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Your sum is not correct. Your $E[S]$ includes the $\frac14$ component applied recursively, which doesn't make sense. EDIT: David's answer (which appeared during this one) is correct. I'll address your second question instead. The $E[Y]$ for the four numbers which end the drawing is obviously $\frac14$. :) Hopefully you can figure out why. For the ...


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We have $$E(S)=\sum_{k=1}^{13}E(S\mid\hbox{first draw is $k$})P(\hbox{first draw is $k$})\ .$$ But if the first draw is $k=1,\ldots,9$ then the expected sum is $k$ plus the overall expected sum; if the first draw is $k=10,\ldots,13$ then the expected sum is just $k$. So $$E(S)=\frac{1}{13}\sum_{k=1}^9(k+E(S))+\frac{1}{13}\sum_{k=10}^{13}k\ ,\tag{$*$}$$ and ...


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One of the most classical examples might be the following: assume that $(X_i)_{1\leqslant i\leqslant n}$ is i.i.d. exponential with parameter $\lambda$, then the likelihood is $$\ell(x_1,\ldots,x_n)=\lambda^n\mathrm e^{-\lambda(x_1+\cdots+x_n)},$$ hence the MLE for $\lambda$ is $$\hat\lambda=\frac{n}{X_1+\cdots+X_n}.$$ For every $i$, $E(X_i)=1/\lambda$, ...


1

But if I sum up the $\tilde{f}$ the error would be really high $n\epsilon$... No. Actually, using the fact that every $x$ is nonnegative, one gets: $$(1-\varepsilon)f(\ )\leqslant\bar f(\ )\leqslant(1+\varepsilon)f(\ )\implies\frac1{1+\varepsilon}\sum_xx\bar f(x)\leqslant\sum_xxf(x)\leqslant\frac1{1-\varepsilon}\sum_xx\bar f(x)$$


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Your sum equals: $$\sum_{j=0}^{+\infty}\frac{(j+5)^3}{32\cdot 2^j}\binom{j+4}{j}.\tag{1}$$ Since: $$\sum_{j=0}^{+\infty}\binom{j+4}{j}x^j = \frac{1}{(1-x)^5},\tag{2}$$ by differentiating three times both terms of $(2)$, you get the sums $\sum_{j\geq 0}\binom{j+4}{4}j^k\,x^j$ with $k\in\{0,1,2,3\}$, hence recombining these pieces you get: ...


1

Since you are exploring a categorical variable (success-failure), the appropriate test is a chi-square test. You have to perform separate analyses for the four tasks. For each task, build a $2\times8$ contingency table where the 8 columns represent the different method (single or combined techniques) and the 2 rows represent successes and failures. Then fill ...


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For 1) if you've simulated the excess infections under the null then you have numerically estimated the sampling distribution. Just calculate the fraction of the simulation results that are $\geq 3.3$ or $\leq -3.3$ This is the two-sided p-value for your observed placebo excess. For 2) you are not using your simulation, but the $\mathcal{N}(0,12.3)$ ...


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If the old data are $x_1$ to $x_N$, then the old variance is given by $$\sigma^2_{\text{old}}=\frac{1}{N}\sum_1^N x_i^2 -\overline{X}_{\text{old}}^2\tag{1}$$ and the new variance is given by $$\sigma^2_{\text{new}}=\frac{1}{N+1}\sum_1^{N+1} x_i^2 -\overline{X}_{\text{new}}^2.\tag{2}$$ You can see that $\sum_1^N x_i^2$ can be recovered from ...


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The degrees of freedom used in each case is different, and they also depend on the specific type of test you are doing. Without a description of the meaning of the data and how you collected it, it is difficult to determine whether your calculations are valid. If your observations for the two groups are not paired, even though the per-group sample sizes ...


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You can exploit the formulas $$\overline x=\frac1n\sum x_i=\frac{S_1}n,$$ $$\sigma^2=\frac1n\sum(x_i-\overline x)^2=\frac1n\sum x_i^2-\overline x^2=\frac{S_2}n-\overline x^2.$$ Every time you get a new value, update $n$, the sum of $x_i$ ($S_1$) and the sum of $x_i^2$ ($S_2$). From these you can compute the current mean, variance and standard deviation. ...


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You can use a truncated maximum likelihood estimate assuming 800=$p\%$ top values of the population and $x_i$ are the 800 observed values: Let $N=\lceil \frac{800}{p\%}\rceil \;\;x^-=\min\{x_i\},\;\; L(\mu,\sigma)=\left\{\Phi\left(\frac{x^--\mu}{\sigma}\right)^{N-801}\right\}\prod\limits_{i=1}^{800}\phi\left(\frac{x_i-\mu}{\sigma}\right)$, where ...


