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I wonder if there is an analytical solution to $~\displaystyle\int_{-\infty}^{\infty}\frac{e^{-x^2}}{(x+a)^2+b}dx,~$ where $a, b>0$. Depends what you mean by that. If you are willing to accept imaginary error functions as being “analytical”, then the answer is yes. Alternatively, one might try and expand $\dfrac1{(x+a)^2+b}$ into a binomial series, ...


1

It is often seen that someone writes things like "$\theta\sim N(\mu,\sigma_0^2)$ and $\mu\sim N(0,\sigma_1^2)$" when they ought to write "$\theta\mid\mu\sim N(\mu,\sigma_0^2)$ and $\mu\sim N(0,\sigma_1^2)$", i.e. the conditional distribution of $\theta$ given $\mu$ is that normal distribution. Now think about the conditional distribution of $\theta-\mu$ ...


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Hint: define $Q(x) = x^2/2$. $$x\phi(x) = \frac 1{\sqrt{2\pi}}Q'(x)\exp(-Q(x)) $$


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It seems that you want the score to be a linear function of the rating. One first finds the slope from two given points, such as $(99,5 )$ and $(1,1)$: $$m=\frac{5-1}{99-1}$$ and then forms the equatin using the slope and one of the points, such as $(1,1)$: $$ s = m(r-1)+1 $$ Then round to the nearest multiple of $0.25$, by rounding $4s$ to the nearest ...


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Yes and no: If you successfully transform your data so that it looks normal, then the ANOVA will apply to the transformed mean only, not the "back-transformed mean" (this is called transformation bias). This caution will apply to any estimates of moments of a distribution. However, percentiles are not affected by such transformations. Since the median and ...


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Like Andre Nicolas suggested, you can check in a table: you are told that the value -10 represents the 2 percentile. A table would give you an approximate match between percentile and standard deviation. Without a table, a rule-of-thumb you can use is the $68-95-99.7$% rule, and the symmetry of the normal deviation. By symmetry, around $47.5$% of the values ...


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My Suggestion: $Var(d^TZ)=E(d^T(Z-\mu)(Z-\mu)^Td)=d^T\cdot E((Z-\mu)(Z-\mu)^T)\cdot d$


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Replace x with $x+1$ in the expression of $f_{_X}$, and factor all constants outside the integral. Then use $$\int_1^\infty\frac{\ln x}x\cdot\exp\bigg[\bigg(\frac{\mu-\ln x}a\bigg)^2~\bigg]dx~=~\frac{a^2}2\cdot\exp\bigg[-\bigg(\frac\mu a\bigg)^2~\bigg]+\dfrac{\sqrt\pi}2a\mu\cdot\bigg[1+\text{erf}\bigg(\dfrac\mu a\bigg)\bigg].$$ More to the point, ...


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It appears to me that what is meant is simply that although the probability that the parameter is near $0$ is high, the probability that it is exactly $0$ is $0$.


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Use the asymptotic expansion $$\Phi(x) = 1-\frac{e^{-x^2/2}}{\sqrt{2\pi}}\left(\frac{1}{x}+ \ldots\right)$$ If you can select $a_n$ and $b_n$ such that as $n \rightarrow \infty,$ $$\frac{e^{-(a_n x + b_n)^2/2}}{(a_n x +b_n)\sqrt{2\pi}}\sim \frac{e^{-x}}{n}, $$ then, as desired, $$\lim_{n \rightarrow \infty}\Phi(a_nx+b_n)^n =\lim_{n \rightarrow ...


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Given your problem as stated, it appears that assuming independence would be incorrect, since you don't know if the unknown location is essentially right next to your known location or on the other side of the globe. Since they are Gaussain, you can model the sum as a sum of correlated gaussian random variables, in which case, inference on the correlation ...


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On the first page of the cited document, $X_1$ and $X_2$ were previously defined to be two (distinct) independent, identically distributed random variables. For your purposes, the "identically distributed" part is not important, but the "independent" part is. On the second page, where $X_1$ and $X_2$ are considered to be normal variables, there's still the ...


1

By the strong law of large numbers we have $$ \frac{\sum_{j=1}^n X_j^2}{\sqrt{n}\sqrt{2n}}=\frac{1}{\sqrt{2}}\frac1n\sum_{j=1}^n X_j^2\to \frac{1}{\sqrt{2}}{\rm E}[X_1^2]=\sqrt{2} $$ almost surely, and hence the result follows from Slutsky's theorem.


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As clearly shown by Guy in his comment, the substitution $x=\alpha u$ leads to $$\int\limits_{-\infty}^{+\infty} \frac{n-1}{\alpha}\Phi\Big(\frac{x}{\alpha}\Big)^{n-2}\phi\Big(\frac{x}{\alpha}\Big)^2dx=(n-1)\int\limits_{-\infty}^{+\infty} \Phi(u)^{n-2}\phi(u)^2du$$ which is independant of $\alpha$. If you need to go further, could you clarify, at least for ...


