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The standard undergraduate text for real analysis is Rudin's Principles of Mathematical Analysis (affectionately referred to as "Baby Rudin" since he wrote it when he was quite young). Another text I enjoyed was Serge Lang's Undergraduate Analysis. I think they're both have their pros and cons but are ultimately both fine books for first learning some ...

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Let $y_i = x_i - \bar{x}$ then by Cauchy-Schwarz we have the following: $\sum_{i=1}^n |y_i| \frac{1}{n} \leq \|y\|_2 \sqrt{\sum_{i=1}^n \frac{1}{n^2}} = \sqrt{\frac{1}{n}\sum_{i=1}^n |x_i - \bar{x}|}$

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1-) The meaning of ∫P(Q)*dQ is the probability of finding the value of Q, according to the limits of the integration. 2-) The use of dR in the proof is fundamental to simplify the description of the shield of the sphere. Since we are talking about a sphere, nothing more appropriate than use spherical coordinates. In cartesian coordinates, it would be very ...

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Given that all $X_i$ are at most $s$, all values of $X_1$ from $1$ to $s$ are equiprobable, so the expected value of $X_1$ is $\frac{s+1}2$, and the expected value of $T_1$ is $2\cdot\frac{s+1}2-1=s$. The event that the maximum is $t$ is the event that all $X_i$ are at most $t$ minus the event that all $X_i$ are at most $t-1$, and the probabilities for those ...

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We get the same answer for any continuously distributed random variable. For let $M$ be a median (medians need not be unique). The probability that $X_i$ is $\gt M$ is $\frac{1}{2}$. The probability that the minimum is $\gt M$ is therefore $\left(\frac{1}{2}\right)^3$. Remark: Things can break down if the distribution is not continuous. For example, let us ...

