New answers tagged

0

We have to show if $X\sim \mathcal N(\mu, \sigma^2)$ and $Z=\frac{X-\mu}{\sigma}\sim \mathcal N(0,1)$ then $P(X\leq w)=P(Z\leq \frac{w-\mu}{\sigma})$. $Z=\frac{X-\mu}{\sigma}\Rightarrow Z\cdot \sigma+\mu=X$ $P(X\leq w)=P(Z\cdot \sigma+\mu\leq w)=P(Z\cdot \sigma\leq w-\mu)=P(Z\leq \frac{w-\mu}{\sigma})$.


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Of course, it depends on what high means when you say $\ldots$ "When $m$ and $V$ are very high, we get ridiculously high number." Nonetheless, I could not replicate your concerns. I did two simulations ($N = 10\,000$ each) using MATLAB, one with $m$ varying from $-70$ to $70$, and another with $v$ varying from $0.5$ to $30$. The two figures show the ...


1

(a) Yes, the sum of $n$ iid Geometric Distributed Random Variables has a Negative Binomial distribution, and that is the right moment generating function for the given one. (b) is okay, and see also (d) below. (c) Well, $\mathsf e^{4t/3}$ is the moment generating function for a Degenerate Distribution. In this case ...


0

If $A$ is a matrix with $\left(U,V\right)^{T}=A\left(X,Y\right)^{T}$ then $\mathbb{E}\left(U,V\right)^{T}=A\mathbb{E}\left(X,Y\right)^{T}$. If $\Sigma$ is the covariancematrix of random vector $\left(X,Y\right)^{T}$ then $A\Sigma A^{T}$ is the covariance matrix of random vector $\left(U,V\right)^{T}$. $\left(U,V\right)^{T}$ has normal distribution, so its ...


1

In linear regression with Gaussian (and heteroscedastic) noise, our model assumes that for $n$ observations of data, for each $i \in [n]$, $$Y_i = \beta X_i + \epsilon_i,$$ where $\epsilon_i$ is our ERROR term for the $i$th observation (note that residual $e_i$ is an estimator of $\epsilon_i$) Such that $\epsilon_i \sim N(0,\sigma^2_i).$ NID means ...


1

To find expressions like $P(X\geq x)$ where $X$ has normal distribution observe that $$P(X\geq x)=P(\sigma U+\mu\geq x)=P\left(U\geq\frac{x-\mu}{\sigma}\right)$$ where $U$ has standard normal distribution. So actually for $z=\frac{x-\mu}{\sigma}$:$$P(X\geq x)=1-\Phi\left(z\right)$$ You can find this $z$ by substituting. From here tables come in. Also ...


1

For every $x\in\mathbb R$:$$\Phi\left(-x\right)=1-\Phi\left(x\right)$$ Substituting $x=z_a$ leads to$$\Phi\left(-z_{a}\right)=1-\Phi\left(z_a\right)=1-a=\Phi\left(z_{1-a}\right)$$ Conclusion: $$-z_a=z_{1-a}$$ Or equivalently (but a bit nicer): $$z_a+z_{1-a}=0$$


1

I think it must be proved that $\mu=\mathcal N(0,\sigma^2)$ but for convenience I will also preassume that $\sigma=1$ If $\phi$ denotes the characteristic function then:$$\phi(t)=\phi\left(\frac{t}{\sqrt2}\right)^2$$ Note that this can be repeated to arrive at $\phi(t)=\phi(\frac{t}2)^4$ and can be repeated again. Actually with this it can be shown that ...


-1

Hint: without relying on the CLT, you can use characteristic functions. Let $f$ be the characteristic function of the law $\mu$. You can easily show that $f$ satisfies this functional equation which can be solved to find that $f$ is the characteristic function of $\mathcal{N}(0,1)$.


3

Just normalized X to the N(0,1) and used a Z table! $$P(X<7)=P(X-3/\sqrt4<7-3/\sqrt4)=P(Z< \frac{7-3}{\sqrt4})$$ Just use a Z table now for this number.Same for the others, let me know if this helped.


3

You are partially correct. You need to integrate the PDF so you get the CDF, then it's: $$P(X<7)=F(7)$$and$$P(X≥7)=1-F(7)$$ Note: $F$ here is the CDF. Normally a lowercase $f$ indicates the PDF. As you have a normal distribution though, you can get around integrating that expression, and instead use the table for the normalized version as Steven Ramos ...


3

If $f(x)$ is the probability density function, then $P(X<X_0)$ is an integral of $f(x)$: $$P(X<X_0) = \int_{-\infty}^{X_0} f(x)\,dx$$ But for the probability density function $f$ for the normal distribution this integral is difficult to express in terms of commonly used functions. It is usually tabulated or calculated numerically. For the other ...