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As eigenjohnson suggested, taking the logarithm is a reasonable way to deal with numbers of different scales (if none of the values are exactly equal to $0$). However, you want the numbers to remain in $[0,1]$, and logarithm will not do that. I suggest raising them to a small power $p>0$. This stretches the neighborhood of $0$: for example, here is ...


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First it'll be easier if we convert everything to ounces or pounds. Let's go for ounces so the mean is 20 oz, and the weight limit is 25 oz. If X is a normal variable with mean 20 oz and standard deviation 3 oz, then for one fish, the probability that it is heavy enough is P(X > 25). Excel 2010 has a command NORM.DIST which calculates probabilities with ...


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Method 1: Optionally apply a median filter to the histogram to reduce the noise. Then apply a threshold filter to find the location of the peaks. You can (optionally) use a weighted mean to get the locations as accurately as possible. After that, apply a least square match of a normal distribution with an unknown standard deviation. Method 2: Apply a ...


3

Let $h$ be an arbitrary bounded Borel function then $$\eqalign{ \mathbb{E}(h(Z))&=\frac{1}{\sqrt{2\pi}}\int_{|y|\leq a}h(y)e^{-y^2/2}dy+ \underbrace{\frac{1}{\sqrt{2\pi}}\int_{|y|>a}h(-y)e^{-y^2/2}dy}_{y\leftarrow-y}\cr &=\frac{1}{\sqrt{2\pi}}\int_{|y|\leq a}h(y)e^{-y^2/2}dy+ \frac{1}{\sqrt{2\pi}}\int_{|y|>a}h(y)e^{-y^2/2}dy\cr ...


3

$\textbf{My Question is:}$ (...) are $Y\mathbf1_{|Y|\le a}$ and $-Y\mathbf1_{|Y|>a }$ independent ? Of course not. Let $U=Y\mathbf1_{|Y|\le a}$ and $V=-Y\mathbf1_{|Y|>a }$, then $[U=0]=[|Y|\gt a]\cup[Y=0]$, $[V=0]=[|Y|\leqslant a]$, and $[U=V=0]=[Y=0]$. What does all this tell you about the possibility that $(U,V)$ is independent? An extended ...


0

Hint: the variance of $\sum a_iX_i = a\cdot X$ is $$ a\cdot \left( \begin{array}{ccc} \sigma^2 & 0& 0&\dots & 0 \\ 0 & \sigma^2 & 0 & \dots & 0 \\ \vdots &&&&\vdots \\ 0 & 0 & 0&\dots & \sigma^2 \end{array} \right) a $$


0

Assuming covariances are all 0's, $\mathbf{Var} T_1 = \frac{5 \sigma^2}{9}$ and $\mathbf{Var} T_2 = \frac{2 \sigma^2}{9}$.


1

Central Limit Theorem says $\sqrt{n}(\bar X-\mu) \stackrel{d}{\rightarrow} N(0,\sigma)$, as $n\rightarrow\infty$. Or you can think of $\bar X$ approaches $N\big(\mu,\frac{\sigma}{\sqrt{n}}\big)$, so the normal distribution shrinks around the mean $\mu$ and the distribution approaches zero rapidly outside any fixed interval around $\mu$. Your statement lost ...


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You need to look at a more precise statement of the central limit theorem. For i.i.d. random variables $X_i$ with mean $0$ it states that $\frac{X_1 + ... + X_n}{\sqrt{n}}$ (note the division by $\sqrt{n}$ rather than $n$) approaches a normal distribution with variance the variance of $X_i$. In particular, even if each $X_i$ is supported in $(a, b)$, this ...


1

Hint Assume without loss of generality that $\alpha \geq 0$. Then, $$X_t = \alpha W_t^2 + \beta t \geq \beta t.$$ In particular, there exists $x \in \mathbb{R}$ such that $\mathbb{P}(X_t<x)=0$. Can such a process be Gaussian?


0

Let us compute the value of $E((Z-z)^+)$ for every real number $z$, where $Z$ is standard normal, so that you will be able to deduce the exact result you are interested in. Consider the standard normal PDF $\varphi$ and the standard normal CDF $\Phi$, and note that, by definition, $$ E((Z-z)^+)=\int_z^\infty(u-z)\varphi(u)\mathrm du. $$ Since ...