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The step in question employs a trick called "differentiating under the integral." The idea is to introduce a parameter in the integrand, and express the integrand as a derivative with respect to this parameter, then change the order of integration and differentiation (assuming certain regularity conditions hold). Explicitly, suppose we let $f(y,\alpha) = y^... 1 The conclusion is correct, but the derivation isn't quite correct, because you applied the variance sum rule that holds for independent variables, but$\bar X$and$X_1$aren't independent. One way to resolve this is to take the contribution$\frac2nX_1$out of$2\bar X$and combine it with$-X_1$; then you have two independent variables and can apply the ... 0 You may also want to look at the paper linked here: http://papers.ssrn.com/sol3/papers.cfm?abstract_id=1937879 It details how to consistently estimate the signal and noise variances (as well as measurement parameters, if you have those) using an EM algorithm. The application may not be generalizable to yours, though, so beware. 1 Assume$H_0$is true, then you can compute the law of your test statistic (you must have done this) and then you come up with the expression of that critical region. By definition of critical region $$P_{H_0} (X \in CR) = \alpha$$ Where$P_{H_0}$means that you are computing this probability assuming$H_0$true,$\alpha$is the level of the test (roughly ... 1 Your calculation is confusing. What is$P(X^2=-1,-1)$supposed to be?$X^2$is a random variable with only one possible value, and that values is$1$.$X$has a value of$1$with a probability of$1$(because there is a$0.5$probability that$X$is$1$-in which case$X^2$is$1$, and a$0.5$probability that$X=-1$, in which case, also,$X^2=1$) 0 In addition to the oversight already mentioned in the comments, the numbers of options for the national blocks should be multiplied, not added (see multiplication principle). So the probability is $$\frac{3!\cdot4!\cdot3!\cdot6!}{13!}=\frac1{10010}\approx10^{-4}\;.$$ 1 Since all you have are the means and the covariances, the most natural thing to do would seem to be to assume a multivariate Gaussian with those parameters, plug in$x_2,\ldots,x_n$and normalise to get the conditional distribution for$x_1; the estimate would then be the mean of that distribution (which is the same as its mode and median, since it's normal)... 0 This is late but I'm wondering if I understand things correctly as well. The way I see it, the sample mean is obtained from other sample means and is used to predict the population mean and variance, as well as the mean and variance for other samples. Anyway, I think that it's N-1 because a sample of size 1 has a mean, itself, but doesn't have a variance. ... 1 Let me try to understand what you are doing: You sum up the individual values of the vector, you divide each value by the sum, and voila... they sum to 1. First the sum: $$S(x) = \sum_i x_i$$ Then the described normalization: $$x' = x / S(x)$$ So $$S(x') = \sum_i x_i' = \sum_i \frac{x_i}{S(x)} = \frac{1}{S(x)} \sum_i x_i = \frac{1}{S(x)} S(... 1 There is no reason why you should expect 1 - ratios to lead to a vector where the sum of the elements is 1. In particular, if you have the initial array [a,b,c,d] with a + b + c + d = 1, then 1 - [a,b,c,d] = [1-a, 1-b, 1-c, 1-d] (where I use equality in the numpy sense). When you sum these elements together, you get 1-a + 1-b + 1-c + 1-d = 4 - (a + ... 0 Here's the simplest thing I can think of: Suppose you have K>1 tasks. A tasks was attempted N_k^t times at time t and successfully solved n_k^t\le N_k^t times at time t. The "success rate" of task K is therefore s_K^t:=n_k^t/N_k^t at time t. Now the simplest ranking would be to order these s_K^t; smaller rank for higher success rate (... 1 Let's see what we have: P(A_{calc})=\frac{1}{4} P(A_{prob}|A_{calc})=\frac{3}{2}p, where p=P(A_{prob}|A_{calc}^C) (complementary probability). We can calculate this p, since P(A_{prob}|A_{calc})+P(A_{prob}|A_{calc}^C)=1. We have 1.5p+p=1\Rightarrow p=\frac{2}{5}. Now, we need to calculate P(A_{calc}|A_{prob}). Using Bayes' theorem, we have ... 1 If the discrete process is a Markov process with no state (that is, if the number of success events is simply the result of N independent trials, then the number of successes is described by a Bernoulli distribution with p = \mu/N, and indeed this is well approximated by a Poisson process if N is large and \mu N is not too small. The exact answer ... 0 The Normal Approximation to the Binomial Distribution can be Defined as: P(a \leq X \leq b)\approx \phi(\frac{b+\frac{1}{2}-\mu}{\sigma})-\phi(\frac{a-\frac{1}{2}-\mu}{\sigma}) So to find what you're looking for is P(X \geq a)= 1-\phi(\frac{a-\frac{1}{2}-\mu}{\sigma}) The rest is just computing values. a=60 \mu=np=(100)(0.5)=50, \sigma=... 1 You probably meant that a \textit{pair} of independent variables are highly correlated (0.9). Basically, although technically it's fine, this may indicate that both variables measure (almost) the same factor. Hence, generally speaking, bring the same information to the model, as such may cause redundancy. The fact that both variables are significant may ... 0 First, we can considerably simplify your analysis of the k=2 case. By comparing an arbitrary number of samples from the distribution with the given value v, you're effectively finding the cumulative distribution function F(v) at that value. The strategy is to switch if this is less than \frac12, and the success probability is 1 if the two values in ... 2 min! is the argmin, i.e., f(x)=\min! is identical to \arg \min f(x). See also http://mathoverflow.net/questions/182112/a-question-about-some-notation-involving-the-exclamation-mark and the link there, which discusses your example. 1 Your reasoning is correct. Example: If X_1, X_2, \ldots, X_n are iid from the density$$ f_\theta(x) = {c\over 1 + |x-\theta|^3},$$where c is a normalizing constant, then by the SLLN the sample mean converges almost surely to E_\theta(X)=\theta, hence \bar X is a consistent estimator for \theta. However, the distribution lacks a second moment so \... 2 The probability that some two men choose the same number is the complement of the probability that no two men choose the same number. The number of ways to choose the three men's numbers such that they are all different is$$\binom{20}{3}\binom{17}{3}\binom{14}{3}\binom{11}{3} = 1140\cdot 680\cdot 364\cdot 165$$The number of ways to choose the same ... 2 Note that we have$$\begin{align} \lim_{x\to n^-}\sum_{k=0}^x p^k(1-p)^{n-k}&=\lim_{x\to n^-}(n-x)\binom{n}{x}\int_0^{1-p}t^{n-x-1}(1-t)^x\,dt\\\\ &=\lim_{y\to 0^+}\left(y\frac{\Gamma(n+1)}{\Gamma(n+1-y)\Gamma(y+1)}\int_0^{1-p}t^{y-1}(1-t)^{n-y}\,dt\right) \tag 1 \end{align}$$Now, let's examine the limit,$$L=\lim_{y\to 0^+}\left(y\int_0^{1-p}t^{... 0 My instinct is that the RHS will actually be0 \cdot \infty$since your integral will no longer converge. So, rather than being an incorrect equation, it'll just become an ill-posed one. :D 0 I don't think I have enough of your problem to answer that question, but if you look at Huber's Robust Statistics or Keener's Theoretical Statistics, the sections on M-estimators might help. For example, in Keener, the maximum likelihood estimator is characterized as when the derivative of the likelihood equals 0. They are able to do proofs using a Taylor ... 0 Let's write this out: The CLT says that if$X_j$are iid with finite expected value$\mu$and finite variance$\sigma^2$, then the sequence $$Y_n = \sqrt{n}\cdot\left({\frac{X_1+\ldots+X_n}{n}-\mu}\right)$$ converges in distribution to$N(0, \sigma^2)$. Now$n$here is not actually a variable, it is merely an index. So saying that$Y_n/n$converges in ... 1 This is a "fake" proof by contradiction ,ie you don't need to assume something is false to run the proof. You are just saying$x_1 + ... x_n \leq n (\max\{x_1,...,x_n\}) $0 I take it that at least one same sex group must be there. It is enough to distribute the girls into$20$slots, the boys automatically get remaining slots. Count the unfavorable ways using [ Choose slots for pattern ]$\times$[ permute ]$3-3-2-1-1:\; \binom43^2\binom42\binom41^2\frac{5!}{2!2!}3-2-2-2-1:\; \binom43\binom42^3\binom41\frac{5!}{3!} ...

