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0

You're on the right track; you can simplify using $\operatorname{Cov}(aZ_1,bZ_2)=ab\operatorname{Cov}(Z_1,Z_2)$, also the fact that $\operatorname{Cov}(Z_1,Z_2)= 0$ since $Z_1$ and $Z_2$ are independent, and the fact that $\operatorname{Cov}(Z_1,Z_1) = \operatorname{Var}(Z_1)=1$, and so on. But there is a more direct method. If $X$ denotes the column vector ...


0

Of course it is normal. The integral (if taken as the Ito integral) is the limit of a linear combination of brownian motions which are normally distributed (To see this, discretize the integral). Therefore, the integral is normally distributed. You're second claim is just an application of the fundamental theorem of stochastic calculus. So, you're all good.


0

You're not doing this well. I think that A~N(30,7.75) and B~N(20,7.75). Let C = A - B. Find the distribution of C. Irrational choice is P(C<0).


0

Your notation is mildly confusing for a reason I will explain further below. You have $X_t = \sqrt{t} Z$ where $Z\sim\mathrm{N}(0,1)$ and $0<s<t$, so that $X_t \sim \mathrm{N}(0,t)$. It follows that $X_t-X_s = \sqrt{t}Z-\sqrt{s}Z$ $=(\sqrt t - \sqrt s)Z$, so the distribution of this random variable is $\mathrm{N}(0,(\sqrt t - \sqrt s)^2) = ...


0

Is the standard deviation of $600$ correct? If so, here are some hints: a) $z=\frac{630-650}{600}$. You need to find the area to the left of this. b) $z=\frac{630-650}{\frac{600}{\sqrt{70}}}$. Again, area to the left.


0

I'm pretty sure you are supposed to use the normal approximation to the normal, as suggested by @Karl. I hope you used continuity corrections to give that the best chance to work. Here are exact binomial answers from R statistical software to compare with your approximate answers: dbinom(100, 100, .9) # (a) 2.65614e-05 # ...


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: ...


0

$a)$: The null and alternative hypothesis should be: $H_0: \mu = 0.85, H_1: \mu > 0.85$ $b)$: It is not necessary to get an exact $P$-value, you can estimate it with $df = n-1 = 6-1 = 5, \alpha = 0.05$, and the test is a one-tail test ( right tail ), then $P$-value $< 0.005 < 0.05 = \alpha $ because $4.9 > 4.032$ which corresponds to the ...


0

You use $z=\frac{\overline{x}-\mu_{\overline{x}}}{\sigma_{\overline{x}}}=\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$ when you are dealing with a sampling distribution. That is, when your data points are means of samples of size $n$. You use $z=\frac{x-\mu}{\sigma}$ in a normal distribution to tell you how many standard deviations a given $x$ is from ...


0

Suppose you have $X\sim N(\mu,\sigma^2)$ then you would use $Z=\frac{X-\mu}{\sigma}$. The $Z$ score tells you how many standard deviations $X$ is from the mean. If you took a sample of size $n$ then you might want to look at the distribution of the sample mean denoted $\overline{X}$. The central limit theorem says that $\overline{X}\sim ...


-1

An exacter inequality can be obtained with Maple by with(Statistics): X := RandomVariable(Normal(0, 1)): phi := PDF(X, t); $$ {\frac {\sqrt {2}{{\rm e}^{-1/2\,{t}^{2}}}}{2\sqrt {\pi }}} $$ Phi := CDF(X, t); $$\frac 1 2+ \frac 1 2\,{\rm erf} \left(\frac 1 2\,t\sqrt {2}\right) $$ Optimization:-Maximize(4*phi*(1-2*Phi)/(1+(1-2*Phi)^2)^2); $$ [-{ ...


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

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

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

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, ...


0

Assuming $\sigma \gt 0$, from $Y=10(2+\mu+\sigma X)$ you could say $X = \dfrac{Y-20-10 \mu}{10\sigma}$. So $P(Y \gt 0)$ is equivalent to $P\left(X \gt \dfrac{-2- \mu}{\sigma} \right) = 1 - \Phi\left(\dfrac{-2- \mu}{\sigma} \right)$.


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

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. ...


0

I think you are using words that you don't understand. The sum of multiple dice does not follow a normal distribution. See my explanation for the sum of two $n$-sided dice case in my answer here. Supposing you remove the word "normal," you can see how the mean and standard deviation interact by the linearity of expected value. The expected value of a ...


1

Since the sequence $(X_n)_{n\geqslant 1}$ is independent, an application of the Borel-Cantelli lemma shows that $X_n\to 0$ almost surely is equivalent to the convergence of the series $\sum_{n\geqslant 1}\mathbb P(|X_n|>\varepsilon)$ for each fixed $\varepsilon$. Thus, $X_n\to 0$ almost surely is and only if $\sum_{n=1}^\infty\mathbb ...


0

I think I got the answer. I started by assuming the function $f(x)$ derivative $f'(x)$ kinda looks like the Gaussian function, and its double derivative $f''(x)$ looks like the original function $f(x)$. $$f''(x)=f(x)$$ So I asked Wolfram|Alpha's help (yup I cheated) and the general solution is (ignoring constants) $$f(x)=e^x±e^{-x}$$ The $f(x)=e^x-e^{-x}$ ...


