New answers tagged

0

If $X$ and $Y$ are independent Poisson with parameters $\lambda, \mu$ respectively, and $z \ge 0$, $$P(X - Y = z) = e^{-\lambda-\mu} (\lambda/\mu)^{z/2} I_{|z|}(2\sqrt{\lambda \mu})$$ where $I_z$ is the modified Bessel function of the first kind of order $z$. For $z\ge 0$ this can be obtained from the series $$ e^{-\lambda-\mu} \sum_{y=0}^\infty \dfrac{\...


1

Let's first address the question of finding solutions to the equation $$\Pr[X \le x] = x, \quad 0 \le x \le 1,$$ for $X \sim \operatorname{NormalDistribution}(\mu,\sigma^2)$, or equivalently, $$\Phi\left(\frac{x - \mu}{\sigma}\right) = x$$ where $\Phi$ is the CDF of the standard normal distribution. This of course has no closed form solution for $x$, but ...


1

For better understanding I would like the to rephrase the question. If two random variables have normal marginal densities are they jointly normally distributed? The answer is NO. This enlightening counterexample appears in the famous book "Counterexamples in Probability and Statistics" by Joseph Romano and Andrew Spiegel. Let $$g(x,y)=(1/2\pi)\exp\...


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It is certainly not true. Suppose $X\sim N(0,\sigma^2)$ and $$ Y = \begin{cases} \phantom{-}X & \text{if } -c<X<c, \\ -X & \text{otherwise.} \end{cases} $$ Then $Y\sim N(0,\sigma^2)$. The correlation between $X$ and $Y$ depends on $c$, and for one special value of $c$ it is $0$, and for all others it is not. (That it is not normally ...


1

No, for an extreme example, take $X \sim N(0, 1)$, and consider the random vector $(X, -X)$, clearly, $X$ and $-X$ are (perfectly) correlated while they are not jointly normally distributed. A simple way to verify this is by noting $X + (-X) \equiv 0$ is not normally distributed. On the other hand, given $(X, Y)$ are jointly normal, then $X + Y$ must also ...


1

Setup hint: Start with finding a suitable matrix $A^{2\times2}$ such that $AA^T=\Sigma$. Then write $X=AU+\mu$ where $U=(U_1,U_2)^T$ is a random vector such that $U_1,U_2$ are iid and have standard normal distribution. Then $X$ has the distribution that you mention and:$$\Pr\left(X\in C\right)=\Pr\left(\left(X-\mu\right)^{T}\left(X-\mu\right)\leq R^{2}\...


3

$X_i-\mu$ is a standard normal random variable, hence $$ \mathbb{E}[|X_i-\mu|]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}|x|e^{-\frac{x^2}{2}}\;dx=\sqrt{\frac{2}{\pi}}\int_0^{\infty}xe^{-\frac{x^2}{2}}\;dx=\sqrt{\frac{2}{\pi}}$$ Therefore the sum is equal to $$\frac{\sqrt{\pi}}{2n}\cdot n\sqrt{\frac{2}{\pi}}=\frac{1}{\sqrt{2}}$$ for all $n$. Even without ...


0

Let $z=\frac{x-\mu}{\sigma}$, so your first equation gives $$z = \pm\sqrt{-\log(2\pi\sigma^2)} \tag{1} $$ and your second equation gives $$\mu=\Phi(z)-\sigma\,z\tag{2} $$ where $\Phi$ is the standard normal CDF. Substituting (1) into (2), we obtain (two branches of) $\mu$ as a function of $\sigma$: $$\mu_a (\sigma) = \Phi\left(\sqrt{-\log(2\pi\sigma^2)}\...


2

If $X\sim \chi^2_n$ then $\Pr(X\ge 0) =1$. But $\Pr(X_1^2+X_2^2 - X_3^2 \ge 0 ) <1$, because the probability that $X_1^2+X_2^2$ is small and $X_3^2$ is large is not $0$.


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Short answer If you have a calculator, the invNorm function on the TI-84 or similar commands on other calculators will take a probability $p$ as input and output the upper bound of the interval (starting from the left tail) by the number of standard deviations away it is from the mean (a negative output is left of the mean and a positive output is right of ...


