# Tag Info

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Assume that $\exists \mathbb{E}\left(e^{Xt}\right),\mathbb{E}\left(e^{Ys}\right)$ in a neighbourhood of $0$. Then from independence: $$\mathbb{E}\left(e^{t(aX+bY)}\right)=\mathbb{E}\left(e^{atX}\right)\mathbb{E}\left(e^{btY}\right)$$ Now differentiating and using the properties of mgf: $$\mathbb{E}(aX+bY)=\left.\frac{\partial}{\partial ... 0 The probability of the first one is equal to one out of 2^{24}=4^{12}=8^{6}=16^3=4096 The probability of the second one is equal to one out of 0.8^{33} The power can be solved with a calculator. For individually seperate probabilities that are expected to happen simultaneously, the total probability is equal to the product of the individual ... 2 For example you calculated$$P(R\text{ or }B|D_1)$$as 3/10. That is wrong. It should be$$P(R\text{ or }B|D_1) = \frac{1}{10}+\frac{3}{10} = \frac{4}{10}.$$There rest are probably similar mistakes. The very last line you gave is correct. 0 This would be completely appropriate. Just state that the two wedges are part of a larger category. 0 We have$$G_2 = \frac{1}{n(n-1)}\sum^n_{i=1}\sum^n_{j=1}|X_i - X_j|^2$$We are squaring the brackets, so there is no need for absolute value. We can multiply that.$$\frac{1}{n(n-1)}\sum^n_{i=1}\sum^n_{j=1}(X_i^2 - X_iX_j+X_j^2)$$We also can assign the sums to each variable. It should be noted, that \sum_{i=1}^n\sum_{j=1}^nX_i=n\sum_{i=1}^nX_i. ... 1 The test is one-sided:$$H_0 : \mu = \mu_0 \quad \text{vs.} \quad H_a : \mu < \mu_0$$where \mu_0 = 1 is the hypothesized mean. The test statistic is$$T = \frac{\bar x - \mu_0}{s/\sqrt{n}} \sim t_{n-1},$$where$$\bar x = \frac{1}{n}\sum_{i=1}^n x_i = 0.938182$$is the sample mean,$$s = \left(\frac{1}{n-1} \sum_{i=1}^n (x_i - \bar x)^2 \right)^{1/2} ...

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Thanks Andre, I found it. For any event A, let IA be the indicator random variable of A, i.e. IA equals 1 if A occurs and 0 otherwise. Then

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The Chebyshev inequality is $$\mathbb{P}(|x - \mu| \geq a) \leq \frac{\sigma^2}{a^2}$$ .Substituting $$a = k\sigma$$gives the answer.

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When do you reject $H_0$? You reject it when your p-value is less than the stipulated $\alpha$-level. So the computed p-value is precisely the smallest $\alpha$-level that would been needed, because any higher $\alpha$-level would lead to a rejection, and any lower $\alpha$-level would lead you to fail to reject, as then the p-value would have been higher.

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Moment is a very commonly used word in physics and it measures the turning affect of a force around some point. The moment of a force around any point is the product of the magnitude of the force and the perpendicular distance between the point and the force. In Statistics, moments are used to understand the various characteristics of a frequency ...

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you know the moment generating of Binomial i.e. $(pe^t+1-p)^n$ now compute $$lim_{n \to \infty}[1+\frac{np(e^t-1)}{n}]^n$$ letting $np\to \lambda$. This should remind you a well known limit. Let us know. Bye.

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What we have: $2$ people with only blue eyes (not curly hair) $1$ person with blue eyes and curly hair $3$ people with only curly hair (not blue eyes) Let's take the case $N_1 = 2$ and $N_2=1$. When does this occur? Let's define the following events: $A_1$ is the event where at the first test you get not blue eyes and curly hair. $P(A_1) = 3/6$ because ...

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you are wrong in the standard error of the t distribution, the formula is: $$\frac{\bar X - \mu}{s/\sqrt n}\sim T(\mathcal v)$$ So the standard error of the mean equals: $$SE(\bar X)=\frac{s}{\sqrt n} = 1.5$$ With this you build your confidence interval using the t. Note when n>30 you can use the normal approximation as well.

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Your situation is a bit unclear. But from what I can understand, I think you are confusing a linear change of variable with a convolution. Those two are different. In a linear change of variable, you are not computing an integral. In this case, you want $$f_Y(y) = \frac{f_X(2y)}{\left|\frac{dy}{dx}\right|}.$$ If you are asking why you have to divide by ...

