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1

Since $\frac{\sum X_i-np}{\sqrt{np(1-p)}}$~$N(0,1)$, so $\sum X_i$~$N(np,np(1-p))$. Then you get $(\sum X_i-\sum Y_j)$~$N(np-mq,np(1-p)+mq(1-q))$

0

I would have thought that the variance of the sum was $3^2+10\times 1.5^2 +20\times 0.1 \times 1.5^2$. You would then need to take the square root for a standard deviation. The problem seems to be with your $0.0015$. My guess is that the units of power should be megawatts (MW) rather than milliwatts (mW), and the square of this for variance and ...

1

Hint1: In a symmetric distribution the mean equals the median. Hint2: Since the triangle distribution is symmetric, you can infer that $b-a=c-b$ according to the notation in Wikipedia, where $a$ is the lower bound and $b=$ median and mean value (due to symmetry).

0

Actually the average IQ is 100 and its standard deviation is 15. Intelligence tests are scored in such a way the resulting IQ distribution conform to these properties. http://en.wikipedia.org/wiki/Intelligence_quotient

2

Since it is the "middle" $95\%$, meaning with with equal tails of $2.5\%$ on each side, the mean must be $\frac{60+140}{2}$. (Recall that the normal distribution is symmetric about the mean.) From tables of the standard normal, the point which has $2.5\%$ in the right tail is $1.96$ standard deviation units from the mean. So if $\sigma$ is the population ...

0

Imagine that you work in a factory that makes thousands of cupcakes each day. The cupcakes are supposed to weigh 125g each and the distribution of the weight is supposed to be Normal with mean 125 and standard deviation 15. Now imagine that each day you randomly select 20 cupcakes from all of the cupcakes that were made that day. This is a sample of size ...

0

It's correct. Note that you are using that the CDF of $Z$, $F_Z(z)=P(Z\leq z)$, is injective to conclude that $$F_Z(-1)=F_Z(-6/\sigma) \;\;\Longrightarrow \;\; -1=-6/\sigma.$$

0

A Monte Carlo simulation easily gives you access to the mean and variance of the variable you've been simulating, and to the probability of this variable being higher/smaller than a given value. You can also try to guess if the distribution you see can possibly correspond to a standard law, and use some tests to confirm it.

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Conditioning on $n_1=n$ means that when the condition is applied, $n_1$ is constant and no longer random, so no random variable is correlated with it. You then let $n$ vary over all possible values and integrate against the probability of each of these values occurring.

0

For small sample sizes, usually the Clopper-Pearson exact CI is appropriate, although it should be emphasized that it will frequently have a coverage probability well in excess of the desired confidence level. http://en.wikipedia.org/wiki/Binomial_proportion_confidence_interval

0

I might be missing something, but in an $\overbrace{n \times n \times \cdots \times n}^m$ Latin hypercube, there are $n$ distinct symbols each occurring exactly $n^{m-1}$ times. Thus, a random cell will contain any given symbol with probability $\tfrac{n^{m-1}}{n^m}=\tfrac{1}{n}$.

1

You don't need to draw the curve. You need to determine the mean $\mu$ and standard deviation $\sigma$ of the Normal Distribution since $\mu, \sigma$ define it completely. You then use $\mu, \sigma$ to determine the 75th percentile.

1

$$E \left[ f\left(\frac X {|X|}\right) g(|X|) \right]= \frac 1 {Z^n}\int f \left(\frac x {|x|}\right) g(|x|) \exp \left(-\frac 12 |x|^2\right) dx_1\cdots dx_n$$ now note $x = r x'$, with $|x'| = 1$ and $r>0$, and $M(r,n)$ beeing the size of the sphere of $\mathbb R ^n$ of radius $r$ yields $$E \left[ f\left(\frac X {|X|}\right) g(|X|) \right]= \frac ... 0 You are right about X_1^2+\cdots+X_n^2, and there's really nothing more you need to say about it, unless the definition of the \chi^2_n distribution that you're working with is something other than the distribution of the sum of squares of independent standard normally distributed random variables. For the distribution of X/\|X\|, I think I might ... 0 I took an easier approach and I think I got it. Just to be a bit more general I'll be using (\mu_1, \mu_2 )  as the means of (\epsilon_1, \epsilon_2)  even though the question stated they were zero. Write the standard normal PDF and CDF by \phi and \Phi. Define$$Pr(\epsilon_2 > \alpha) = 1 - F_2 (\alpha) $$and multiply and divide by this term: ... 1 (z+3)^2 is a noncentral \chi_1^2, so (z+3)^2-8 is a noncentral \chi_1^2 with a location and scale shifts. 0 You need a normalization factor for the integral. Basically the right hand side is the weighted sum of some conditional densities (Gaussians with possibly different mean and variance) then the normalization should be the sum of weighing factors. Namely it should read:$$ f(x_1 | x_2 > \alpha) =\frac{\int_\alpha^\infty g(x_1 | x_2 = u) \; f_2 ...