3

Such $k$ does not exist. We have that \begin{align*} |X+k|-|X-k| &=\frac{(|X+k|-|X-k|)(|X+k|+|X-k|)}{|X+k|+|X-k|}\\ &=\frac{|X+k|^2-|X-k|^2}{|X+k|+|X-k|}\\ &=\frac{4kX}{|X+k|+|X-k|}. \end{align*} Hence, $|X+k|-|X-k|>0$ if and only if $X>0$ when $k>0$ and $|X+k|-|X-k|>0$ if and only if $X<0$ when $k<0$. The probability does not ...


2

Hint: If $k>0$ then:$$|x-k|<|x+k|\iff x>0$$ In words: to be more close to $k>0$ instead of $-k$ (on the same distance of $0$ as $k$, but on the other side) it is necessary and sufficient that $x>0$. edit: Consequently $P(|X-k|<|X+k|)=P(X>0)=0.5\neq0.7$. Conclusion: $k$ cannot be positive. Likewise it can be shown that $k$ cannot ...


0

As discussed in the comments, I do not believe this problem can be solved as stated. A sample is just a sample and might well fail to perfectly respect the properties of the underlying distribution. To be more precise: Let $p$ be the probability that a given unit is defective. using the given normal distribution we see that $$p\sim 0.01513014$$ (Note: ...


0

You can prove this by induction on $n$, starting at $n=0$. Hint: If $N\ge0$ then integration by parts shows that $$\int_{-\infty}^\infty t^Ne^{-t^2/2}\,dt=\frac 1{N+1}\int_{-\infty}^\infty t^{N+2}e^{-t^2/2}\,dt.$$ Now that that's been done for you, you should derive a similar formula for $\Bbb E[|X|^{2n+1}]$.


1

$$\int_{-\infty}^{\infty}x^{2n}e^{-\frac{1}{2}x^{2}}dx=2\int_{0}^{\infty}x^{2n}e^{-\frac{1}{2}x^{2}}dx=-2\int_{0}^{\infty} x^{2n-1}de^{-\frac{1}{2}x^{2}}=$$$$2\int_{0}^{\infty}e^{-\frac{1}{2}x^{2}}dx^{2n-1}-2\left[x^{2n-1}e^{-\frac{1}{2}x^{2}}\right]_{0}^{\infty}=2\left(2n-1\right)\int_{0}^{\infty}x^{2n-2}e^{-\frac{1}{2}x^{2}}dx$$


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Hint : Do a proof by induction using integration by parts by writing: $x^{2n+2} \exp(-\frac{x^2}{2}) = (-x^{2n+1})(-x\exp(-\frac{x^2}{2}))$


0

Answer on 15. (A) If $x>0$ then inequality $\frac53x<x$ would allow us to conclude that $$\frac53<1$$ (dividing by positive $x$ is permitted and does not change the sign) which is not true. So it cannot have positive solutions. (B) We can rewrite this as $$x(x-1)(x+1)<0$$ Then $x$ is a solution of it if it makes an odd number of these $3$ ...


2

Partial answer: As said, before extending the work, lets work out the ordinary case first. Let $X_1 = e^{\sigma Z_1}, X_2 = e^{\sigma Z_2}$ where $Z_1, Z_2$ are i.i.d. standard normal. So $$ \begin{align} &~ E[\min\{e^{\sigma Z_1}, e^{\sigma Z_2}\}] \\ =&~ E[e^{\sigma Z_1}\mathbf{1}_{Z_1 < Z_2}] + E[e^{\sigma Z_2}\mathbf{1}_{Z_2 < Z_1}]\\ ...


1

This is easily shown to be impossible for any real-valued random variable $X$: if we require $$Y = e^X \sim \operatorname{Normal}(\mu,\sigma^2),$$ then there must be some value of $X$ for which $e^X < 0$. But this is impossible if $X \in \mathbb R$.


2

Here is the current edition of the question: If $X$ is a symmetric $n$-dimensional random vector with mean $0$ then is it true that: \begin{align*} & X \text{ follows a multivariate normal law} \\ & \text{iff} \\ & \|X\| \text{is a chi random variable with $n$ degrees of freedom?} \end{align*} Let $X=(X_1,\ldots,X_n)$. As phrased above, ...


1

HINT: You made a mistake in this step: $$P((y-2)^2>a^2z)=0.9 \implies P\left(3\frac{(y-2)}{3}>a\sqrt{\frac{z}{7}}\sqrt{7}\right)=0.9$$ What you should have found was: $$P((y-2)^2>a^2z)=0.9 \implies P\left(3\left|\frac{(y-2)}{3}\right|>\left|a\right|\sqrt{\frac{z}{7}}\sqrt{7}\right)=0.9$$ If we simplify this and assume $a\geq 0$, then we get: ...


