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0

For 1) if you've simulated the excess infections under the null then you have numerically estimated the sampling distribution. Just calculate the fraction of the simulation results that are $\geq 3.3$ or $\leq -3.3$ This is the two-sided p-value for your observed placebo excess. For 2) you are not using your simulation, but the $\mathcal{N}(0,12.3)$ ...


0

Then you have heteroscedasticity. The estimated values for the standard errors of the parameters are biased. And additional you can´t use the t-Distribution and the F-Distribution for testing the parameters.


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For help with the 'rigorous mathematics', there is a proof of the affine property here. Just substitute your $0$ for $\mu_x$, $I$ for $\Sigma_x$ and $\mu$ for $b$ and you should get the same result :) For your 2nd question about the intuition for the spectral decomposition, we begin by observing that $C = UDU^T$, and that $U$ provides an orthogonal basis ...


2

If the sample consists of just one person, then the number of smokers is $0$ or $1$, with respective probabilities $1-p$ and $p$, so you have a random variable whose expected value is $p$ and whose variance is $p(1-p)$. If the sample size is $n$, then the number of smokers is the sum of $n$ random variables with that distribution, so the expected number is ...


2

If each of A and B must have $5$ pencils at least, give them each $5$ pencils. That leaves $15$ pencils. We can give $0$ of these to A, or $1$, or $2$, and so on up to $15$. The other questions are done in exactly the same way.


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For the first inequality, you could (prove and) use the identity $$E(\mathcal O^n)=\int_\mathbb R(\mathbf 1_{x\gt0}-\Phi(x)^n)\,\mathrm dx,$$ and the fact that, for every $x$, the sequence $(\Phi(x)^n)$ is decreasing. (Hint: Start from the identity in your post and integrate by parts, using the functions $u=x$ and $v=\Phi(x)^n$.) The second inequality ...


3

Not always--otherwise every sum of normal random variables would be normal, and this ain't so. Canonical (counter)example: Assume that $\xi$ is standard normal and that $\eta=\sigma\xi$, where $\sigma=\pm1$ is symmetric Bernoulli and independent of $\xi$. Then $\eta$ is standard normal but $\xi+\eta$ is not normal since $P(\xi+\eta=0)=P(\sigma=-1)=\frac12$ ...


1

and note that since the mean is zero by symmetry, you can calculate the variance quite simply. with parameter $a$ for the distribution, the variance is: $$ v = \frac {I_a(x^2)}{I_a(1)} \\ $$ where $$ I_a(f(x)) = \int_{-\infty}^{\infty} f(x) exp(-ax^2)dx $$ a straightforward integration by parts gives: $$ I_a(x^2) = \frac1{2a}I_a(1) $$ so for $v=1$ we ...


2

We are given $Z$ is the sum of 2 independent normals so $Z$ will be normal. We just have to straighten out the means and variances. Evidently we can measure power in $W$ or dBW. Never heard of dBW but it is just $W$ measured on a log scale. (Like measuring strength of earthquakes on the Richter scale.) So we have $$\text{dBW}=10\log_{10}W.$$ We can ...


0

$$ \int_0^\infty e^{-x^2/2} \Big( x\,dx\Big) = \int_0^\infty e^{-u}\,du. $$


5

\begin{align} \int^{\infty}_{-\infty}\int^{\infty}_{-\infty}e^{-\frac{1}{2}(x^2+y^2)}dxdy &=\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}e^{-\frac{1}{2}x^2}e^{-\frac{1}{2}y^2}dxdy\\ &=\int^{\infty}_{-\infty}e^{-\frac{1}{2}y^2}\underbrace{\left[\int^{\infty}_{-\infty}e^{-\frac{1}{2}x^2}dx\right]}_{constant \ w.r.t.y}dy\\ ...


6

Here are the steps that help make it obvious, \[ \int_0^\infty xe^{\frac{-x^2}{2}} dx = \lim_{\beta \to \infty}\int_0^\beta xe^{\frac{-x^2}{2}} dx \] Let $u=\frac{-x^2}{2}$ then, \[ \frac{d}{dx}u = -x \Rightarrow du = -x\ dx \Rightarrow-du=x\ dx \] So now after adjusting the limits, we have \[ \lim_{\beta \to -\infty}-\int_0^\beta e^{u}\ du = \lim_{\beta ...