1

$$\Gamma\left(\frac{m+1}{2}\right) = \frac{m-1}{2}\cdot\frac{m-3}{2}\cdot\dotsc\cdot\frac{1}{2}\cdot\Gamma\left(\frac{1}{2}\right),$$ and $\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$. So it cancelled against $\Gamma\left(\frac{1}{2}\right)$.


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Cluster $r$ is a set of points $\{x_1,x_2,\ldots, x_{n_r}\}$. To compute $D_r$, calculate the square of the Euclidean distance between every possible pair of points of cluster $r$, and add up all these numbers. This may look more familiar to you: $$D_r = \sum_{i=1}^{n_r} \sum_{i'=1}^{n_r}\|x_i-x_{i'}\|^2.$$ Then, we claim that the following identity holds ...


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I assume you mean that $E\left[ {\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{X}} \right]=E\left[ X|D \right]$. So first, note that if $E\left[ XY|D \right]<\infty $ and $X,Y$ are both measurable with respect to $D$, then $E\left[ XY|D \right]=YE\left[ X|D \right]$. Now, by definition:      ...


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Since $f_{x}=\frac{a-2y}{(a-x-y)^2}$ and $f_{y}=\frac{2x-a}{(a-x-y)^2}$, f does not have any critical points in the interior of the closed region bounded by the lines $x=0$, $y=x$, and $y=\frac{a}{2}$. On the boundary of this region, $f(x,y)=0$ on the line $y=x$; $f(x,y)=-1$ on the line $y=\frac{a}{2}$; and on the line $x=0$, ...


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If x < y <= a/2, then x - y is always negative but a - x - y is always positive. That means f(x,y) must be a negative number. So the best we can do is 0 - epsilon. Notice that the closer you get to x = y, the closer the numerator approaches zero while the denominator is bounded below by a. So the max of this function does not exist (its supremum is ...


1

One way of modelling what is going on here is via the General Linear Model. In your case, if we label the objects $1$ through to $5$, then supposing the $i$-th object $(i = 1, \ldots, 5)$ is given a score $a_i$, $b_i$, $c_i$ and $d_i$ in each of the four characteristics you have, and the total/overall score of the object is $Y_i$, we model the ...


1

Since $0\le x<y\le a/2$, we have that $x-y<0$ and $a-x-y>0$. Hence $f(x,y)<0$ everywhere on the given domain. However, for $x+y<a$, we have $\lim\limits_{x\rightarrow y}f(x,y)=0$. This means that $\sup f(x,y)=0$, but this supremum is never achieved.


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Thats what Wolfram Alpha says: http://www.wolframalpha.com/input/?i=maximize+%28x-y%29%2F%28a-x-y%29+on+0%3C%3Dx%3Cy%3C%3Da%2F2 So the maximum is basically where $$x-y\rightarrow 0$$


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This is a heuristic, and as such it comes with advantages, disadvantages, and assumptions. Let's go over the idea, see its assumptions, and see the advantages and disadvantages, and one possible (relatively easy) improvement. Start simple For the moment, let's forget that there are two ways of talking about how far through a cycle we are. Instead of both ...


1

Directly using the distribution function: $$P(X \geq 2\alpha \beta)= 1 - P(X < 2\alpha \beta)=1 - F(2\alpha \beta)\\=1-\frac{1}{\Gamma(\alpha)\beta^\alpha}\int_0^{2\alpha\beta}s^{\alpha-1}e^{-\frac{s}{\beta}}\, ds.$$ Let $u = s/\beta$. Then $$P(X \geq 2\alpha \beta)= 1- \frac{1}{\Gamma(\alpha)}\int_0^{2\alpha}u^{\alpha-1}e^{-u}\, du>0.$$ So ...


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 ...


0

Wouldn't GRR reflect the fecundity of a population while NRR reflects the difference between fecundity and mortality?


1

You are confusing a probability distribution with the distribution of a sample. The sample is what it is, you do not weight them. If, in fact, a certain point is more probable than another, then it should, on average, come up more often in a sample. For a simple example, consider a die where the sides are $\{1,1,1,2,3,4\}$ If I rolled the die $n$ times, I'd ...


1

I believe the best strategy for a problem of this kind would be to proceed in two steps: Fit a continuous time Markov chain model to the data by estimating the (infinitesimal) generator $Q$. Using the estimated generator and the Kolmogorov backward equations, find the probability that a Markov chain following the fitted model transitions from state $i$ to ...