0

In Maple, you can do it like this (which uses the Remez algorithm): with(numapprox): Digits:= 20: Fapp:= minimax(F,x=0..7,[7,6],1,'maxerr'); Fapp := ...


1

HINT Required rational approximation for $\operatorname{erfc}$ function can be constructed analytically, on the basis of known Maclaurin series $$\sqrt\pi ze^{z^2}\operatorname{erfc}z = v,$$ ...


2

This isn't a definitive answer, but hopefully points you in the right direction. The short of it is, I don't know how the coefficients were derived, but it is likely related to some Pade, Remez, interpolating, or least-squares approximant. Read on for details. The function we're talking about is $$ Q(x) = \frac{1}{\sqrt{2\pi}}\int_{x}^\infty e^{-t^2/2}\;dt ...


0

The definition of an error function is: $$\text{erf}(x) = \frac{\sqrt{2}}{\pi}\int_{0}^x e^{-s^2}ds$$


1

Given a uniform prior and (independent) observations from a Normal distribution then the resulting posterior is a truncated normal distribution. However, in this case the observations are drawn from a truncated prior which makes it more complicated. First, you can 'ignore' the integral in the denominator since this is just a constant assuring that the ...


2

Yes, that's correct. If $X \sim N(0,\sigma^2)$ is Gaussian with mean $0$ and variance $\sigma^2$, then the random variable $Y:= X/\sigma$ is also Gaussian and $$\mathbb{E}(Y) = \frac{1}{\sigma} \mathbb{E}(X)=0$$ and $$\mathbb{E}(Y) = \frac{1}{\sigma^2} \mathbb{E}(X^2)=1.$$ Hence, $Y \sim N(0,1)$. This shows $$\underbrace{X}_{\sim N(0,\sigma^2)} = ...


0

For now, I am going to omit the $n$ subscripts, for it seems that they are not involved in the resulting expressions. Suppose $X_{m\times p}, \beta_{p\times 1}$, hence $\mu_{m\times 1}, M_{m\times m}$. I believe you meant $\epsilon \sim N_m(0, I)$ otherwise $Z$ would not be a univariate normal. In that case, let $$W = ...


1

Well, correlated or not, the distribution of X1+X2 is going to be normal, so I don't see any big difference. Only the variance of this is going to change according to the correlation.


1

Well, one interpretation is that the thing with z score equal to 3.4 is in the $99.96631\%$ percentile. In other words, there is a $0.9996631$ that other possible values are lower than this one. Another way to look at this is to say that there is a $0.0003369293$ to observe something as or more extreme than this one. Remember, the normal distribution is ...


2

Some of the following conditions can be left out, but let's keep it easy. Let $F$ as CDF of random variable $X$ be continuous, strictly increasing and taking values in $(0,1)$. Then $F:\mathbb R\to (0,1)$ is a bijection, so has an inverse. For $u\in(0,1)$ we find:$$F(X)\leq u\iff X\leq F^{-1}(u)$$ hence:$$P(F(X)\leq u)= P(X\leq F^{-1}(u))=F(F^{-1}(u))=u$$ ...


2

If $f_{X\mid Y}(x\mid y)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(x-y)^2}$ then $$\begin{align}\mathsf E(\mathsf e^{\imath\theta X}\mid Y) ~=~& \tfrac 1{\surd (2\pi)}\int_\Bbb R \mathsf e^{\imath\theta x-(x-Y)^2/2}\operatorname d x \\[1ex] ~=~& \tfrac 1{\surd (2\pi)}\int_\Bbb R \mathsf e^{(-x^2+2(\imath\theta+Y) x-Y^2)/2} \operatorname d x \\[1ex] ...


1

Notice that $Z \sim Norm(0,1)$ because $$0 = E(Y^*) = E[(1/2)X^* + (\sqrt{3}/2)Z] = 0 + (\sqrt{3}/2)E(Z)$$ and $$1 = V(Y^*) = V[(1/2)X^* + (\sqrt{3}/2)Z] = (1/4) + (3/4)V(Z).$$ Also, $X^* > 0$ and $Z > 0\,$ imply $(1/2)X^* + (\sqrt{3}/2)Z > 0.$ which takes care of the $\ge$ in the long display ending with $1/4.$ The following simulation in R may ...


0

A first point that is not very clear in your question: (1) are $f_1$ and $f_2$ random variables ? (2) or are they densities $f_i(x)=K_i exp(-(x-m_i)^2/(2 \sigma_i^2))$ ? In this case, it would mean a mixture of densities. My answer works for case (1) and the very particular case where $f_1$ and $f_2$ are random variables following a $N(0,1)$ ...


0

The bimodal data you have may be a mixture of normal components, but that mixture is not normal. Thus it may be enough for you to use ordinary tests of normality. Most software packages incorporate such tests. I will show you briefly how R statistical software can be used for this purpose. First, some bimodal data. I am simulating a mixture of two normal ...