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We are dealing with multivariate hypergeometric distribution. Give the groups numbers $1,2,3,4,5$. Let $\mathbf{X}=\left(X_{i,j}\right)$ denote a $5\times2$ matrix of random variables. Here $X_{i,1}$ denotes the number of girls in group $i$ and $X_{i,2}$ denotes the number of boys in group $i$. Let $S$ denote the collection of $5\times2$ matrices $\mathbf{... 0 Hint (full answer later, as usual): You have a group of 20 people, and you need to find the odds of choosing a group of 4 in which all are boys. The beauty of this type of problem is that it lends itself to visualization on paper well. Form a group one person at a time, repeating the question "What are the odds of choosing a boy from the remaining people?" ... 0 Are your sure it's 88 people altogether, not 66? I get four disjoint categories$A \cap B = AB$with 11,$BA^c$with 6,$AB^c$with 10, and$(A\cup B)^c$with 39. Then 11 + 6 + 10 + 39 = 66. 0 Here's one way. This will work only if you understand matrix algebra and the geometry of$n$-dimensional Euclidean space. The model says$y_i = \alpha_0 + \sum_{\ell=1}^k \alpha_\ell x_{\ell i} + \varepsilon_i, \quad i=1,\ldots,n $where$y_i$and$x_{\ell i}$are observed; The$\alpha$s are not observed and are to be estimated by least squares; The$\...

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The number of people who miss their flight has binomial$(n=120,p=.3)$ distribution, so the mean is $np$ and the variance is $np(1-p)$ and the SD is $\sqrt{np(1-p)}$. These work out to: mean = $120\cdot0.3=36$, SD = $\sqrt{120\cdot0.3\cdot 0.7}=\sqrt{25.20}=5.02$.

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We certainly CAN use whatever outlier bound we wish to use, but we will have to justify it eventually. In the not-so-recent past, it was typical to expect distributions to be Gaussian. With that assumption, ±1IQR is too exclusive, resulting in too MANY outliers, ±2IQR is too inclusive, resulting in too FEW outliers. ±1.5IQR is easy to remember, and is a ...

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The terms used in the question is not accurate and here is my interpretation. You have demonstrate how to express the mean of a autoregressive model of order $1$ ($AR(1)$ model) in terms of the given parameters. Here we should have assumed that the process/time series has the same mean throughout the all $n$ (or a stronger assumption of statationarity). And ...

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A null hypothesis that is rejected at the $1\%$ level is necessarily rejected at the $5\%$ level, or equivalently, a p-value that is less than $1\%$ is necessarily less than $5\%$. But a null hypothesis that is rejected at the $5\%$ level may not be rejected at the $1\%$ level, i.e. a p-value that is less than $5\%$ may not be less than $1\%$. The ...

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When you perform a hypothesis test, you get a number out at the end, called the p-value. This is the probability that, under the null hypothesis, you could get results at least as far from the expectation as the ones you actually got. For example - you flip a coin 10 times, get 8 heads, your null hypothesis is that P(H) = 0.5, the p-value is the probability ...