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Random trials/Monte Carlo simulations are notoriously slow to converge, with an expected error inversely proportional to the square root of the number of trials. In this case it is not hard (given a programming language that provides big integers) to do an exact count of cases. Effectively the outcomes are partitions of the 15 balls into some number of ...


4

All in terms of binomial coefficients, where $\binom{n}{k}$ is interpreted as the number of combinations of $k$ items from a pool of $n$: ...


2

$$ \mathbb E \bar X = \mathbb E\frac{X_1+\cdots+X_n}{9} = \frac 1 9\left( (\mathbb E(X_1)+\cdots+\mathbb E(X_9)\right) = \frac 1 9 (5+\cdots+5) = 5. $$ $$ \operatorname{var}\bar X = \operatorname{var}\frac{X_1+\cdots+X_n}{9} = \frac{1}{81}( \operatorname{var}(X_1)+\cdots+\operatorname{var}(X_9)) = \frac{1}{81}(9 + \cdots+9) = 1. $$ Therefore $\mathbb ...


1

In order to know that the Rao-Blackwell theorem is applicable, you have to know that $\max$ is sufficient, i.e. the conditional distribution of $X_1,\ldots,X_n$ given $\max$ does not depend on $\theta$. The conditional distribution of anything given $\max$ is a function of $\max$. Since $\mathbb E(2X_1)= \theta$, the law of total expectation implies ...


2

The answer is: $$\frac{365!}{320!365^{60}}\frac{60!}{32!3!3!2!^{11}}\frac{1}{2!11!}$$


0

Whether to use a $t$-test or $z$-test depends on whether the standard deviation of the population from which the sample is drawn is known. In your case, you are testing whether the average lifespan $\mu_d$ of doctors is less than the average lifespan $\mu_g$ of the general population; i.e., your hypothesis is $$H_0 : \mu_d = \mu_g, \quad \mathrm{vs.} \quad ...


2

If $\max(X_i) = t$ then one of the list of $\{X_i\}$ is equal to $t$ and the other $n-1$ are all less then $t$. The chance that $X_1$ is that one variable is $\tfrac 1 n$. If it is, the conditional expected value is $t$. If it isn't the value is uniformly distributed on $[0, t)$, and the conditional expected value of that is $t/2$. Double the answer ...


1

The confidence interval/hypothesis test relationship goes that if 0 is not in the C% confidence interval for a difference, then there is a significance difference at the (100-C)% significance level. So if 0 is not in your 85% confidence interval, then there is a significant difference at the 15% confidence level. I suspect the person who wrote the question ...


1

I think this should work: let's first calculate the probability that the people enter the room one at a time to form a sequence $T_1, T_2, D_1, D_2, \ldots, D_{11}, S_1, S_2, \ldots, S_{32}$, where $T_i$ is the $i$th triplet of birthdays, $D_i$ is the $i$th double, and $S_i$ is the $i$th singleton. For $T_1, T_2$, we'll have $(\frac{365}{365} \cdot ...


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The best choice might depend on which type of book you would prefer. In my opinion: If you want to privilege clarity, I would suggest you "Models for Probability and Statistical Inference: Theory and Applications" by James H. Stapleton: this is a relatively short but clear and comprehensive book on probability and statistical inference, with a lot of ...


0

1) There are two common definitions of the Wilcoxon rank-sum statistic, that have been around since the start (each appears in one of the first two papers that relate to the Wilcoxon tests). One of those is the sum of the ranks in the smallest sample, which sounds like the one you're used to. The Wilcoxon rank-sum statistic and the Mann-Whitney U statistic ...


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Proof by induction: For every complete graph $K_{2n}$ with $2n$ vertices, there is a labeling of the edges with ${-1,+1}$ such that every vertex has $n$ edges labeled $+1$ and $n-1$ edges labeled $-1$. The claim is trivially true for the null graph since there are no vertices. For the graph $K_2$, label the single edge $+1$. Take a graph $K_{2n}$ labeled ...


3

Solution of Question 1: This is an occupancy problem with $n=30$ boxes and $k=15$ balls. Let's first consider the expected number of empty boxes. That is much easier to obtain. The exact answer is $30(1-1/30)^{15}=18.04.$ This is approximately $30/\sqrt e.$ See the answer by Mr.Spot to this question: Making 400k random choices from 400k samples seems to ...