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I hope this isn't just semantics and that I'm reading you're question correctly, but the expression you write is less of a "fundamental property" of $\mathbb{E}[X\vert Y]$ than it is a definition. Intuitively, $\mathbb{E}[X\vert Y]$ denotes the expectation of X conditional on the "$\sigma$-algebra" generated by the random variable $Y$ (which we denote ...

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The Chebyshev Inequality, Wikipedia version, tells us that for any positive real $k$, $$\Pr\left(\left|X-\mu\right|\ge \sigma k\right)\le \frac{1}{k^2}.$$ Taking $\mu=\frac{4}{3}$, $\sigma=\frac{2}{3}$ and $k=1$, we find that $$\Pr\left(\left|X-\frac{4}{3}\right|\ge \frac{2}{3}\right)\le \frac{1}{1^2}.$$ So the inequality tells us that ...

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Take a look only on one of these restaurants, hence the number of students that enters it distributed Binomial with $n=200$ and $p=1/2$. Denote it by $Y$. Then use the continuous correction and the Normal approximation to compute the probability of interest: $$P(Y> 120) = P(Y\ge 120.5) \approx 1 - \phi \left(\frac{120.5 - np}{\sqrt{npq}} \right),$$ ...

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This is a problem of functional optimization. $T$ depends on $F$ and $p$ is given (or at least could be known using polls). Let us first follow @Ian's comment by assuming that $F$ is absolutely continuous. This means that $\mathrm d F(y)=f(y)\mathrm dy$, where $f$ is the probability density of the price. As $T$ depends not on a single variable, but on all ...