1

The question asks you to verify that $E[\bar{X}^2]\neq \mu^2$. The expansion $E[\bar{X}^2]$ gives: $$ E\Big[\bar{X}^2\Big]=\frac{1}{n^2}E\Big[\Big(\sum_iX_i\Big)^2\Big]=\frac{1}{n^2}E\Big[\sum_iX_i^2+2\sum_{i<j}X_iX_j\Big]. $$ Independence only gives $E[X_iX_j]=E(X_i)E(X_j)=\mu^2$ for $i<j$. Independence doesn't make expectations of the cross terms go ...


1

$$ E\bar{X}_n^2 = var(\bar{X}_n)+E^2\bar{X}_n = 1/n+\mu^2 >\mu^2. $$ So the simplest unbiased estimator would be $$ \bar{X}^2_n-1/n $$


2

Assuming $|[X]|$ means absolute value of floor of $X$, that is $1$ if and only if $X \in [-1,0) \cup [1,2)$. So $$P(X<0||[X]|=1) = \dfrac{P(X \in [-1,0])}{P(X \in [-1,0)) + P(X \in [1,2))}$$ Also useful will be symmetry, so $P(X \in [-1,0)) = P(X \in [0,1))$.


2

We start with the Pearson differential equation: $$f'(x)=-\frac{\left(a_1 x+a_0\right) }{b_2 x^2+b_1 x+b_0}f(x),$$ Define $g(x)=b_2 x^2+b_1 x+b_0$ (using a trick in Diaconis et al.(1991)). Consider (f g)'(x), also written $(f(x) g(x))'=f'(x) g(x) +f(x) g'(x)$. We have $$(f g)'(x)=(-a_0 + b_1 - a_1 x + 2 b_2 x) f(x)$$ We assume that the distribution has ...


1

The new density is again truncated normal at $t$ with new parameters $\frac{\sigma^2\mu_0+\sigma_0^2 x}{\sigma^2+\sigma_0^2},\sigma^2\sigma_0^2/(\sigma^2+\sigma_0^2)$. The new normalising constant is $$\Phi\left( \frac{t-\frac{\sigma^2\mu_0+\sigma_0^2 x}{\sigma^2+\sigma_0^2}}{\sigma^2\sigma_0^2/(\sigma^2+\sigma_0^2)}\right )$$ So the interesting thing ...


1

$M$ and $x$ are independent? $\int_0^1Mx=M\int_0^1x=M/2$ so $I\sim N(0,1/4)$.


0

Standard deviation and mean give a probability mass distribution, which is continuous. That means that you can't talk about discrete things (like saying "there are exactly $x$ elements greater than some cutoff $n$) just on the basis of mean and standard deviation. As such, you can't say with certainty that "there are no elements less than the lower bound $a$ ...


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I doubt there can be a definite answer unless we find the person who wrote it there and asked him or her. I am guessing that $G$ refers to Gaussian. Could $\mu+C(\sigma)$ be a confidence interval? and the person might want to know what is the image of this interval under exponential map?


2

If you have seen the Moment Generating Function, we can solve this by using it. We have $$ \mathbf{E}(e^{tZ}) = e^{\frac{t^2}{2}}.$$ By equating the coefficient of $t^4$, we have $$\frac1{24}\mathbf{E}(Z^4) = \frac18.$$ This gives $\mathbf{E}(Z^4)=3$.


2

$$\mathbb{E}[Z^4]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}z^4e^{\large-\frac{z^2}{2}}dz=\frac{2}{\sqrt{2\pi}}\int_{0}^{\infty}z^4e^{\large-\frac{z^2}{2}}dz$$ set $\frac{z^2}{2}=u$, we have $z\,dz=du$ and $z^3=2\sqrt{2}u\sqrt{u}$, thus $$\mathbb{E}[Z^4]=\frac{4}{\sqrt{\pi}}\int_{0}^{\infty}u\sqrt{u}\,e^{-u}du=\frac{4}{\sqrt{\pi}}\Gamma\left(\frac{5}{2}\...


2

Integrate $\int_0^\infty x^4 e^{-x^2/2}\; dx$ by parts using $u = x^3$, $dv = x e^{-x^2/2}\; dx$. Or change variables with $x = \sqrt{t}$ and use properties of the Gamma function.


1

The variable $X$ is not independent of variable $X$. Thus the correlation coefficient is unequal to zero. The variance of the sum of two dependent, normally distributed random variables (X,Y) is $\sigma_{X+Y}^2=\sigma_X^2+\sigma_Y^2+2\rho \sigma_X\cdot \sigma_Y$ Setting Y equal to X. $\sigma_{X+X}^2=\sigma_X^2+\sigma_X^2+2\rho_X \cdot \sigma_X\cdot \...