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The standard significance level is $5\%$, which indicates a $5\%$ chance of concluding that a difference exists when there actually is no difference at all. Thus, lower levels of significance reduce the risk of an incorrect conclusion. The $P$-value on the other hand indicates the probability of obtaining such an extreme value, assuming the null hypothesis ...

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Your results are not statistically significant. If you limit the number of trials, the expected winnings is finite. For $n$ trials we have $$E(W_n) = \sum_{k=1}^n2^{-k}2^k = n.$$ So you should see expected winnings of $10$ for $10$ trials, $100$ for $100$ trials, etc. However, $$E(W_n^2) = \sum_{k=1}^n2^{-k}2^{2k} = \sum_{k=1}^n2^{k} = 2^n - 2$$ ...

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No. Take $(X_n)_{n\in\mathbb{N}}$ s.t. $X_n$ does not converge in distribution to $X$. Take $g:\mathbb{R}\to \mathbb{R}$ s.t. $g(x)=0$ for all $x\in\mathbb{R}$. Then $g$ is continuous and $g(X_n)$ converges in distribution trivially to $g(X)$.

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I have found the answer, the steps are: $$E(X^r) =\frac{\Gamma(\alpha + \beta) \lambda^\alpha}{\Gamma(\alpha) \Gamma(\beta)} \int^1_0 x^{\alpha + r - 1}(1-x)^{\beta-1}[1-(1-\lambda)x]^{-(\alpha+\beta)} dx$$ Using binomial expansion, i.e. $(1-ax)^{-b} = \sum^{\infty}_{k=0} \frac{\Gamma(b + k)(ax)^k}{\Gamma(b)k!}$ the integral is rewritten as $$E(X^r) =\... 0 Let x be the expected number of flips to get (HT)^5 and y the expected number of further tosses required if we are starting with XH where we cannot use part of X to form the first part of the required run (eg X is empty or ends in H). The first toss is H or T, so x=\frac{1}{2}(1+y)+\frac{1}{2}(1+x). Hence y=x-2. How consider the ... 0 Hint Let X , Y are two absolutely continues random variables with joint density function f_{X,Y} and Z=X+Y , we have$${{F}_{Z}}(z)=P\,(Z\le z)=P\,(X+Y\le z)=\iint\limits_{x+y\le z}{{{f}_{X,Y}}(x\,,y)\,dy\,dx}=\int_{-\infty }^{+\infty }{\,\int_{-\infty }^{z-x}{f(x\,,y)\,dy\,dx}}$$let y=t-x\,, therefore$${{F}_{Z}}(z)=\int_{-\infty }^{+\infty }{\...

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When working with sum of i.i.d random variables, it is often convenient to use characteristic functions as $\phi_{X+Y}(t) = \phi_X(t)\phi_Y(t) = \phi_X(t)^2$. For example, if $X,Y$ are Binomial($n,p$) then $\phi_{X+Y}(t) = \phi_X(t)^2 = \left((1-p+pe^{it})^{n}\right)^2 = (1-p+pe^{it})^{2n}$ Hence $X+Y$ is Binomial($2n,p$). You can use the same reasoning ...

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If I´m understand it right you want to calculate $P(Y\leq 1175)$, where $Y$ is normal distributed as $Y\sim\mathcal N(1200,35^2)$. You can transform the random variable $Y$ to $Z=\frac{Y-\mu}{\sigma}=\frac{Y-1200}{35}$. Then $Z$ is standard normal distributed: $Z\sim \mathcal N(0,1)$. Thus we have $P(X\leq 1175)=\Phi\left(Z\leq \frac{1175-1200}{35} \right)=... 1$C_n$is an interval with random endpoints, denoted$a$and$b$. Both the endpoints are functions of your sample$X_1, X_2,\ldots, X_n$, and the joint distribution of the$X$'s is parametrized by$\theta$, hence the subscript on$P_\theta$. The parameter$\theta$that governs this joint distribution is nonrandom, and generally unknown (and the mission of the ... 1 You should interpret the statement Experience suggests that the standard deviation is more stable than the mean as the assertion that even when the machine is not working properly, the standard deviation of a package weight remains unchanged. Judging from the context, it's a statement about the population$\sigma$, not the sample standard deviation. So the ... 1 Recommendations for use of the Yates correction differ depending on textbook and field of application. If you are supposed to do a Yates correction for$2 \times 2$tables, then it does not matter whether that$2 \times 2\$ table was 'derived' from a larger table. (Some people feel, I believe with good reason, that the Yates correction is too 'conservative'--...

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