Tag Info

New answers tagged

0

$$P(X > 3) = \frac{1}{2}\left(1 + \mathrm{erf}\left(\frac{\mu - 3}{\sigma \sqrt{2}}\right)\right),$$ so $\mu = 3 + \sigma \sqrt{2}\, \mathrm{erf}^{-1}(2 c - 1)$, where $c = 0.8413$. $(X + Y)/2$ has normal distribution with $\mathsf{E} (X+Y)/2 = \mu$ and $\mathrm{Var}((X + Y)/2) = \sigma^2 / 2$. Therefore, $P((X + Y)/2 > 3) = \frac{1}{2} (1 + ...


1

Since $X$ has a symmetic distribution about $0$, you have $P(X-0 \le x) = P(0-X \le x)$, i.e. $P(X \le x) = P(-X \le x)$. Now consider the cumulative distribution function $P(Z \le x)$. This is $P(XY \le x)=P(XY \le x|Y=1)P(Y=1)+P(XY \le x|Y=-1)P(Y=-1)$ which is $P(X \le x|Y=1)P(Y=1)+P(-X \le x|Y=-1)P(Y=-1)$ which by the symmetry and independence is ...


2

If $X$ and $Y$ are independent and $Y$ takes values in $\{-1,1\}$ while the distribution of $X$ is symmetric about $0$, then $XY$ has the same distribution as $X$. Hint: condition on $Y$.


2

Set $$f_n(s) := \sum_{j=1}^n s_j 1_{[t_{j-1},t_j]}(s).$$ Then your calculation shows $$\Phi_{X^m}(s) = \exp \left(- \frac{1}{2} \sum_{k=1}^m \langle f_n, e_k \rangle^2 \right).$$ Letting $m \to \infty$, we obtain $$\Phi_X(s) = \exp \left(- \frac{1}{2} \sum_{k \geq 1} \langle f_n, e_k \rangle^2 \right). \tag{1}$$ Since $(e_k)_{k \geq 1}$ is an ONB, we ...


1

Since $X$ is a vector $$\mathbb{E}X=\mathbb{E}[A+BZ]=A$$ and $$Var(X)=\mathbb{E}[(X-\mathbb{E}X)(X-\mathbb{E}X)']=\mathbb{E}[BX(BX)']$$ $$=\mathbb{E}[BXX'B']=B\mathbb{E}[ZZ']B'=BIB'=BB'$$


7

If we set $$I := \int_{\mathbb{R}} \exp \left(- \frac{x^2}{2} \right) \, dx,$$ then $$I^2 = \int_{\mathbb{R}} \int_{\mathbb{R}} \exp \left( - \frac{x^2+y^2}{2} \right) \, dx \, dy.$$ Introducting polar coordinates, i.e. $$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} r \cos \varphi \\ r \sin \varphi \end{pmatrix},$$ yields $$I^2 = ...


0

Hint: Change variables into the integral $\displaystyle\int_{-\infty}^{+\infty} e^{-x^2}\,dx$, square that, then transform into a polar integration.


2

The "easiest" way is to use a change of variables to change your integral into a multiple of $$\int_{-\infty}^{+\infty}e^{-u^2}\,du$$ and use the famous fact that that last integral equals $\sqrt{\pi}$.


1

There are two answers depending on how restrictive your setting is. As the OP asks for a general principle I will consider the following easier example: Let $\Omega =\{0,1,2\}$, $P = (\delta_0+2\delta_1+\delta_2)/4$ and $Z:\Omega \to \mathbb{R}$ with $Z(\omega ) = \omega$. We know that we may obtain $Z \stackrel{d}{=} Z'$ with $Z' = X'+Y'$ where $X'$ and ...


3

There is a mistake in the computation of the characteristic function. By definition, $$\varphi_Z(t) = \mathbb{E}e^{\imath \, t Z}, \qquad t \in \mathbb{R}.$$ Therefore, $$\begin{align*} \varphi_{\frac{Y}{\sqrt{\lambda}}}(t) &= \mathbb{E} \exp \left( \imath \frac{t}{\sqrt{\lambda}} Y \right) = \exp \left( \lambda \left[ \exp \left(- ...


2

You are using a binomial model of the number of errors. For a binomial distribution for errors with $N$ cases and an error rate probability $p$ the mean number of errors is $\mu=Np$ and the standard deviation of the number of errors is $\sigma=\sqrt{Np(1-p)}$. Plugging in $N=80$ and $p=0.07$ (which is 7%) gives $\mu=5.6$ and $\sigma=2.2821$. For the second ...


3

We consider $m_n=\max\limits_{j=1}^n|a_j|$, $s_n^2=\sum\limits_{j=1}^na_j^2$, $W$ some random variable distributed like every $W_j$, and we simplify the quantities involved in Lindeberg's condition as follows; $$\begin{align} \sum_{j=1}^nE( X_{nj}^2 \cdot \mathbf 1\{ |X_{nj}|\gt \varepsilon \sigma_n \}) &= \sum_{j=1}^n \frac{a_j^2}{n} E\left( W^2 \cdot ...