-1

For $np$ and $nq$ to increase $n$ must increase. $n$ is the number of independent trials, so it should be clear that the more independent trials made, the more accurate your approximation is. The probability histogram approximates a normal curve pretty accurately when $np$ and $nq$ are greater(or equal to) $5$. However bigger is better! If $np$ and $nq$ ...


0

$\newcommand{\E}{\operatorname{E}} \newcommand{\var}{\operatorname{var}}$ It appears to me that what you have in mind is this: \begin{align} X & = G + U \\ Y & = G + V \\ & X,Y,G\text{ are independent}, \\ \text{and} & \text{ $X,Y,G$ are normally distributed (i.e. Gaussian)} \end{align} and you want $\E(XY)$. \begin{align} \E(XY) & = ...


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[ ...


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 ...


0

If $X\sim N(220000, 160000^2)$ describes the profits of the project, then the probability of a loss greater than $80000$ is $\mathbb{P}(X\leq -80000)$. I.e. the probability of a profit less than $-80000$. Now this can be calculated by transforming $X$ into a standard normal random variable $Z$, and looking this probability up in a table as @Ian suggested. Or ...


0

$$-80,000 - 220,000=-300,000$$ I.e. an $\$80,000$ loss is $\$300,000$ below the average. $$\dfrac{-\$300,000}{\$160,000} = -1.875 $$ I.e. this is $1.875$ standard deviations below the mean. The probability that a standardized normally distributed random variable is less than $-1.875$ is $\Phi(-1.875)\approx 0.030396$ if I can believe the software I'm ...


0

The usual technique is to convert a normal variable $X$, which has mean $m$ and standard deviation $\sigma$, to a normal variable with mean $0$ and standard deviation $1$, usually denoted by $Z$. You do this by defining $Z=\frac{X-m}{\sigma}$. Here $m=220000$ and $\sigma=160000$. Then your inequalities change. Let's say your inequality was $X \leq x$, then ...


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) = ...


2

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

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

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

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.


0

You can use moment matching techniques to approximate the maximum of two skew-Normal variables by another skew-Normal distribution. See "A New Statistical Max Operation for Propagating Skewness in Statistical Timing Analysis".


0

(1) Re-sampling is certainly a possibility, as suggested in a previous answer. Also. possibly a permutation test or a bootstrap confidence interval, which are different kinds of resampling than suggested there. (If you can figure out how to do one of these, this might be your best bet.) (2) If there are few (if any) ties within or between the two samples, a ...


0

This works: $G_{intermediate}(\theta_i) = \frac{G(\theta_i) - 2I_0}{I_1}$ $G_{norm}(\theta_i) = I_0 + \frac{I_1}{\max(G_{intermediate}(\theta_i))}G_{intermediate}(\theta_i)$.


0

let $A_i$ indicate that sample $i$ had at least $178$ level of glucose. The probability $\mathbb{P}\{A_i\}$ can be calculated. Denote it by $p$. We now need to find the probability that out of $A_i,i=1...100$ independent variables, $10$ will be true. So we have $100$ Bernoulli variables with success probability $p$. This is the Binomial distribution with ...


0

I've got it! $j = 360 / (\sigma \sqrt{2 \pi} erf(\frac{180}{\sigma\sqrt{2}}))$. (Not quite a "symbolic" representation, but I've gotten rid of that pesky -- read, harbinger of imprecision -- decimal point.)


0

Computations involving multivariate densities can quickly become a mess. Your best bet is to reduce to the case of independent normals by applying the same transformations as you did in deriving $E(\mathbf X)=\mathbf m$. So define $\mathbf Y:=\mathbf{X-m}$, and set $\mathbf Y=V\mathbf Z$ where we've decomposed the covariance matrix $\mathbf \Sigma$ of ...


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 ...


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 ...


0

General comments. The distribution of toe lengths of the two species together is called a 'mixture' distribution. The mixture of two independent normal distributions is not normal (if means and variances differ), but may be difficult to distinguish from normal. If the means are sufficiently far apart relative to the variances, the mixture of two normals is ...


1

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


0

You're almost there. In your expression $$ \frac{1}{(2\pi)^{d/2}(\operatorname{det}(\Sigma))^{1/2}} \left(\int_{R^d}\mathbf{y} \exp(-\frac12 \mathbf{y}^T \Sigma^{-1}\mathbf{y})d \mathbf{y}+ \int_{R^d}\mathbf{m} \, \exp(-\frac12 \mathbf{y}^T \Sigma^{-1}\mathbf{y})d \mathbf{y}\right)\\ $$ the second integral equals $\mathbf m$ because you can pull the ...


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.)


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. ...


0

You are expected to find or make a curve like you find by searching "standard normal distribution". Your $Z$ is the horizontal coordinate. Then color in the range of $Z$ called out in each part of the question.


1

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


0

Try characteristic function: we know the CF for $N(\mu,\sigma^2)$ is $$\phi(t)=\exp\left\{i\mu t-\dfrac{1}{2}\sigma^2t^2\right\}$$ \begin{align} \mathbb{E}[e^{itZ}]&=\mathbb{P} [X=0]+\mathbb{E}\left[\exp \left\{it\sum\nolimits_{j=1}^kY_j\right\}\cdot\mathbb{1}_{X\geq 1}\right]\\ &=\mathbb{P} ...



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