2

Yes, you're essentially correct. There's one important thing you've implied by saying "random" but haven't stated explicitly, which is that the random variables must be independent. (It is often stated that the variables must be identically distributed as well as independent, but this can be relaxed somewhat.) Also, there's a caveat that the mean and ...

1

Here is a quick example in c++ of an implementation. Let me know if it helps. If not I will remove the answer :).

1

Yes, in general the probability of a random variable $X$ being in the interval from $x_0$ to $x_1$ is given by: $$P(x_0 < X < x_1) = \int_{x_0}^{x_1} f(x)\;dx$$ This represents the area under the graph of the pdf $f(x)$. Also note that it doesn't make sense with a continuous distribution to talk about the probability $P(X=x)$. This is zero because ...

2

By definition, we have $$f(x) = \frac{1}{\sqrt{2\pi \sigma^2}} \exp \left(- \frac{(x-m)^2}{2\sigma^2} \right).$$ Hence, $$\int_{-\infty}^0 e^{rx} \cdot f(x) \, dx = \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^0 \exp \left(rx - \frac{(x-m)^2}{2\sigma^2} \right) \, dx.$$ Now write \begin{align*} rx - \frac{(x-m)^2}{2\sigma^2} &= ... 0 Maybe V=\cos(\Theta) ? \arcsin (V) = \pi/2 - \Theta for \Theta \in [0,\pi) \arcsin (V) = \Theta - 3\pi/2 for \Theta \in [\pi, 2\pi) So \arcsin (V) is a uniform distribution on (-\pi/2, \pi/2), am I correct? 0 The sample variance (S^2) satisfies that(n-1)S^2/\sigma^2 \sim \chi^2(n-1)$$Then Var(S^2)=Var\left(\frac{(n-1)S^2}{1}\chi^2(n-1)\right)=\frac{\sigma^4}{n-1} In your case \sigma^2=1. you can find more here almost at the bottom of the page 0 First, for the calculation we will need the well-known result that$$ \int_{-\infty}^\infty e^{-ax^2} \; dx = \sqrt{\pi/a} $$when a>0. Also, in general we have that if X is a random variable with density function f(x) then, for any function h of X we have$$ \DeclareMathOperator{\E}{E} \E \left( h(X) \right) = \int_{-\infty}^\infty ...

0

In the link is a derivation of the chi-squared distribution, that does not however make use of moment generating functions, see Wikipedia. The moment generating function (mgf) of the chi-squared distribution can be found here or here but I do not know of a proof that uses the mgf to prove the above statement.

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Yes, because you can't choose numbers not within 1 standard deviation, the numbers you can choose have higher probabilities. The probability of numbers within the interval is the same, just scaled by $1/0.623$, or the probability a number is within 1 standard deviation.

1

Let $f(x)$ be the density function of the original normal $X$. The probability that you "keep" a number is the probability that the number obtained is between $\mu-\sigma$ and $\mu+\sigma$. This is approximately $0.6826$. The resulting truncated distribution $Y$ has density function which is $0$ outside the interval $[\mu-\sigma,\mu+\sigma$. Inside the ...

2

$P[X>16\vert X>10]$ is the proportion of that part of the probability mass of $X$ which is above $10$ which is also above $16$. Draw the pdf of $X$. Lightly shade that part which is above $10$ (this is the right hand part of the distribution). Now, heavily shade the part which is above $16$. The conditional probability is the proportion of the ...