2

The normal distribution is maximally uncertain on the real line. Precisely, this means that the normal distribution has the highest entropy of all distributions on the real line. In this way, if your distribution has a mean and standard deviation, and support equal to the real line, and you know nothing more, then from an entropy point of view, the best ...


2

The reason the Normal is a reasonable fit for a lot of phenomena can be explained via the Central Limit Theorem: Roughly, it says that a sample average of iid random variables $X_i$ with mean $\mu$ and standard deviation $\sigma$ from any distribution will approach a Normal distribution: $$ \lim_{n\rightarrow \infty}P\left( \frac{\sum^n (X_i-\mu)}{n} \in ...


0

No. Random physical processes don't necessarily have to be distributed like the normal distribution. The text you referenced is talking about how in physical processes will give a distribution similar to the normal distribution. Think of throwing a dart at a dartboard and measuring the distance it lands from the bullseye (ignore the actual rules of darts ...


1

It's not a necessary condition. There are natural processes that are well modeled by other distributions such as the Poisson distribution for waiting times. Some processes are naturally normal such as a random walk. Other processes converge to normality in the limit. For example, the average of many trials of a uniform distribution on an interval, is ...


0

Four speculative scenarios follow: Perhaps the context of the problem in the text or lectures could provide a clue which (if any) might be intended. 1) Uniform. Perhaps, somehow, "constant" is supposed to suggest a uniform distribution. To have mean 35, that would have to be $X \sim Unif(0, 70),$ so $P(X \le 45) = 45/70 = 0.6429.$ 2) Normal. Some yards may ...


3

Hints: Since $(W_t)_{t \geq 0}$ is a Brownian motion, it has independent normally distributed increments; in particular, $W_3-W_2, W_2-W_1,W_1$ are independent random variables which are Gaussian with mean $0$ and variance $1$. This implies that $(W_1, W_2-W_1,W_3-W_2)$ is (jointly) Gaussian with mean $\mu=0$ and covariance matrix $$C = \begin{pmatrix} 1 ...


3

The paper "On The Bound of Proximity of the Binomial Distribution to the Normal One" - Nagaev, Chebotarev (2010) has improvements to $C$ specifically for the Binomial Distribution. Theorem 2 on page 3 gathers together the results in the paper to show that $C$ can be taken to be .4215 in general. The paper notes that an Esseen (1956) paper demonstrates that ...


1

If $\mu$ were known, then $$\frac{1}{\sigma^2}\sum_{i=1}^n (X_i - \mu)^2 = \sum_{i=1}^n \left(\frac{X_i - \mu}{\sigma}\right)^2 = \sum_{i=1}^n Z_i^2 \sim Chisq(df=n),$$ where $Z_i$ are iid standard normal. In your case, $\mu$ is not known, and it needs to be estimated by $\bar X.$ The random variable $$\frac{1}{\sigma^2}\sum_{i=1}^n (X_i - \bar X)^2 \sim ...


2

Because with the factor of $2$ included, the corresponding Gaussian random variable has variance and standard deviation $1$, whereas without it, the random variable has variance $1/2$. This $2$ floats around in all sorts of connections between probability and other subjects. One example is in connections between Brownian motion and PDEs; for instance, if ...


1

It is probably just convention. I suppose that another reason is that if we set $f : \, x \mapsto e^{-\frac{x^2}{2}}$ then $\frac{\mathrm{d}}{\mathrm{d}x}f(x) = x\cdot f(x)$. No $2$ appears in the "polynomial" part.


1

The entire "normal distributions" aspect of this problem is a red herring. If $X_1,\ldots,X_n$ are i.i.d. normally distributed random variables with mean $0$, then the joint distribution of $(X_1,\ldots,X_n)$ is a multivariate normal distribution centered at the origin, which has complete spherical symmetry. Now, each sequence ...


1

If you want a normal-type distribution, you could consider a folded normal distribution. Essentially, if $X$ is a normally distributed random variable, then the random variable $Y=|X|$ follows a folded normal distribution. In particular, if $X$ has mean $0$, then $Y=|X|$ follows a half normal distribution, which decreases monotonically. Thus $1-Y$ increases ...


0

Suppose that $X_M, Y_M \in \mathbb{C}^{M+1}$ are vectors whose components $X_{Mi}, Y_{Mi}$ are iid $\mathbb{C}\mathcal{N}(0,1)$ random variables. I want to show that $$\lim_{M \to \infty} \frac{X_{Mi}\overline{Y_{Mj}}}{M} = 0$$ for all $i, j$. First suppose $M + 1 > \max(i, j)$ (so the vectors actually contain an $i$th and $j$th element). Then ...