1

If $X \sim N(\mu_X ,\sigma_X)$ and $Y \sim N(\mu_Y ,\sigma_Y)$ then $$X + Y \sim N(\mu_X + \mu_Y,\; \sigma_X^2 + \sigma_Y^2 + 2\sigma_{X,Y})$$ We have $$X + X \sim N(\mu_X + \mu_X,\; \sigma_X^2 + \sigma_X^2 + 2\sigma_X^2)$$ Indeed $$2X \sim N(2\mu_X ,\; 4\sigma_X^2 )$$ Note $$2\sigma_{X,Y}=2\operatorname{Cov}(X,Y) $$ $$2\sigma_{X,X}=2\operatorname{Cov}(X,...


0

$$\mathbb{P} (Y<y)=\mathbb{P} ({{e}^{X}}<y)=\mathbb{P} (X<\ln y)=\frac{1}{\sigma \sqrt{2\pi }}\int_{-\infty }^{\ln y}{\exp \left( -\frac{{{(x-\mu )}^{2}}}{2{{\sigma }^{2}}} \right)}\,dx$$ set $u=e^x$, or $x=\ln u\,$, we have $$\mathbb{P} (Y<y)=\frac{1}{\sigma \sqrt{2\pi }}\int_{0 }^{ y}\frac{1}{u}{\exp \left( -\frac{{{(\ln u-\mu )}^{2}}}{2{{\...


1

I consider the two variable case and take into account that the means of $X_1$ and $X_2$ are 0. The pdf of the univariate untruncated distribution of $X_1$ is $f_{X_1}(x_1)=\frac{1}{\sqrt{2\cdot \pi \sigma_1^2}}\cdot e^{-\frac12 \left[\frac{x_1^2}{\sigma_1 ^2}\right]} $ The cdf of the univariate untruncated distribution of $X_1$ is $F_{X_1}(x_1)=\int_{-\...


1

The two word answer is "polar coordinates". In more detail, let $f:S^{n-1}\to\Bbb R$ be a continuous function. Then $$ \eqalign{ \Bbb E[f(X)] &=\int_{\Bbb R^n}f(x_1/z,\ldots,x_n/z)(2\pi)^{-n/2}e^{-z^2/2}\,dx_1\cdots dx_n\cr &=(2\pi)^{-n/2}\int_0^\infty\left[\int_{S^{n-1}} f(u)\,\sigma_{n-1}(du)\right]e^{-r^2/2}r^{n-1}\,dr\cr &=c_n\int_{S^{n-1}} ...


1

Yes: the normal distribution cares only about the distance between the mean and the point. Note that in $g(x \mid x_p)$, the mean is $x_p$ now, not $x$, so the distance between the mean in the two cases are the same and the variance in the two cases are the same. (In this symmetric situation, the algorithm is commonly called just Metropolis, rather than ...


1

Let $X_1,..,X_n$ i.i.d $\mathcal{N}(\mu, \sigma^2)$ each one,hence by definition $$ \sum_{i=1}^n\left(\frac{X_i - \mu}{\sigma} \right)^2 \sim \chi^2_n \, . $$ Now, in order to estimate the variance of the Normal distribution, you are using (variation of) $$ S^2 = \frac{1}{n}\sum_{i=1}^n(X_i - \bar{X})^2, $$ thus $$ \sigma^2 S^2 = \frac{\sigma^2}{n}\sum_{i=...


1

The pdf of $Y$ is obtained by taking the joint pdf of $(X,Y)$ and marginalizing $X$ out. That is: $$f_Y(y)=\int_{-\infty}^\infty f_{X,Y}(x,y) dx.$$ The joint pdf of $(X,Y)$ is the product of the conditional pdf $f_{Y|X}(y|x)$ and the pdf of $X$, $f_X$. (If this seems weird to you, it is basically analogous to the familiar identity $P(A \cap B)=P(A \mid B) ...


2

Let's define $$ f_z(\beta):=\int_{-\infty}^z \phi(x)\,\Phi(\beta\, x)\,dx $$ For $\beta=1$ $$ f_z(1)=\int_{-\infty}^z \phi(x)\,\Phi( x)\,dx=\int_{-\infty}^z \Phi( x)\,\Phi(x)'\,\,dx=\frac12 \Phi(z)^2 $$ For any other $\beta$, I am almost sure that there is not an elementary expression. One can try some manipulation to put it in a different form, but still ...