3

Hint. Start with the usual conditional probability formula, $$P(A\mid B)=\frac{P(A\cap B)}{P(B)}\ .$$ In this case $A\subseteq B$, so $A\cap B=A$. See if you can finish the calculation from here.

0

Firstly, we establish some notation. We will denote the $n$th vector of the sequence thus: $X^n$, rather than thus: $X_n$ as in the original post. We will denote by $N$ the common dimension of $X$ and the $X_n$'s, i.e. $X, X_1, X_2, \dots \in \mathbb{R}_{N \times 1}$. For every $k \in \left\{1, \dots, N\right\}$, $X_k$ will denote the $k$th component of ...

0

For b) you need that the sample variance $S^2$ has the chi-squared distribution with $n-1=8$ degrees of freedom. You need to determine the points $\chi^2_{8;0.05}$ and $\chi^2_{8;0.95}$ from the chi squared distribution which can be found in the tables at the back of your book (I guess). They are equal to: $$a=\chi^2_{8;0.05}=2.733$$ and ...

0

(a)The distribution of $\bar{X}$ for $n=9$ in Normal with mean $\mu_n=8.78$ (the same as $X$) and variance $\sigma_n^2=\frac{0.16}{n}$. That follows from the additive properties of the normal distribution. Now using that the probability density function of the normal distribution is $$f(x)=\frac{1}{\sqrt{2\pi\sigma^2}}\cdot e^{-\frac{(x-\mu)^2}{2\sigma^2}}$$ ...

1

If you write $F(u)=\int_0^u e^{-\frac{x^2}{2}}dx$, then $F$ is $\mathcal C^1$ with derivative $e^{-\frac{u^2}2}$, and $$\int_t^{t+\varepsilon}e^{-\frac{x^2}{2}}dx =F(t+\varepsilon)-F(t) .$$ So $\int_t^{t+\varepsilon}e^{-\frac{x^2}{2}}dx$ is equivalent to $\varepsilon \, F'(t)=e^{-\frac{t^2}2}\,\varepsilon$ as $\varepsilon\to 0$. Your denominator seems to be ...

0

Hint: for $X\sim \mathcal{N}(\mu,\sigma)$ and $Y:=e^X$, then $Y$ is lognormal distributed with mean $E[Y]=e^{\mu+\frac{\sigma^2}{2}}$. If you don't know this result, it would be a good exercise.

0

To plot the histograms, try something like this: Nbins = 256; figure; subplot(3,1,1); hist( vec(im(:,:,1)), Nbins ); title( 'Channel R' ); subplot(3,1,2); hist( vec(im(:,:,2)), Nbins ); title( 'Channel G' ); subplot(3,1,3); hist( vec(im(:,:,3)), Nbins ); title( 'Channel B' ); To estimate the probability that a pixel is within some range of values, ...