1

Clearly, for $u<0$, $F_U(u) = \mathbb{P}(\max\{0,X\} \leq u)= 0$. Now, for $u\geq 0$, \begin{align} F_U(u) &= \mathbb{P}(\max\{0,X\} \leq u) \\ &=\mathbb{P}(X<0)\mathbb{P}(\max\{0,X\} \leq u|X<0) + \mathbb{P}(X\geq 0)\mathbb{P}(\max\{0,X\} \leq u|X\geq0)\\ &= \mathbb{P}(X<0) + \mathbb{P}(X\geq 0) \mathbb{P}(X \leq u |X\geq0)\\ ...


2

$F_{U}\left(u\right)=P\left\{ \max\left\{ X,0\right\} \leq u\right\} =P\left\{ \omega\in\Omega\mid X\left(\omega\right)\leq u\wedge0\leq u\right\} $ and $F_{X}\left(u\right)=P\left\{ X\leq u\right\} =P\left\{ \omega\in\Omega\mid X\left(\omega\right)\leq u\right\} $ If $u<0$ then $\left\{ \omega\in\Omega\mid X\left(\omega\right)\leq u\wedge0\leq ...


0

Let $\mathcal{F}$ be the class of Lipschitz densities with Lipschitz constant bounded by $C>0$. Topologically, we regard $\mathcal{F}$ as a subspace of $L^{1}[0,1]$. We first show that $\mathcal{F}$ is closed in $L^{1}$. Suppose $(f_{n})_{n=1}^{\infty}\subset\mathcal{F}$ converges to $f\in L^{1}$. Passing to a subsequence if necessary, we may assume by ...


0

For the first inequality, you could (prove and) use the identity $$E(\mathcal O^n)=\int_\mathbb R(\mathbf 1_{x\gt0}-\Phi(x)^n)\,\mathrm dx,$$ and the fact that, for every $x$, the sequence $(\Phi(x)^n)$ is decreasing. (Hint: Start from the identity in your post and integrate by parts, using the functions $u=x$ and $v=\Phi(x)^n$.) The second inequality ...


0

The presence of $\sigma^2$ in your formulae is correct. I suspect a normalization $\sigma = 1$: this is a point to check in the reference-unfortunately I have no access to it-. In any case I would define some context. Let $Y=X\theta+\epsilon$ be a regression line with $$E[\epsilon]=0,~~\operatorname{Cov}(\epsilon)=E[\epsilon\epsilon^T]:=\sigma^2 I_n,$$ ...


3

Pretend there is a fifth student that sits on the empty sofa. Now there are two possibilities. The fifth student can sit the same place, then you are asking for a derangement of the four students. There are (closest integer to) $\frac {4!}e=9$ of these. Otherwise, the fifth student can move, and you need a derangement of five. There are(closest integer ...


1

Some context: a density means a nonnegative integrable function on $[0,1]$ with integral equal to $1$. Total boundedness is understood in the $L^1$ norm. Let $\mathcal F_C$ be the set of $C$-Lipschitz densities. I'll prove that it is totally bounded in the uniform norm $\sup|f|$, which will imply total boundedness in the $L^1$ norm (since the $L^1$ norm ...


0

I'll work straight from the definition. $T_1$ is sufficient for $\theta_1$ if $\theta_2$ is known. I take that to mean that the conditional distribution of the data given $T_1$ does not change when $\theta_1$ changes but $\theta_2$ remains fixed. Similarly the conditional distribution of the data given $T_2$ does not change when $\theta_2$ changes but ...


0

This question is similar to Motivation behind standard deviation? but since you specifically ask for a parallel between standard deviation and distance, here goes. The Euclidean distance, unlike for example Manhattan distance, is compatible with an inner product. The inner product of two vectors $x,y$ is $x\cdot y = \sum x_i y_i$ (also known as the dot ...


0

A simple solution, ignoring the effect on larger numbers, is to take your two numbers $x,y$ and compute $(e^x - e^y)/e^x$ instead of $(x-y)/x$. If the numbers are small this will return a value somewhere close to the range $[-1,1]$ instead of giving wild answers.


0

A relatively simple way to adjust for the number of reviews is to divide all rankings for all locations by 5, so your scores are normalized rankings $r_i$, then calculate for each location, $i$, calculate the average normalized ranking, $\bar r_i$ and assign each location the following score: $y_i=\frac{\bar r_i-b}{\nu_i n_i+2}$ - this is a modified version ...


0

I will assume the balls and boxes are indistinguishable. The first problem is: If I distribute $15$ balls among $30$ boxes, what is the probability that at most $10$ boxes contain a ball? First, the fact that there are $30$ boxes does not matter, since they are indistinguishable. So we only need to consider the problem as if there were $15$ boxes. Second, ...



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