1

Presumably $X$ and $Y$ are jointly normal. Then $$E[X|Y] = \mu_x + \dfrac{\rho \sigma_x}{\sigma_y} (Y - \mu_y)$$ and then $$ E[X e^Y] = E[E[X|Y] e^Y] = \ldots $$ Now use the moment generating function for $Y$: $$E[e^{tY}] = \exp(t \mu_y + t^2 \sigma_y^2/2)$$ and its derivative with respect to $t$, which is $E[Y e^{tY}]$.


0

A function $f:\mathbb R\to \mathbb R$ is right-continuous if $\forall x\in\mathbb R,\ \lim_{t\to x^+}f(t)=f(x)$. The c.d.f. of a probability $\mathbb P$ is $$F_{\mathbb P}(x):=\mathbb P\left((-\infty,x]\right)$$ For $\sigma$-additive probabilities over a $\sigma$-algebra $\mathcal E$ the following properties hold: a. $A\subseteq ...


3

a) $Y^{2}=X^{2}$ so that $\mathbb{E}Y^{2}=\mathbb{E}X^{2}=1$. $X$ is symmetric and consequently $Y$ is symmetric. That implies $\mathbb{E}Y=0$ so that $\text{Var}Y=\mathbb EY^2=1$. Then: $$\rho\left(X,Y\right):=\mathbb{E}\left(\frac{X-\mu_{X}}{\sigma_{X}}\right)\left(\frac{Y-\mu_{Y}}{\sigma_{Y}}\right)=\mathbb{E}XY$$ Now work this out. b) If $a$ ...


3

Hint: By standardizing, you have $\;2\,\mathsf P (Z< -\frac 3\sigma) = \mathsf P(Z<\frac 3\sigma)$ for $Z\sim \mathcal N(0,1^2)$ Further, a standard normal distribution is symmetric around the zero value, so $\mathsf P(Z<-\frac 3 \sigma) = \mathsf P(Z>\frac 3 \sigma) \quad = 1 -\mathsf P(Z<\frac 3 \sigma)$


0

There are certain properties of Gaussian/normal distributions that make them appealing, beyond the simple stuff like the central limit theorem. For example, the Gaussian/normal has the maximum entropy for a given mean and variance. This says that the Gaussian/normal distribution provides the maximum overall "variation" in the entropy sense, given standard ...


2

The Gaussian can be viewed as the "best guess" of a distribution, given that we only know that it is a distribution, and we know its mean and its variance. For instance, suppose I have a deck of 52 cards, and I tell you to pick a card "at random". If you had no prior knowledge as to how I would choose my card, what probability of selection would you assign ...


1

Denote $\mathbf Z_n=(Z_1,Z_2,...,Z_n)'$ then $$\mathbf S_n=A\mathbf X_n$$ where $A$ is the lower triangular part of $\mathbf 1_n\mathbf 1_n'$. So, by definition $\mathbf S_n$ is jointly normal with zero mean and covariance matrix $\Sigma=AA'$ s.t. $\Sigma_{i,j}=i\wedge j$. Consequently, $$\widetilde{\mathbf S}_n\equiv diag[\epsilon]\times\mathbf S_n ...


0

An elementary proof need not be easier, it just requires less theorems and techniques. Let's see an elementary proof: $$X_n \sim N(0, \sigma_n) \Rightarrow P(X_n \in A) = \int_A \frac{1}{(2 \pi \sigma_n^2)^{1/2}} \exp\bigg\{-\frac{x^2}{2 \sigma_n^2}\bigg\}\, dx$$ therefore since $X_n \to X$ everywhere, you have that $P_n \xrightarrow[]{*} \bar{P}$ (where ...


0

If you are trying to learn original hyper-parameters that e.g. make your new hyper-parameters positive (e.g. through exponential applied to the given original hyper-parameters, where the original can be positive or negative), then the only thing you need to do is compute the new gradient/Hessian/whatever derivatives you want with respect to the original ...


2

There is an easier approach. Note that $\Phi(x)$ is a continuous increasing function going from $0$ to $1$. Let $Y=\Phi(X)$, so $Y$ is in the interval $(0,1)$. Then $$F(y)=\Pr(Y \le y) =\Pr(\Phi(X) \le y) = y$$ so $f(y)=1$ when $y \in (0,1)$ and $E[Y]=\int_0^1 y \,f(y) \, dy =\frac12$. This works for any continuous distribution.


1

It is a good idea to plug in $f(x)$. $$\mathbb{E}[\sin^2 X]=\sqrt{\frac{2}{\pi}}\int_{0}^{+\infty}\sin^2(x) e^{-x^2/2}\,dx =\color{red}{\frac{\sinh 1}{\exp 1}}=0.432332358\ldots.$$ The middle equality follows from expanding $\sin^2(x)=\frac{1-\cos(2x)}{2}$ as a Taylor series, then integrating termwise. As an alternative, you may prove that: $$ ...