1

The expected value of a normal distribution is the mean, in your case $\mu$, because the normal is symmetrical from let to right. For any probability density function $f(x)$ this can be calculated by integrating $x\times f(x)$ across its domain. The normal is defined as having the pdf: $$f(x)=\frac{1}{\sqrt{2\pi}}\exp{(\frac{-x^2}{2})}$$ The mean of that ...


1

You can describe the number of wins after $n$ games with $X_n \sim Bin(0.5, n)$, i.e. a binomial distribution with probability of success in a single event $p = 0.5$. The number of losses is always going to be $n - X_n$ (assuming no draws). So you want to find the probability $P(X_n - (n - X_n) \ge 10) = P(2X_n - n \ge 10) = P(X_n \ge \frac{n + 10}{2})$ ...


0

Note that $h'(x)+1=\frac{\phi(z)}{\Phi(x)^2}f(x)$, where $f(x)=\frac{\phi'(x)}{\phi(x)}\Phi(x)-\phi(x)+\frac{\Phi(x)^2}{\phi(x)}$. Also knowing that $\phi'(x)=-x\phi(x)$, we can show that $f'(x)=\Phi(x)\left(1+\frac{x\Phi(x)}{\phi(x)}\right)$. Since, it is already shown that $h(x)+x\ge 0$, we know $\left(1+\frac{x\Phi(x)}{\phi(x)}\right)\ge 0$. Thus, $f(x)&...


1

Some confusion here: $X$ is a count, and .33 is a proportion. You are trying to use the normal approximation to the binomial distribution. The binomial random variable $X$ is a count. Specifically, you have the distribution Binom(n = 500, p = .38), You're right about n being large enough to use a normal approximation. The average number of voters out of 500 ...


6

The condition $(X+2)(X−4) \leq 0$ is equivalent to $-2 \leq X \leq 4$. Therefore, $$ P[(X+2)(X−4) \leq 0] = P[X \leq 4] - P[X < -2]. $$


0

$$\max \{\mu +\sigma x\,,\,\,0\}=\left\{ \begin{align} & \mu +\sigma x\,\,\,\,,\,\,\,\,x>-\frac{\mu }{\sigma } \\ & 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,\,x\le -\frac{\mu }{\sigma } \\ \end{align} \right.$$ $$\mathbb{E}[E]=\int_{-\infty }^{+\infty }{\max \{\mu +\sigma x\,,\,\,0\}}\,\varphi (x)\,dx=\int_{-\frac{\mu }{\sigma }}^{+\...


3

The $k$th moment of a Binomial$(n,p)$ random variable $X$ is $$\mathbb{E}(X^k)=\sum_{j=1}^k{k\brace j}j!{n\choose j}p^j,\tag1$$ where ${k\brace j}$ is the Stirling number of the second kind. For large $n$, we have $j!{n\choose j}\approx n^j-{j\choose 2}n^{j-1}$. Taking the two leading terms in (1) gives the approximation $$\mathbb{E}(X^k)\approx \left[n^...


1

The cumulative distribution function for $Y$ is: $$ F_Y(y)=\operatorname{Prob}(Y<y)=\operatorname{Prob}(-\sqrt{y}<z<\sqrt{y})=F_Z(\sqrt{y})-F_Z(-\sqrt{y})=2F_Z(\sqrt{y})-1 $$ Now to get the density differentiate with respect to $y$, and of course to differentiate $2F_Z(\sqrt{y})$ you use the chain rule and the fact that the derivative of $F_Z(x)$ ...


0

Recall two useful facts about (generalized) normal random variables: If two random variables $u,v$ have a joint normal distribution and $u^{\prime},v^{\prime}$ are random variables obtained by applying a $2\times 2$ matrix to the random vector $(u,v)^T$, then $u^{\prime}$ and $v^{\prime}$ also have a joint normal distribution. If two random variables $X,Y$ ...


0

If we assume perfect normality, then we want $$\Pr(Z\gt (500-500)/50),$$ that is, $\Pr(Z\gt 0)$, where $Z$ is standard normal. This is exactly $0.5$, since $\Pr(Z=0)=0$. The problem said account amounts are normally distributed, so $0.5$ is the technically correct answer to the problem as stated. But the proposed answer of $49.6\%$ indicates that they are ...