1

If $X$, $Y$ and $Z$ are not independent, this means that the covariance matrix $\Sigma$ is not diagonal. The covariance matrix in your case is defined as follows: $$\Sigma = \left[\begin{array}{ccc}\mathbb{E}[(X-\mu_X)^2] & \mathbb{E}[(X-\mu_X)(Y-\mu_Y)] & \mathbb{E}[(X-\mu_X)(Z-\mu_Z)]\\ \mathbb{E}[(Y-\mu_Y)(X-\mu_X)] & ...


1

(As Did commented "Characteristic functions Characterize Distributions".) Assume that $a'X\sim N(a'\mu,a'\Sigma a)$ for every $a\in\mathbb{R}^p$. (Here $a'$ represents the transpose of the vector $a$). This is equivalent to the fact that the characteristic function $\Phi$ of $a'X$ is given by $$ ...


1

Some starting points: The ten largest airports in the world average 70,0124,224 passengers per year (2013; ATL, PEK, LHR, HND, ORD, LAX, DXB, CDG, DFW, CGK)$^1$ or 192,121 passengers per day. Assume that 1 per 1,000 passengers are bad guys (adjust as appropriate). It would be reasonable to assume that $A$ = Number of passenger $A$rrivals at the airport's ...


1

Ok, spectral techniques work, even if they produce a huge constant $M$. Consider the matrix $B$ in which $b_{ii}=1, b_{ij}=\operatorname{sign}(a_{ii})\cdot\frac{a_{ij}}{a_{ii}}$. The off-diagonal entries are Cauchy distributed, hence, by the Gershgorin theorem: $$\mathbb{P}\left[|\det B|\geq \left(1-\frac{n-1}{\lambda n}\right)^n\right]\geq ...


3

If displacement and velocity are familiar from your calculus course, recall that if you put time on the horizontal axis and the velocity of a moving object on the vertical axis, then the (signed) area under the velocity graph between times $t_1$ and $t_2$ is the net displacement of the object during that time interval. In probability and statistics, we ...


0

All the variables $Z_i$ are linked by some $Y_i$, so they are not independent (you do not have $f(z_1,z_2,z_3)=f(z_1)f(z_2)f(z_3)$ as usual under independence), but assuming the $\bar{Y_i}$ are i.i.d. normally distributed, you have a multivariate joint distribution with $\rho_{i,j}=+-0.5$. In particular: $\rho_{1,2}=0.5$ $\rho_{2,3}=0.5$ ...


1

To get you started, define a random variable $X$ by $$X = \frac{(n-1)S^{2}}{\sigma^{2}} \sim \chi_{n-1}^{2}$$ Then $$S = \sqrt{\frac{\sigma^{2}X}{n-1}}$$ and $$\mathbb{E}[S] = \mathbb{E}\left[\sqrt{\frac{\sigma^{2}X}{n-1}}\right]= \sqrt{\frac{\sigma^{2}}{n-1}}\mathbb{E}[\sqrt{X}]$$


0

We have that $-X\sim N(-\mu,\sigma)$, so regarding the variance we can indeed just sum up to get $V(Y-4X)=V(Y)+V(4X)=V(Y)+16V(X)=3.2+16\cdot 0.8=16$. That the variance from the "negative" X adds positively is from the fact that the Normal distribution is symmetric.


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So I think we all agree the book gave a wrong answer. For the extended questions, yes, I believe $P(W > 0)$ where $W = Y - 4X \sim N(10,16)$ would tell you the probability that the large bag will be more than four times the weight of the (single) small bag, both selected at random. If you instead select four small bags, it seems to me a reasonable ...


2

Hint. The minimum will be greater than $1$ if and only if $Z_1>1$ and $Z_2>1$. And the random variables are independent, so. . . ?


0

In your edit, you wrote that you were looking for a "$z$ score for the SAT score". Note that the question asks for the SAT score itself. If what you're seeking is a $z$ score, more often than not, the answer is going to be something like $-5 < z < 5$. (That's probably a pretty wide interval, but statistics classes like to have students look at ...


2

Hints: for constants $a,b$: $$E[aX+b] \equiv aE[X]+b;$$ $$\rm{Var}[aX+b] \equiv a^2 Var[X].$$ Now, rew-write $\frac{X-\mu}{\sigma}$ as $$\underbrace{\left(\frac{1}{\sigma} \right)}_{a}X+\underbrace{\left(-\frac{\mu}{\sigma}\right)}_{b}.$$ Finally, note that $$\rm{SD}[aX+b] \equiv \sqrt{\rm{Var}[aX+b]} \ .$$


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We want $X,Y$ in terms of $Z_1,Z_2$ so that we can use the change of variables transformation and show that the joint density of $Z_1,Z_2$ can factor into separate densities of $Z_1$ and $Z_2$ to show they're independent. Note that $\frac{Z_2}{Z_1}=\tan(2\pi Y)$ and since we know $\cos^2(x)+\sin^2(x)=1$, then we get $\frac{Z_1^2}{\log X}+\frac{Z_2^2}{\log ...


