\sqrt{2\pi}} e^{-\frac{a^2}{2}} \frac{1}{\sqrt{2\pi}} e^{-\frac{(y-a)^2}{2}}da}} = ...

2

The integral of any function over a set with measure $0$ is equal to $0$.

1

For every $\theta$, the change of variable $t=F_\theta(x)$ yields $$E(\psi_\theta(X)^2)=\int_{-\infty}^\infty\Phi^{-1}(F_\theta(x))^2\mathrm dF_\theta(x)=\int_0^1\Phi^{-1}(t)^2\mathrm dt.$$ The RHS does not depend on $\theta$ (and is equal to $1$) hence, differentiating, one gets ...

0

The $k$th moment is $$\int_{-\infty}^\infty z^k \varphi(z)\,dz$$ where $$\varphi(z) = \frac 1 {\sqrt{2\pi}} e^{-z^2/2}.$$ If $k$ is odd then we have $(-z)^k = -(z^k)$. We also have $(-z)^2 = z^2$, so that $\varphi(-z)=\varphi(z)$. Then let $u=-z$ so that $-du=dz$, and we have \begin{align} \int_{-\infty}^\infty z^k \varphi(z)\,dz & = ...

0

Hint 1: Show that if $x\mapsto f(x)$ is even, then $x\mapsto x^kf(x)$ is odd if $k$ is an odd positive integer. Hint 2: What is the net signed area under the graph of an odd function over an interval centered at $0$?

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Question: Let $X \sim N(\mu, \sigma^2)$. Find $E[e^{-X}]$. Answer: Let $Z = -X$, so $Z \sim N(m,\sigma^2)$, where $m = -\mu$. You seek $E[e^Z]$ which is just the mean of a Lognormal random variable, with Normal parent $N(m,\sigma^2)$. Thus: $$E[e^{-X}] = E[e^Z] = e^{m +\frac{\sigma ^2}{2}} \quad \text{ where } m = -\mu$$ If you need help deriving the ...

0

This is one of the basic steps in the Expectation Propagation algorithm. See section 3.2 of A family of algorithms for approximate Bayesian inference, where the formulas you want are given by (3.32,3.33). Keep in mind that the variance of the likelihood might turn out to be negative, so SD2 would be imaginary. This is why it is better to write the ...

0

We have $$N(f(\sigma)) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{f(\sigma)}exp \left(-\frac{s^2}{2}\right)ds.$$ Thus, using the chain rule: $$\frac{\partial N(f(\sigma))}{\partial \sigma} = \frac{\partial N(f(\sigma))}{\partial f(\sigma)}.\frac{\partial f(\sigma)}{\partial \sigma} = ... 0 By the central limit theorem, the distribution of the mean of many independent and identically distributed random variables is approximately normal. When samples from some distribution are taken, the individual samples are considered as coming from independent but identically distributed random variables, so it is often assumed in calculations that the ... 0 If you have a discrete distribution like Binomial or poisson and if you are getting an approximation to these in normal distribution that is called Normal approximation 0 Use the normal distribution by creating the z-score$$ z = \frac{X-\mu}{\sigma} = \frac{X-103}{4.5} $$and then compute$$ P\left(z<= \frac{93-103}{4.5} = -\frac{10}{4.5}\right) $$1 For a two-sided confidence interval [1.9,2.1] with coverage probability 0.99, the corresponding quantile is 0.995, thus the correct z-score is 2.57583, not 2.3, and the desired n satisfies$$\frac{2.1 - 2}{2/\sqrt{n}} = 2.57583.$$1 It should be clear that A=\left\lbrace Z\leq 106\right\rbrace \subset \left\lbrace Y\leq 106\right\rbrace=B -- if 11 fit then so do 10. What you want is the probability of \left\lbrace Y\leq 106\right\rbrace\setminus\left\lbrace Z\leq 106\right\rbrace=B\setminus A. Since \begin{equation*} \mathbf{P}(B)=\mathbf{P}(B\setminus A)+\mathbf{P}(B\cap A), ... 1 The answer is about right, but there are two errors that happen to cancel! The probability a tube is bad (too big or too small) is roughly 2(0.0062), that is, 0.0124. And you need to multiply by 1000, not by 100, giving expected value about 12.4. 1 First of all, define a normalized variable$$ Z={X-\mu\over \sigma} $$Then, on the normal distribution table, you need to look for a probability equal to about 0.8. The linked table gives z_0\approx0.84 so we have$$ P(Z<0.84)\approx0.80 $$Thus to get the right number of hotdogs,$$ {x_0-16036\over100}=z_0=0.84 $$so$$ x=16036+840=16876\approx16900 ...