1

In general, you are asking the following question. If we know the distribution of $Y$, what is $E(Y|Y > x).$ The fundamental idea is to take the part of the distribution of $Y$ to the right of $x,$ re-scale that part so that it sums (or integrates) to unity, and then find the mean of the new conditional distribution. Methods and results differ greatly ...


2

I can't say what went wrong in the computation of the variance, but $S_n$ is $B\bigl(n,\frac{1}{2}\bigr)$-distributed, and the variance of a $B(n,p)$ distribution is $n\cdot p(1-p)$, so here we have $\operatorname{Var} S_n = \frac{n}{4}$ and hence $\operatorname{Var} Z_n = \frac{1}{4n}$. For a $N(0,1)$-distribution, we have $$P(\lvert X\rvert \geqslant c) ...


1

Remember that the normal distribution is parametrized by its mean and variance which are usually denoted by $\mu$ and $\sigma^2$, respectively. The notation $f(x;\theta) $ generally means $ f $ is a function of $ x $ parametrized by $\theta $. $ N $ is referring to the normal density function.


0

Hints For (i), you have both $X,Y \sim \mathcal{N}(55, 1.25)$ and $X,Y$ are independent. Then, $$ \mathbb{P}[X > 56, Y > 56] = \mathbb{P}[X > 56] \times \mathbb{P}[Y > 56] $$ but $X,Y$ have the same distribution, can you take it from here? For (ii), look at $\mathbb{P}[|X-Y| > 1]$ using the joint distribution between $X$ and $Y$, or ...


1

I have good news for you: You are confused. Firstly, what is a sample? You have $X_1,\ldots,X_n$. That is one sample, not a set of $n$ samples. If you base a histogram on $X_1,\ldots,X_n$, and $n$ is large, then with high probability, the histogram will look like the density of the "original" distribution. However, that is only one sample, and it is the ...


1

That is exactly it! You said, "it appears averaging any random set of samples from an entire given distribution and then taking a histogram of that set just results in that initial distribution". The law of large numbers is the assumption that the limit of those sample means and variance converges to the actual mean and variance respectively, CLT is that ...


1

Your post is a bit confusing--to me, at least. The original question alone is interesting. Along with just the assumption that scores are normal, it contains enough information to give a good answer. (It is not clear whether items 1-3 are part of the question, or your own research towards an answer. They are also worthwhile information for many purposes, but ...


1

At one point you say $Z=X^2+Y^2$ and at another you say $Z=\dfrac{X^2+Y^2}2$. I'll go with the latter version. Use polar coordinates: \begin{align} \Pr(Z>z) & = \iint\limits_{(x,y)\,:\,\frac{x^2+y^2}2>z} \frac 1 {2\pi} e^{-(x^2+y^2)/2} \,d(x,y) \\[10pt] & = \frac 1 {2\pi} \int_0^{2\pi} \int_{\sqrt{2z}}^\infty e^{-r^2/2} \, r\, dr\, d\theta. ...


1

$$\mathbb{P}[X^2+Y^2\leq R^2]=\frac{1}{2\pi}\iint_{x^2+y^2\leq r^2}e^{-\frac{x^2+y^2}{2}}\,dx\,dy=\frac{1}{2\pi}\int_{0}^{2\pi}\int_{0}^{R}\rho e^{-\rho^2/2}\,d\rho\,d\theta$$ so: $$\mathbb{P}[X^2+Y^2\leq R^2]=1-e^{-R^2/2}$$ and $X^2+Y^2$ has an exponential distribution with parameter $\lambda=\frac{1}{2}$.


0

This is a chi distribution with $n$ degrees of freedom, so called because it is the square root of a chi-square distribution. Its density for $x\gt 0$ is $$\frac{2^{1-n/2}x^{n-1}e^{-x^2/2}}{\Gamma(n/2)}.$$


0

The latter definition is a PDF. It's the continuous version of the distribution $b_j$ of observations for state $j$. The discussion at the bottom of page 7 (and continues to top of page 8) says: There are two forms of output distributions we will consider. The first is a discrete observation assumption where we assume that an observation is one of $L$ ...


1

The mutual information of two normal variables with correlation factor $\rho$ is $$ I(X;Y) = -\frac{1}{2} \log(1-\rho^2) \tag{1}$$ See proof in any Information Theory textbook - eg. In our case, letting $X=A+B$ and $Y=A+C$, we have $\sigma_X^2 = \sigma_A^2+\sigma_B^2$, $\sigma_Y^2 = \sigma_A^2+\sigma_C^2$, $\sigma_{XY}=E(X Y) = \sigma_A^2$. Hence $$ ...



Top 50 recent answers are included