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4

Nobody is making assumptions in this matter; rather, someone is drawing conclusions. And those conclusions are not based simply on the marginal distributions. Conventionally, one has $$ X_1,\ldots,X_n \overset{\mathrm{i.i.d.}}\sim N(\theta,1), $$ $$ \bar X = \frac{X_1+\cdots+X_n} n \sim N\left(\theta,\frac 1 n \right) $$ This gives use the mean vector and ...


2

Well, let's explore the expectation directly: $$E\left[e^{e^y}\right] = \int_{-\infty}^\infty e^{e^x} e^{-(x-\mu)^2/(2\sigma^2)}\, dx = \int_{-\infty}^\infty \exp\left( e^x-\frac{(x-\mu)^2}{2\sigma^2}\right)\, dx.$$ Now take $x \to +\infty$; it is clear that $e^x \to \infty$ much faster than $-x^2 \to -\infty$, so it dominates in the exponential. ...


2

We have to check that the integral $$\int_0^{+\infty}\exp\left(e^y-(y-\mu)^2/(2\sigma^2)\right)\mathrm dy$$ is divergent. There is a huge problem at $+\infty$ because $\lim_{y \to +\infty}e^y-(y-\mu)^2/(2\sigma^2)=+\infty$.


2

See: the Maxwell-Boltzmann distribution with $$a=\sqrt{\frac{kT}{m}}=1,$$ where $m$ is the particle mass and $kT$ is the product of Boltzmann's constant and thermodynamic temperature. (The pdf in question describes the speed distribution of particles in idealised gases.)


2

I have a horrible feeling about this answer but... Let the average height of be the random variable $X$. We know that $\sigma = 2.5$ inches. By the central limit theorem for a sample of $100$ we have $\overline{X} \sim N(\mu,\frac{2.5^2}{100})$ The condition "find the probability that the difference between the sample mean and the true population will ...


2

(B) is correct and here is one formal demonstration. I'll write $C$ below for covariance and $V$ for variance. Also, let $S=\sum_{i=1}^{98}X_i$ and $T=\sum_{i=3}^{100} X_i$. Due to independence, we have $$ V(S)=\sum_{i=1}^{98}V(X_i)=98,\quad V(T)=\sum_{i=3}^{100}V(X_i)=98. $$ Moreover, using $C(X_i,X_j)=0$ for $i\neq j$, we have $$ ...


2

Since the $X_i$ are i.i.d. and have unit variance, we have $$\DeclareMathOperator{Cor}{Cov} \Cor(X_i,X_j) = \begin{cases} 1 & \text{if } i = j, \\ 0 & \text{if } i \neq j. \end{cases}$$ Therefore $$ \begin{align*} \Cor\left(\sum_{i=1}^{98} X_i, \sum_{j=3}^{100} X_j\right) &= \sum_{i=1}^{98} \sum_{j=3}^{100} \Cor(X_i,X_j) \\ &= \sum_{i=1}^{98} ...


1

Even with some confusion about notation and terminology, you are on the right track. Here is a corrected version of what you have. You want $Z = \frac{\sqrt{n}(\bar X - \mu_0)}{\sigma} = \frac{\sqrt{85}(98-100)}{20} = -0.922.$ From normal tables $P(Z < -0.922) \approx 0.1783.$ This is the P-value. Because the P-value exceeds 5% we do not to reject. ...


1

It is well-known that the random variable $X/Y$ has the two-sided Cauchy density $\frac{1}{\pi(1+x^2)}$ for $-\infty<x<\infty$. Thus $P(X/Y<1)=\int_{-\infty}^{1}\frac{1}{\pi(1+x^2)}dx=0.5+\int_{0}^{1}\frac{1}{\pi(1+x^2)}dx$ and so $P(X/Y<1)=0.5+\frac{1}{\pi}\hbox{arctg}(1)=0.5+0.25=0.75$. Note: A much simpler way is to consider the random ...