2

I agree with the result of @Did. The "extra n" appears in the right place by carefully following the delta method. From the given info, $\sqrt n[B_n-\beta ]\to N(0,\Sigma)$ where $B_n= \begin{bmatrix} (1/n)\sum_1^n{X_i^4}\\ (1/n)\sum_1^n{X_i^2} \\ \end{bmatrix}$, $\beta=(3,1)^T$ and $ \Sigma=\begin{bmatrix} 96 & 12 \\ 12 & 2 \\ \end{bmatrix} $. ...


0

In general, a Gaussian random variable with mean $\mu$ and standard deviation $\sigma$ can be generated form a normal random variable $\zeta$ by $X = \mu+\sigma \zeta$. So you have $$R_1 = 0+5\zeta_1,\\ R_2 = 0+3\zeta_2.$$ Therefore, since $\zeta_1 \sim \zeta_2$, you have $R_1+R_2 \sim 8 \zeta$.


0

So $S_t\to0$ almost surely. But on the other hand $S_t>0$ almost surely for any $t\ge0$. So my question is, how do these two properties correspond to each other and how is it (intuitively speaking) possible for those two properties to be true at the same time? Consider $f(t)=\mathrm e^{-t^2}$ for every real $t$. Then $f(t)\gt0$ for every $t$ and ...


4

Let $X \sim N(0,1)$ be a normal distributed random variable with mean $0$ and variance $1$. Then, $$\begin{align*} \mathbb{P}(X \geq r) &= \frac{1}{\sqrt{2\pi}} \int_r^{\infty} \exp \left(- \frac{x^2}{2} \right) \, dx \\ &\leq \frac{1}{\sqrt{2\pi}} \int_r^{\infty} \frac{x}{r} \exp \left(- \frac{x^2}{2} \right) \, dx \\ &= \frac{1}{\sqrt{2\pi} r} ...


1

you have in Y the sum of a normal distribution and rest just constants. We know that the sum of iid $N(\mu,\sigma^2)$ variables is $N(n\mu,n\sigma^2)$ distributed (see http://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables). Therefore we get: $E(Y)=\frac{\beta^2n\alpha}{\beta ...


1

Answer: Using the linearity of expectations $Y={\frac {{\beta}^{2}\sum _{i=1}^{n}X_{{i}}}{n{\alpha}^{2}}}-{\frac { \beta}{na}}+1 $ $E(Y) = E\left({\frac {{\beta}^{2}\sum _{i=1}^{n}X_{{i}}}{n{\alpha}^{2}}}-{\frac { \beta}{na}}+1\right)$ $E(Y) = \frac{n\cdot\beta^2\cdot\alpha}{n\cdot{\alpha}^2\cdot \beta}-\frac{\beta}{n\cdot\alpha} + 1$ If you simplify, ...


3

Hint: $$ \Pr \! \left( \frac{X + Z}{2} > Y \right) = \Pr(X + Z - 2 Y > 0). $$ Any linear combination of normal random variables that are jointly normally distributed is also a normal random variable (so its probability density function is completely described by its mean and variance).


0

The definition of a probability (and of a sigma algebra) necessarily implies that the CDF is always monotonically increasing with respect to any of its variables. So, yes, it is also the case if Σ is degenerate. Edit : A less theoretical answer consists of stating that if this was not the case, then you could find strictly negative probabilities. If $a_{1} ...


0

This article (much of which was written by me) explains how to find the MLE of $\Sigma$. Sufficient conditions for existence and uniqueness of the MLE are not necessary conditions for the existence of the MLE. Thus it may be unnecessary to prove that the sufficient conditions hold.


0

1) Determinant consists of products and sums of elements of $\Sigma$. "Differentiable w.r.t. $\Sigma$" actually means "diff'ble w.r.t. its elements". 2) There is also a restriction on $\Sigma$, namely $\Sigma$ is symmetric and positive definite. This means $\det(\Sigma) > 0$. At the boundary, $\det(\Sigma) = 0$, so... Also, if you're not quite ...


0

We have: $F(z)=P(X+Y\leq z)=\int_{-\infty}^{\infty}\int_{-\infty}^{z-x}f_{x,y}(x,y)\,dx\,dy$ (see CDF of sum of dependent random variables) The marginal distributions: $F(x)=P(\max x_k\leq x)=P(x_1\leq x,...,x_n \leq x)=\Phi(x)\cdots\Phi(x)$ (same for y-max, also see ...


2

Even the distribution of the maximum of two correlated normal random variables is a rather complicated thing. Do you really think this has a closed form?



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