2

I would substitute $u(x)=-\frac{1}{2}x^2$ $\Rightarrow \frac{du}{dx}=-x\Rightarrow -du=x \ dx$ $ -\int_0^{-\infty} e^{u} \ du $ The upper limit has been adjusted. The limits can be switched: $ \int_{-\infty} ^0 e^{u} \ du $


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Isn't it obvious that $$\int_0^\infty xe^{-x^2/2}=-e^{-x^2/2}\Bigg|_0^\infty=1\qquad?$$


5

We have $\lim_{y\to-\infty}\phi(y)=0$ and $\lim_{y\to\infty}\phi(y)=1$. Moreover, the function $\phi$ is continuous. Thus, by the Intermediate Value Theorem, the function $\phi$ is surjective.


0

We have $$ X \in N(111, 5) $$ The probability that a single match will last between 110 and 118 minutes is $$ p = P(110 \le X \le 118) = F_X(118) - F_X(110) $$ if $F_X$ is the cummulative dist. function. The probability that at least 30 (out of 64) lasts between these values is $$ 1 - \sum_{i=0}^{30} {64 \choose i}p^i(1-p)^{64-i} $$


1

Find the probability $p$ that a single match will last between 110 and 118 minutes using the normal distribution. Then use the cumulative binomial distribution to find the probability of more than 30 succeeding.


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HINT: Do you know how to use normCDF on your calculator? If you can't use a calculator, do you know how to use a z-score table and calculate the z-score by hand?


0

First of all, if $X$ and $Y$ are independent and Gaussian, then their sum is also Gaussian. There is no doubt about it. You question is a bit different. You will have a transformation to one of your random variables before the convolution. This means first, both of the domains of the PDFs should be the same. When you take Log of a random variable distributed ...


1

If I understood correctly you need to find the mean and the variance of $\hat \sigma^2$? Not dealing with the fact why did you use this estimate, one can do the following: $$\mathbb{E}\left[\hat \sigma^2\right]=\mathbb{E}\left[\frac{1}{N} \sum_{n=0}^{N-1} x_n^2\right]=\frac{1}{N} \sum_{n=0}^{N-1}\mathbb{E}\left[ x_n^2\right]=\mu_2'.$$ where $\mu_2'$ is the ...


0

The answers are all helpful so far, but the original question was on the right track: there is indeed an error function that results. The reason is that the original integral, which Henry correctly pointed out is over t, not $\tau$, is from 0 to $\infty$, and not from $-\infty$ to $\infty$. This changes everything. In what follows, I took the liberty of ...


0

I have a (easy) formula, which approximate the standard normal distribution quiet good: ${\Phi(x) \approx 0,5 \cdot \Bigg( 1 + \sqrt{1-e^{- (\sqrt{\frac{\pi}{8}} \cdot x^2)}} \Bigg)}$ The diagram below shows the values of the Standard normal distribution ($\color{red}{ red}$) in comparision to the values of the approximation ($\color{blue}{ blue} $) . ...


1

See this section on Wikipedia The values in the table come from evaluating the integral $$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^x e^{-t^2/2}\mathop{dt}.$$ It is hard to evaluate this integral, which is why we rely on tables instead.


0

The confidence interval for a regression coefficient $\beta$ follows a $t$ distribution with $ n-p$ degrees of freedom. To calculate it the following steps are necessary: identify the parameter $\hat\beta_i$ for which you want to calculate the CI; specify a confidence level (e.g., 95% 99% etc...), which corresponds to $1-\alpha$); calculate the critical ...


1

One should approach this through characteristic functions. Recall that $X$ is normal $N(\mu,\Sigma)$ if and only if, for every deterministic vector $t$ of size $N\times1$, $$ E(\mathrm e^{\mathrm it'X})=\mathrm e^{\mathrm it'\mu-t'\Sigma t/2}, $$ where $t'$ denotes the transpose of $t$. For every $(A,b)$ of compatible sizes, if $Y=AX+b$, one gets $$ ...


2

The equation $$\lim_{n \to \infty} \frac{1}{\sigma_n \sqrt{2\pi}} \int e^{-x^2/2\sigma_n^2} g(x) \, dx = \frac{1}{\sigma \sqrt{2\pi}} \int e^{-x^2/2\sigma^2} g(x) \, dx$$ shows that $\mu(f_n^{-1}(\cdot))$ converges in distribution to a normal distribution with variance $\sigma^2$. On the other hand, by assumption, $f_n$ converges in $L^2$ to $f$ and this ...


0

Yes it does hold for a $M \times N$ matrix You can use the generalized inverse in place of the normal inverse, where the original inverse is now the left pseudo-inverse, such that $A^{-1}_{left}A=I$ (Property 1: Not that this inverse is not commutative) Applying $A_{left}^{-1}$ to both sides of $y=Ax+b$, we get $A^{-1}_{left}y = A^{-1}_{left}(Ax+b) = ...