1

Your $\sigma(1)=cov(x_{t},x_{t+1})$ should be $cov(m_1\varepsilon _{t-1} + m_{2} \varepsilon _{t-2} + \varepsilon_{t}, m_1\varepsilon _{t} + m_{2} \varepsilon _{t-1} + \varepsilon_{t+1})$ which can be expanded to $$cov(m_1\varepsilon _{t-1},m_1\varepsilon _{t}) + cov(m_1\varepsilon _{t-1},m_2\varepsilon _{t-1}) + cov(m_1\varepsilon _{t-1},m_2\varepsilon ... 1 Note that the \{\varepsilon_t\} are independent so$$\operatorname{Cov}(\varepsilon_s,\varepsilon_t) = \begin{cases} \sigma^2_\varepsilon,& s=t\\ 0,& s\ne t.\end{cases}Hence \begin{align} \sigma(0) &= \operatorname{Var}(x_t)\\ &= \operatorname{Var}(\varepsilon_t + m_1\varepsilon_{t-1}+m_2\varepsilon_{t-2})\\ &= \sigma^2_\varepsilon(1 ... 0 Hint: Given X_n=O_p(n^\lambda) implies \exists C>0 P(|\frac{X_n}{n^\lambda}|\le C)\to 1 and Y_n=o_p(n^\tau) implies \forall \epsilon>0, P(|\frac{Y_n}{n^\tau}|> \epsilon)\to 0. \forall \delta>0 \begin{align*} P(|\frac{X_nY_n}{n^{\lambda+\tau}}|>\delta)&\le P(|\frac{X_nY_n}{n^{\lambda+\tau}}|>\delta, ... 0 In the stattrek example, you are not told the true distribution of the population of students. You only know the average weight and the sd. I take a sample of 50 students and then take the average of these 50 students. I call that \bar X. I need to know what is the probability that the average weight \bar X of a sampled student will be less than 75 ... 0 Since you were given information about all the students taking the test, you know the true population mean and standard deviation. In other words, the problem is trying to get you to relate a score of 70 to the area under the curve. The area under the curve to the left of 70 represents the percentage p of students who got a 70 or worse. In other words, if ... 0 You can do (\sum A_k)(\sum B_i)(\sum Cj) and subtract off the elements you do not want. This is (\sum A_kB_k)( \sum C_j)+(\sum A_kC_k)( \sum B_j)+(\sum C_kB_k)( \sum A_j) but now you have added all the terms with all the three indices matching once and subtracted them three times, so add them in twice as  2\sum A_kB_kC_k See the inclusion-exclusion ... 0 If you count n spheres hitting the single slice, there are about (n/21) \times 10^{8} spheres in D. Under the stated assumptions, D is cut into 10^{8} slices. Since the density of spheres is uniform, you expect about n \times 10^{8} pieces of sphere among all the slides. A single sphere, however, has a diameter about ... -1 As mau mentioned, scale invariance is the key assumption in Benford's Law. I'll fill in the mathematical details on how that gets you to Benford's results for the probabilities (for more details see my "proof" in this article). Imagine writing all numbers in scientific notation, like x\cdot10^y. The x is the only part we care about for Benford's law. ... 1 The problem of estimating the true weights m_i is best studied using a matrix formulation. Consider the "weighing matrix" A = \begin{pmatrix} 1&1&0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix} \, . $$Let m be the vector of true weights. You have two observations per combination of objects, namely b_1 = Am + \xi_1 and b_2 = A m ... 3 There is a much more general result which is true. Proposition Let X and Y be independent, real-valued, integrable and centered random variables. Let \varphi be a non-negative convex function. Then:$$\mathbb{E} (\varphi (X+Y)) \geq \mathbb{E} (\varphi (X)).$$Proof The main tool is a conditional version of Jensen's inequality ... 3 The only way this makes sense if it is with replacement. That is each contestant can chose any number. Here is why: If it were without replacement it would not make sense to assign a new winning number after each try. This makes sense in a scenario with replacement so that each contestant still has a 1 in 50 chance of winning, as earlier attempts do ... 1 You say you take the average of all the averages, but I notice that you have a sample count column. Are these averages over different sample sizes? If so, then you would probably want a weighted average for your aggregate average:$$\text{Aggregate Average} =\frac{\sum_i (\text{sample size})_i(\text{average})_i}{\sum_i (\text{sample size})_i}$$But without ... 3 To find the p.d.f of the ratio \frac{Y}{X+Y}, let us first write its c.d.f. Since X and Y are always positive, their ratio is also positive and, therefore, for 0\leq t\lt1 we can write:  P\left(\frac{Y}{X+Y}\leq t\right)=P\left(Y\leq \frac{t}{1-t}X\right)=\int_{0}^{\infty }\left(\int_{0}^{\frac{t}{1-t}x}f_{X}(x)f_{Y}(y)dy\right)dx  as ... 1 No, in general, if X follows a normal distribution, then in shorthand it is written as$$X\sim N(\mu,\sigma^2).$$Thus, for example, the variance of X_1 is 3 not 3^2. Or as you have written it$$\sigma_1^2 = 3$$not \sigma_1 = 3. 0 From the Cramer-Rao bound, we know that variance of some unbiased estimator is lower bounded by reciprocal the fisher information. That is,$$Var(\hat{\theta})\geq \frac{1}{I(\theta)}.$$Then we can get I(\theta)\geq \frac{1}{Var(\hat{\theta})}. Since the variance of some estimator should be 0<Var(\hat{\theta})<\infty. We clearly have 0< ... 2 Suppose 10,000 people are tested (to avoid decimals) Of these, 32\% or 3200 have the disease, of which 97\% or 3104 test positive 6800 don't have the disease, of which 12\% or 816 test positive Thus P(have the disease | test positive) = \dfrac{3104}{3104+816} 1 The notation is used to say that a random variable follows a specific distribution, in this case the Chi distribution with parameter (n-1). 1 (1.) Let s be defined as S_n = x_1+\dotsb+x_n. Then the distribution of s = S_n is$$P(S_n = k) = \Sigma^{(k)}\prod_{i=1}^np_i^{y_i}(1-p_i)^{1-y_i},$$where \Sigma^{(k)} denotes the sum over all 0,1 valued y_i's such that$$y_1+\dotsb+y_n=k.$$As for an approximation, this is not the one you are looking for, but it might help. If ... 1 Define R to be the event of a positive test result, R^c to be the event of a negative test result, D to be the event that the person is diseased, D^c to be the event that the person is not diseased. We are given that P(R|D)=0.97 and P(R^c|D^c)=0.88, i.e. when a person does have the disease, the test result is positive, and when he doesn't, the ... 1 Suppose X is normal with mean \mu and standard deviation \sigma. Then Z=\frac{X-\mu}{\sigma} is normal with mean 0 and standard deviation 1, and X=\sigma Z + \mu. Then$$E[X \mid X \in [a,b]]=E[\sigma Z + \mu \mid X \in [a,b]] \\ = \mu + \sigma E[Z \mid X \in [a,b]] \\ = \mu + \sigma E \left [Z \left | Z \in \left [ ...

