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6

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


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 ...


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\\ ...


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.


3

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


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 $P(\xi+\eta=0)=P(\sigma=-1)=\frac12$ ...


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 ...


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 $


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

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 ...


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.


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} ...


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 ...


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' ...


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

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


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.


1

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


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 ...


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 $$ ...


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 $\theta-\mu$ ...


1

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


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 ...


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.


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 ...


1

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

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?


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 ...


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 ...



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