1

I will construe this to mean that if $X\sim N(\mu,\sigma^2)$ then $(X-\mu)/\sigma\sim N(0,1)$. That $X\sim N(\mu,\sigma^2)$ we may construe to mean $$ \Pr(X\in A)=\int_A \frac{1}{\sqrt{2\pi}} \exp\left( \frac{-1}{2}\cdot \left(\frac{x-\mu}{\sigma}\right)^2 \right)\, \frac{dx}{\sigma} $$ for every Borel-measurable set $A\subseteq\mathbb R$. Now $$ ...


1

The definition of variance for a random variable $X$, $var(X) = E[(X - \mu)^2]$, since $\mu = 0$, it's obvious that $var(X) = E[X^2]$.


0

You need to utilize the percentiles of the inverse CDF of the gaussian-distribution described as Wikipedia. Divide the quantiles over the amount of numbers wanted and then adapt $\sigma$ so that the outer segments align with 1 and Z ($\mu$ will be (Z-1)/2). So if n=3, that means that each segment is 33%. That means that 1 and Z should be at in the middle of ...


3

$P(Z\ge 4.03)+P(Z\le -4.03)=P(|Z|\ge 4.03)=P(Z^2\ge 4.03^2)=P(\chi ^2_1\ge 4.03^2)$ But $P(Z\ge 4.03)=P(Z\le -4.03)$ so $P(\chi ^2_1\ge 4.03^2)=2P(Z\ge 4.03).$ There's the mysterious $1/2.$ Also $P(Z>4.03)=P(Z\ge 4.03)$ because $Z$ is a continuous rv and so $P(Z=4.03)=0.$ And $ P(Z\ge 4.03)\lt 0.0003 $ does not mean they are comparing it to 0.0003, I ...


0

Answer: B has N(31000,640000) and A has N(18000,360000). $$P(B-A >=15000) = P(B-A) >=15000)$$ $$E(B-A) = E(B) - E(A)$$ $$Var(B+(-1)A) = Var(B) + (-1)^2Var(A) = Var(B)+Var(A)$$ $$E(B) = 31000 , E(A) = 18000, Var(A) = 36\times10^6, Var(B) = 64\times10^6$$ $$Var(B-A) = 100\times10^6$$ $$P(B-A>=15000) = P\left(Z>=\frac{15000 - ...


1

You seem to be missing some information when describing your question, and this missing information is critical to the reason behind your question. Suppose I observe one individual from country $A$. Their income is a single normally distributed random variable with mean $\mu_A = 18000$ and standard deviation $\sigma_A = 6000$. If I observe two randomly ...


0

If $X$ is a random variable, the variance of $100X$ is $100^2$ times the variance of $X$. Likewise the variance of $X/100$ is $1/100^2$ times the variance of $X$. I don't know what that has to do with your problem, but that is a common way for the square of a number to show up as a factor when computing a variance.


2

The reason it is giving half the value is that it should give half the value. If the standard deviation is twice the size, then the graph of the probability function is twice the "width". But the area under the probability function is always 1, so all the heights must be divided by 2 in order to maintain the area at 1. So probability density functions' ...


0

You could use $$e^x=\sum_{i=1}^{\infty}\frac{x^i}{i!}$$


0

Unfortunately, you cannot evaluate it in closed form. It is the Gaussian error function and, in general, it can only be evaluated numerically.


1

For a geometric proof, see Bennett Eisenberg & Rosemary Sullivan, Why Is the Sum of Independent Normal Random Variables Normal?, Mathematics Magazine, Dec. 2008, 362-366, available at http://www.maa.org/programs/faculty-and-departments/classroom-capsules-and-notes/why-is-the-sum-of-independent-normal-random-variables-normal


2

I will try to address the case for $\mu=0$. The integral for this case is: $$\frac{1}{\sigma} \sqrt{\frac{2}{\pi}}\int_0^{\infty} \ln(x)e^{-\frac{1}{2}\left(\frac{x}{\sigma}\right)^2}\,dx$$ With the substitution $\dfrac{x}{\sigma\sqrt{2}}=t$, $$\begin{aligned} \frac{1}{\sigma} \sqrt{\frac{2}{\pi}}\int_0^{\infty} ...


0

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 $\theta-\mu$ ...


3

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


1

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 ...


0

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 ...


0

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 ...


1

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


0

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, ...


0

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 $0$.


1

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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