1

What you have calculated for a and b is the solution for b. This is o.k. To get the formula for the excact probability, you have to use the binomial distribution: $$P(75 \leq X \leq 80)=\sum_{k=75}^{80} {100 \choose k} 0.7^k\cdot 0.3^{100-k}$$ $P(75 \leq X \leq 80) \color{red} = F_{Z}[\frac{80 + 0.5 - > 70}{\sqrt21}] - F_{Z}[\frac{75 - 0.5 - ... 1 $$F(v)=\frac{1}{\sqrt{2\pi}}\int^{v}_{-\infty}\exp(-\frac{x^2}{2})\,dx$$ $$F(0)=\frac{1}{2}$$ the inequality is equivalent to $$|\log F(v)|-|\log(\frac{1}{2})|\leq|v|+|v|^2\Leftrightarrow \log\frac{1}{F(v)}-\log2\leq|v|+|v|^2\Leftrightarrow\log\frac{1}{2F(v)}\leq|v|+|v|^2$$ exponentiating both sides yields $$\frac{1}{2F(v)}\leq\exp(|v|+|v|^2)\Leftrightarrow ... 1 No, the distribution is not necessarily normal if it satisfies the 68-95-99.7 rule. Take a normal distribution, then move all the events that are within (\mu+0.25 \sigma, \mu + 0.75 \sigma), some of them to the mean and some of them to \mu + 0.99 \sigma Do the same for the ones within the range (\mu - 0.75 \sigma, \mu - 0.25 \sigma) in the other ... 1 It is correct. But I wouldn't go from \dfrac{dx}\sigma=dy to dx = \sigma\,dy. Instead, since the integral is$$ \int_{-\infty}^{\mu+x\sigma} \frac{1}{\sqrt{2\pi}} \exp\left(\frac{-1}2\left(\frac{x-\mu}{\sigma}\right)^2\right) \frac{dx}\sigma, $$I'd just put dy in place of \dfrac{dx}\sigma. 0 The assumption of mean 0 is a normalization that must be made because you already have a constant term in the regression. It relates to the issue of identification - that you as the researcher cannot tell the difference between the constant term in the regression and the mean of the error term. Proof: Suppose that \epsilon is not mean 0 Let ... 0 Here's the general idea - someone who has a better background than I do in statistics could probably give a better explanation. So you have this linear regression model:$$Y = \alpha + \beta X + \epsilon$$where$\epsilon$follows a normal distribution with mean$0$. What exactly does random mean? My back ground in statistics is very low level, but I ... 0 Perhaps i misunderstand the question but: The Poisson distribution expresses the probability of a given number of events occurring in a fixed interval of time and/or space if these events occur with a known average rate and independently of the time since the last event and is defined as$\pi(x=k)=\frac{\lambda^ke^{-\lambda}}{k!}$so your interval time is ... 0 The central limit theorem applies to every probability distribution. That's what makes it so powerful. 2 Cramér's Theorem even gives that if the sum$X+Y$of any two independent (not necessarily identically distributed) random variables$X$and$Y$has a normal distribution, then the summands$X$and$Y$themselves must be normally distributed. This is more than enough to cover your case: if$X$is independent of$Y$, it is also independent of$-Y$, and$X-Y$... 1 From the rule of complement which says that$P(A^c) = 1 - P(A)$it follows that:$P(z<-\frac{20.5}{\sigma}~~or~~z>\frac{20.5}{\sigma}) = 1 - 0.8 = 0.2$From the symmetry of the normal distribution then,$P(z<-\frac{20.5}{\sigma}) = 0.1$From here, either we look at a table of probabilities associated with different values of$z\$ or we use calculus ...

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