3

The maximum possible mark of $100$ is about $1.836$ standard deviation units above the mean. There is a probability of about $3.4\%$ that a normally distributed random variable is more than $1.836$ standard deviation units above the mean. So at the upper end of the range at least, the normal with the given mean and standard deviation gives a poor fit. ...


0

Let $X\sim\mbox{N}\left(\mu,\sigma^{2}\right)$. So as usual the PDF is given by $$f_{X}(a)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\left(\frac{a-\mu}{\sigma\sqrt{2}}\right)^{2}}.$$ The mean absolute deviation is \begin{alignat*}{1} \mathbf{E}\left[\left|X-\mu\right|\right] & =\int_{-\infty}^{\infty}\left|a-\mu\right|f_{X}(a)da\\ & =\int_{-\infty}^{\mu}\...


0

Hint: Rewrite $$\displaystyle \int_{-\infty}^{\infty} |x-\mu| \frac{1}{\sqrt{2\pi}\sigma} e^{-(x-\mu)^2/(2\sigma^2)}\,dx $$ as $$\displaystyle 2 \int_{\mu}^{\infty} \frac{1}{\sqrt{\pi}}\left(\frac{x-\mu}{\sqrt{2}\sigma}\right) e^{-((x-\mu)/(\sqrt{2}\sigma))^2}\,dx $$ and use fairly simple calculus to show this is $\sqrt{\dfrac{2}{\pi}}\sigma$


0

Here is a simplified example for $n=2$ which might help you to start out. Assume an archer is shooting an arrow onto an infinite target with center $(0,0)$. The arrow hits at the coordinates $(X,Y)$ where $X$ and $Y$ are independent and $X,Y\sim\mathcal{N}(0,1)$. It should be clear that $$ f_{X,Y}(x,y)=\frac{\exp(-1/2\cdot(x^2+y^2))}{2\pi} $$ since $X$ ...


0

If everything is independent, then this is no longer a two-point problem: the variables $x_i - y_i$ are independent $N(0,2)$. The expectation is not terribly hard to compute, see here. It is increasing (moreover, the distribution itself is increasing, in terms of stochastic order) and, as @Did commented, equivalent to $\sqrt{2n}$.


1

The average will converge a.s. to the expected value according to the strong law of large numbers. Let $\phi$ denote the PDF and let $\Phi$ denote the CDF of standard normal distribution. Then in general: $$\int_{a}^{b}u\phi\left(u\right)du=\frac{1}{\sqrt{2\pi}}\int_{a}^{b}ue^{-\frac{1}{2}u^{2}}du=\left[-\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}u^{2}}\right]_{...


1

So you need to find $\lim_{N\to\infty} \dfrac{\sum_{i=1}^NX_i1_{(X_i>0)}}{\sum_{i=1}^N 1_{(X_i>0)}}$. Divide the numerator and denominator by $N$, then apply SLLN to see that the above limit is $\dfrac{E(X1_{(X>0)})}{P(X>0)}$. Now, $P(X>0)=1-\Phi(-1)=\Phi(1)$. Also, $E(X1_{(X>0)})=E((X-1)1_{(X-1>-1})+E(1_{(X-1>-1)})=E(Z1_{Z>-1)}+...


0

Each user gets a "quality" value between 0 and 1. The probability of getting a number is symmetric about $0.5$, and is given by a normal distribution, which defines the probability of being a given "distance" away from the mean. That "distance" is measured in standard deviations and called the $z$-value. The formula for the $z$-value of a given measurement $...


0

The normal distribution is a "bell shaped curve", that appears often when the variable you are measuring is a result of multiple effect which act more-or-less independently of each other. In a normal distribution more people will be clustered around the average, with a few above and below average. Also about 50% will be above average and 50% below average (...


1

My table, which may be very similar to yours, gives $\Pr(Z\le 0.67)\approx 0.7486$ and $\Pr(Z\le 0.68)\approx 0.7517$. To get a better approximation to the $z$ such that $\Pr(Z\le z)=0.75$, we do a linear interpolation. The number $0.7486$ is $0.0014$ short of $0.75$, while $0.7517$ is $0.0017$ above $0.75$. So a better estimate for the appropriate $z$ ...



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