1

If $X, Y$ are independent and have the standard normal distribution, then $X-Y$ has normal distribution of mean $E(X-Y)=E(X)-E(Y)=0$ and dispersion $\sigma^2(X-Y)= \sigma^2(X)+\sigma^2(Y)=2$. Analogously, $X+Y\sim N(E(X)+E(Y), \sigma^2(X)+\sigma^2(Y))$, i.e. $X+Y\sim N(0,\sigma^2=2)$. In order to get the joint distribution of $(X-Y, X+Y)$ we have to ...


1

As commented http://en.wikipedia.org/wiki/Binomial_distribution. $$\mu=np$$ $$\sigma=\sqrt {np(1-p)}$$ $$B(X\le x)\approx N(X\le x+0.5)$$


1

$x^T A x = tr(x^T A x) = tr(x x^T A)$. By linearity of expectation, $E[x^T A x] = E[tr(x x^T A)] = tr(E[x x^T A]) = tr(E[x x^T]A) = tr( (\Sigma + \mu \mu^T) A) = tr(\Sigma A) + tr(\mu \mu^T A) = tr(A \Sigma) + tr(\mu^T A \mu) = tr(A \Sigma) + \mu^T A \mu$ by the fact that trace is linear and invariant under cyclic permutations and trace of a scalar is ...


1

It's a misprint. Instead of $$Z_j^2 = \left( \theta_j^{\color{red}{2}} + \frac{\sigma}{\sqrt{n}} \epsilon \right)^2 \tag{1}$$ it should read $$Z_j^2 = \left( \theta_j + \frac{\sigma}{\sqrt{n}} \epsilon \right)^2.$$ One way to see that $(1)$ cannot hold true goes as follows: It is well-known that for any random variable $X$ we have $$\text{var} (X) = ...


1

Hint: Use an inverse z-score look up table (or just scan through a standard forward z-score look up table) to figure out how many standard deviations above average gives an upper tail $p$-value of $1/3$. Then subtract this many standard deviations from 20 cm to get the mean.


1

Based on the Normal distribution given , find the probability that a person's height exceeds $180$ cm, call this $P(X>180)$. The expected number of people that can enter the door will be the reciprocal of this, i.e. $1/P(X>180)$


1

Let $Y$ be the number of people who enter until the first one has to bend. Note that $Y$ is discrete. Take $k \in \mathbb{N}$. If $Y=k$ this means that $k$th person to go through the door bends while the first $k-1$ do not, in other words, the height of the $k$th person is $>180$cm while the heights of the first $k-1$ people are $<180$. Assuming ...


1

I'm assuming a very large population of units from which to select, so that the distribution is binomial, and that 'have issues' means 'are defective." Then we have the number of defective units seen $X \sim Bin(n=10, p=.1)$ Thus, (1) $E(X) = np = 1$ as you say. (2) V(X) = np(1-p) = .9, so SD(X) = \sqrt(.9) = 0.9487.$; I think you overlooked a decimal point. ...


1

You are normalising wrong. You should instead write $$Y = \mu + \sigma Z$$ The rest looks fine


1

I think you want to update your beliefs about $\alpha$ and $\beta$ jointly, not separately. Then you can just use Bayes' rule to write $p(\alpha,\beta|Y)\propto p(Y|\alpha,\beta)p(\alpha,\beta)$. Assuming that your joint prior for $(\alpha,\beta)$ is just the product of the marginal priors, I think this works out to: $$ p(\alpha,\beta|Y) = \frac{\sum_{j=1}^2 ...


1

The distance is chi-square distributed. Intuitively, look at the distance between the two normal random variables as an error (for example, distance between randomly sampled point and the mean). We know that this error is distributed chi square. Here's a more rigorous description: ...