0

Normal distributions are always unimodal. It looks like by "bimodal normal data" you mean a Gaussian Mixture Model (GMM) with 2 components (i.e. the PDF of the data is a convex combination of Gaussian PDF's). There are several techniques to estimate the number of components in a Gaussian Mixture Model -- Bayesian Information Criterion, Akiakie Information ...


0

So your integrand is (for fixed values of everything except $\Omega$): $$c_1 \exp(-c_2/\Omega) \Omega^{-m-1} \exp(-c_3 (\ln(\Omega)-\mu)^2).$$ To get a Gaussian-like thing, change variables to $u=\ln(\Omega)$, so $du=d \Omega/\Omega$ giving $$c_1 \exp(-c_2/e^u) e^{-mu} \exp(-c_3 (u-\mu)^2)$$ where the new limits are $-\infty$ to $\infty$. Now merge the ...


1

We want to calculate the probability that the population mean exceeds $\frac{6000}{100}=60$ $P(x>60)=P(\frac{x-\mu}{\sigma/\sqrt{n}}>\frac{60-\mu}{\sigma/\sqrt{n}})=P(\frac{x-\mu}{\sigma/\sqrt{n}}>\frac{60-48}{18/\sqrt{100}})=P(z>\frac{60-48}{18/\sqrt{100}})=P(z>6.667)=1-\Phi (6.667)=\Phi (-6.667)$, Note: z ~ N(0,1). So, by using the R ...


0

Let $X_1, X_2, \ldots, X_{100}$ represent the random baggage weights of each of the $n = 100$ passengers. Then for each $i = 1, \ldots, 100$, each $X_i$ might be assumed to follow a normal distribution with mean $\mu = 48$ and standard deviation $\sigma = 18$. If these weights are all independent, then their sum $$S = X_1 + X_2 + \cdots + X_{100} \sim ...


1

For ordinary simulation, a box-muller/polar marsaglia/ziggurat algorithm with cholesky decomposition allows you to simulate a bivariate normal very efficiently. However as you said you want to apply the central limit theorem, which seems that you want to simulate a lot of random variables and show the convergence. Assume you already have a random number ...


0

Arithmetic mean of sample size follows Normal Distribution with $\mu{}, \sigma{}^2/n$. Calculate the probability of $\bar{x} > 2$ given that $X \sim Normal(\mu = \mu, \sigma^2 = \sigma^2/n)$ (by standardizing it to standard normal).


0

Under $H_0$, $\bar{x}$ follows $N(0,\frac{\sigma^2}{n})=N(0,1)$. So $$P(\bar{x}>2)=1-\Phi(2)$$ where $\Phi$ is the cdf of a standard normal. There is no explicit way to compute $\Phi(2)$, so you have to look it up on a table.


1

It is a one-dimensional normal distribution. https://en.wikipedia.org/wiki/Cram%C3%A9r%E2%80%93Wold_theorem This follows from an alternative definition of multi-dimensional Gaussian distribution. EDIT: see the Wikipedia article (for future reference/anyone new to the question) https://en.wikipedia.org/wiki/Multivariate_normal_distribution#Definition


0

the first formula $$\int_{-\infty }^{\infty } e^{-\frac{x^2}{2}} \left(x f(x)-f'(x)\right)^2 \, dx<\infty$$ it is necesary condition for the converge of the integral using Mellin transfro it very easy to see that $$\int_{-\infty }^{\infty } e^{-\frac{x^2}{2}} \left(x f(x)-f'(x)\right) \, dx=0$$ XE f(x)=Sin(x) or E^-x gives Zero and whith any rational c ...


1

Suppose there are $n$ flowers. The number of flowers still blossoming at time $t$ is a random variable $X(t)$, equal to $$X(t)=\sum_{f=1}^n I(\text{flower $f$ is alive at time $t$})\tag1$$ where the indicator function $I(A)$ equals one if event $A$ is true, zero otherwise. The expectation of $X(t)$ is then $$E(X(t))=n P(\text{flower $1$ is alive at time ...


1

Hint:$$\Phi^{-1}(U)\leq x\iff U\leq \Phi(x)$$ Here $\Phi$ denotes the CDF corresponding with standard normal distribution.


0

Use the inverse transform method: With $X \sim N(0,1)$ and CDF $F$. $1.$ Generate a random number $u$ from $U$ in the interval $[0,1]$ $2.$ Compute the value $x$ such that $F(x) = u$ $3.$ Take $x$ to be the random number drawn from the distribution described by $F$


0

Hint: $\mathbb P(X>Y |Y>0) = \frac{ \mathbb P(X>Y,Y>0)}{ \mathbb P(Y>0)} = 2 \mathbb P(X>Y,Y>0)$ If $X$ and $Y$ are independent, then you can solve this easily. If not, I think you are not given enough information to solve it.



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