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Suppose for simplicity that you have a standard normal $X$ with pdf $f$. One of the main properties of $f$ is that it satisfies $f'=-xf$ which implies $\int_a^b xf\,dx=-\int_a^b df=f(a)-f(b)$. It follows that $$E(X|X\in[a,b])=\frac {f(a)-f(b)}{\Phi(b)-\Phi(a)}$$ Set $b=\infty$ ($\Phi(\infty)=1, f(\infty)=0$) if you want a one-sided bound.

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The "mean" of a continuous probability distribution, P(x), is, by definition, the integral of xP(x). To restrict a normal distribution, $y= Ae^{\frac{(x- \mu)^2}{\sigma^2}}$, between x= a and x= b, with a< b, we have to divide by the probability x is between a and b, the integral of P(x) between a and b. Here, mean is $\frac{\int_a^b xe^{\frac{(x- ... 0 Along the lines of my last comment in your previous question, you could do a reverse convolution type thingy: Suppose we have a renewal process$N(t)$with rate$\lambda>0$. Let$X$be a random variable with the same distribution as the inter-arrival times of$N(t)$. Then$E[X]=1/\lambda$. Fix$\epsilon>0$. We want to add another process$A(t)$... 0 I think you just need to take the values/points (which are above 10MW), sum them up and divide the sum by their count. That's all, no? Whether the distribution is normal or not, that's not relevant here, I think. 0 Rewrite your function as $$2x^2-3y^2-2x = -3(x^2+y^2) + 5x^2-2x = -3(x^2+y^2) + 5\left(x^2-\frac 25x+\frac{1}{25}\right)-\frac{1}{5} = -3(x^2+y^2) + 5\left(x -\frac{1}{ 5}\right)^2-\frac{1}{5}.$$ Let the variable$r=\sqrt{x^2+y^2}$, then you obtain $$F=-3r^2 + 5\left(x -\frac{1}{ 5}\right)^2-\frac{1}{5}$$ with$r\in[0,1]$and$x\in [-r,r]$. If you want ... 1 Your function$F$is defined on a disc of radius one, which is a compact set. Therefore, the maximums and minimums are either on the boundary of the disc, or strictly inside the disc. Case 1: suppose they are on the boundary. In this case, rewrite your problem in polar coordinates as follows: $$F(r,t)=2r^2\cos^2(t)-3r^2\sin^2(t)-2r\cos(t), \quad ... 1 In general what you want to do is calculate the money per hour for taking the toll or in your notation p/t. Then you want to compare that value to the value of your time. If the money per hour calculated is less than the value of your time then it is worth taking, otherwise it is not. In your case the money per hour for the toll is p/t=2/(1/4)=8. So ... 0 Randomness Let's say there are N papers in each stack. If the papers are completely random, then for any paper in the first stack, there is an equal probability of \frac{1}{N} of finding the corresponding paper in any given position in the second stack: It will be equally likely that the test in stack 2 corresponding to the top-most form of stack 1 ... 1 Cool question! With regard to the randomness of stack 2, you are indeed observing Murphy's Law. Given a pile of papers that is randomly stacked, taking some papers out of it leaves a pile that is still randomly stacked. Note that this does not depend on which papers you take out, it is just a consequence of the original pile being randomly stacked. About ... 0 If T = X_1 + 2X_2 were sufficient, then the conditional distribution of (X_1, X_2) given T would be independent of \theta. In particular, the conditional density of X_1 given T = t should be independent of \theta. So let's compute this conditional density. Since X_1 and X_2 are independent, it follows that$$\begin{pmatrix}X_1 \\ ... 2 Inclusion/Exclusion will work. It is easier to find the probability of the complement, that is, the probability that$X$and$Y$are both between$0$and$2$. So our required probability is $$1-\int_0^2\int_0^2 \frac{1}{54}(x^2+y^2)\,dy\,dx.$$ Remark: Inclusion/Exclusion yields$\$\int_{x=0}^3\int_{y=2}^3 k(x^2+y^2)\,dy\,dx+\int_{y=0}^3\int_{x=2}^3 ...

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I do not think that it is possible to calculate probability of two randomly-picked variables. It is the same as saying: What is the probability that x is greater than y? I believe that you can only have probability if you have a domain or set. I may be wrong, but you have a very good question! Thanks!

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