2

Note that the algebraic identity $$\lambda t+\frac{(t-\tau)^2}{2\sigma^2}=\tau\lambda-\frac12\sigma^2\lambda^2+\frac{(t-\tau+\sigma^2\lambda)^2}{2\sigma^2}$$ and the change of variable $s=t-\tau+\sigma^2\lambda$ yield $$\int_{-\infty}^{\infty}\exp\left(-\lambda t\right)\,\exp\left(-\frac{(t-\tau)^2}{2\sigma^2}\right)\mathrm ... 1 It can be done like this:$$\mathbb{E}[X_1 X_3] = \mathbb{E}[X_1 (\rho X_1+\sqrt{1-\rho^2}X_2] = \mathbb{E}[\rho X_1^2 +\sqrt{1-\rho^2}X_1X_2]  = \rho\mathbb{E}[ X_1^2] +\sqrt{1-\rho^2}\mathbb{E}[X_1X_2] $$Here I can use the relation \mathbb{E}(X^2)=\text{Var}(X)+(\mathbb{E}[X])^2 to get$$=\rho + \sqrt{1-\rho^2}\mathbb{E}[X_1X_2].$$Since X_1 ... 2 In your parametrization, the random variables X_1 and X_2 (where X_i=[X_{i1},X_{i2}]) are not correlated, therefore you can treat them independently (for each column of your matirx X), i.e. use the mean and sample variance as the estimators. In additions: If you change your parametrization, and allow a full covariance matrix \Sigma then you can ... 0 There are several things to consider. First, there is a question about which test statistic to use. The z test statistic requires that you know the population standard deviation, which is implausible except when the data are drawn from a Bernoulli distribution. The second question is what distribution to use for your critical values. The choices are ... 0 If n is large enough (in general, n \geq 30, then you can use a z-test. If you don't know \sigma, you can still use s. The t-test is used when n < 30. Based on that information, you should be able to figure out four of these questions. However, two questions here are a little different. Depending on what you mean by "z-test" and "t-test" ... 0 To C. According to the Central Limit Theorem you can use the normal distribution (as an approximation) as long as your sample is big enough (i.e >>30), independently from the initial distribution of the population that you have drawn the sample. So, no it is not required. To B. This condition implies that your probability of success in the sample ... 1 You both agree that A has to hold: you need independence to avoid bias, reduce your true standard error -- and in this case to be able to approximate using the normal distribution (see B, below). Now to C. The population distribution isn't normal: your model is that there is a certain proportion of the population that has a characteristic. So you take an ... 0 In general if X_{1},X_{2} are independent and have pdf F_1, F_2 respectively then:$$P\left\{ \min\left\{ X_{1},X_{2}\right\} \leq x\right\} =1-P\left\{ X_{1}>x\wedge X_{2}>x\right\} =1-\left(1-F_{1}\left(x\right)\right)\left(1-F_{2}\left(x\right)\right)$$Can you take it from here? Also see the comment from Peder on this question. 0 In general, if Y = max({X_1,X_2}), then:$$ F_n(y) = P[(X_1\leq y)\cap (X_2 \leq y)]$$Will that help to formulate your new distribution for Y? 2 Define w=y-Cx, then$$ E[\mathrm e^{\mathrm iu^Tw}|\mathcal F^x] =E[\mathrm e^{\mathrm iu^Ty}|\mathcal F^x]\,\mathrm e^{-\mathrm iu^TCx} = \mathrm e^{-u^T \tilde{Q}u/2} $$is independent of x. Thus w is independent of x and centered normal with covariance \tilde{Q}. 2 For every positive a, the distribution of T=|X|^a has density$$ \frac2{a\sqrt{2\pi}}t^{(1/a)-1}\exp\left(-\tfrac12t^{2/a}\right). $$This follows from the usual change of variables method. For example, the distribution of Z=X^4 has density$$ \frac{1}{2\sqrt{2\pi}z^{3/4}}\exp\left(-\tfrac12\sqrt{z}\right). $$0 We have$$X = 6 - Z_x,\;\;\; Z_x \sim \mathcal{N}(0,\,1)Y = 7.5 - Z_y,\;\;\; Z_y \sim \mathcal{N}(0,\,1)|Y-X|<1$$Using the first two, the 3d relation is written$$|1.5 - Z_y + Z_x|<1 which makes clear that $Z_x$ and $Z_y$ may have identical marginal distributions, but they cannot be the same random variable. Analyzing it further ...

0

Expectation is always linear. So for any two variables $X,Y$, we have $E[X+Y]=E[X]+E[Y]$. And $E[\underbrace{X+X+\ldots+X}_{k \text{ times}}]=E[kX]=kE[X]$ Variance is linear when the variables are independent. In this case, $V[X+Y]=V[X]+V[Y]$. However when the variables are the same, i.e. when we scale, we have $V[kX]=k^2X$. These are true no matter what ...

0

You make a mistake assuming that $X+X = 2X$ like in arithmetic. This does not hold for random variables. As a simple counterexample take $X$ to be a Bernoulli trial with succes probability $0.5$. If $X+X=2X$ then you cannot flip two coins and get heads on one and tails on the other. Summing random variables is given by a convolution.

3

$X_n\overset{p}{\rightarrow} X\Rightarrow X_{n}\overset{d}{\rightarrow }X\Rightarrow \Phi_{n}(x)\rightarrow \Phi(x)$ pointwise where $\Phi_n(\cdot),\ \Phi(\cdot)$ are the characteristic functions of $X_n,X$ respectively. Now, Since $X_n$ are Gaussian, $\Phi_n(x)=e^{jx^T\mu_n-x^TC_nx/2}$ where $\mu_n:=EX_n,\ C_n:=\mbox{Cov}(X_n)$ Then, since $\exp(\cdot)$ is ...

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