1

As @Seyhmus Güngören points out, you have been asked to solve a drill problem that is not likely to be useful in practice. So it may be difficult for you to have an intuitive sense how to proceed. Here is the approach I think you are expected to take. $$0.035 = P(X \le 3) = P(Z \le (3 - 10)/\sigma),$$ where $Z$ is standard normal. From normal tables (or ...


1

What you're looking for is the Rayleigh distribution (distribution of the norm of two centered and independent gaussian RVs) : http://en.wikipedia.org/wiki/Rayleigh_distribution You might also want to look up the $\chi^2$ distribution (distribution of the squared norm) : http://en.wikipedia.org/wiki/Chi-squared_distribution


1

Let $R$ denote the distance of the point $(X,Y)$ from the origin. Draw a circle of radius $a$ centered at the origin. Then, $$P\{R \leq a\} = F_R(a) = \iint_{x^2+y^2\leq a}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy.$$ Convert from rectangular coordinates $(x,y)$ to polar coordinates $(r,\theta)$ hopefully not forgetting the mantra $r\,\mathrm dr\,\mathrm d\theta$ ...


1

First of all, by the definition of the scalar product, $$(B_t-x) \cdot (B_s-x) = \sum_{j=1}^n(B_t^j-x^j) (B_s^j-x^j)$$ where $x=(x^1,\ldots,x^n)$ and $B_t = (B_t^1,\ldots,B_t^n)$. Without loss of generality, $s \leq t$. Choosing $t_1 = s$, $t_2 = t$, the covariance matrix of $Z:=(B_s,B_t)$ equals $$C= \begin{pmatrix} s I_n & s I_n \\ s I_n & t I_n ...


1

Yes, they are norms. My guess is that the book calls them semi-norms because the topology in a locally convex topological vector space is defined through a family of semi-norms (which may or may not be norms.)


1

You use the distribution function: write $Y=X^{1/3}$ $$ P(Y<y)=P(X^{1/3}<y)=P(X<y^3), $$ since the transformation and the distribution function are bijective. You can then differentiate the cumulative function to get the density function.


1

You can also write, given any measurable function $g$, $$E(g(X^{1/3}))=\int_\Bbb R g(x^{1/3})f_X(x)\mathrm{d}x$$ Now, with a change of variable $x=y^3$, and writing $Y=X^{1/3}$, $$E(g(Y))=\int_\Bbb R g(y)3y^2f_X(y^3)\mathrm{d}y$$ Hence the density of $Y$ is $f_Y(y)=3y^2f_X(y^3)$.


1

User Did is spot on. I'll just show you the details: We can reformulate the expression within the square brackets as follows: \begin{align*} 1_{\displaystyle \left\{ \sum_{i=1}^n u_i \cdot Y_i > 0 \right\}} \sum_{i=1}^n u_i \cdot Y_i = \max\left(\sum_{i=1}^n u_i \cdot Y_i, 0 \right) \end{align*} We then have $\sum_{i=1}^n u_i \cdot Y_i \sim N(0, ...


1

Number the bulbs. For $i=1$ to $100$, let $X_i$ be the lifetime of bulb $i$. Let $X=X_1+\cdots +X_{100}$. We want $\Pr(X\gt 50000)$. Let exponentially distributed random variable $T$ have parameter $\lambda$, that is, density function $\lambda e^{-\lambda t}$ for $t\gt 0$. Then $T$ has mean $\frac{1}{\lambda}$ and variance $\frac{1}{\lambda^2}$. (This is a ...


1

Hint: If you want to calculate it using an integral expression, it is actually easier to calculate the variance directly, i.e. $$\text{var}(X) = \mathbb{E}((X-\mu)^2) = \int (x-\mu)^2 \frac{1}{\sqrt{2\pi \sigma^2}} \exp \left(- \frac{(x-\mu)^2}{2\sigma^2} \right) \, dx.$$ To this end, write $$\sqrt{\frac{\sigma^2}{2\pi}} \int(x-\mu) \cdot \left[ ...



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