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(Too long for a comment:) I can offer an explanation showing that dividing by $n$ would give an underestimation of the variance. The sum of squares $\sum (X_i - \overline{X})^2$, where $\overline{X}$ is the sample mean, is smaller than the sum $\sum (X_i - \mu)^2$ where $\mu$ is the true mean. This is the case since $\overline{X}$ is expected to be ...

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Hint: First step: You should recognize that the above functions are pdfs of normal random variables. Second step: Determine the parameters of the normal distributions. For the random variable $X$ they are $μ_Χ=1$ and $σ_Χ^2=2$. Third step: Use the fact that the sum (or difference) of independent normal random variables has again the normal distribution ...

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Since you asked this we have updated the Rayleigh wikipedia entry with the unbiased MLE for the Rayleigh parameter as well as confidence intervals, which are conveniently functions of the $\chi^2$ distribution. In particular: Given a sample of N i.i.d. samples $x_i$ from the Rayleigh distribution with parameter $\sigma$, $\widehat{\sigma^2}\approx ... 2 Hint: For$y\in\mathbb{R}$: $$P(Y\geq y)=(X_1\geq y,\ldots,X_n\geq y).$$ Without assuming anything about the joint distribution of$(X_1,\ldots,X_n)$you can't say more. But if you're willing to assume that$X_1,\ldots,X_n$are independent, then the probability above factors. 2 Try hist(data) to get a histogram. You can also try [f,xi] = ksdensity(data) plot(xi,f) to get a smoothed histogram. You can restrict the domain (since the default is the real number line): support = (0:.01:1)' [f] = ksdensity(data,support) plot(support,f) 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 ... 2 It sounds Like German Tank Problem: http://en.wikipedia.org/wiki/German_tank_problem: In the statistical theory of estimation, the problem of estimating the maximum of a discrete uniform distribution from sampling without replacement is known in English as the German tank problem, due to its application in World War II to the estimation of the number of ... 2 The first set denotes the set of all values for$\theta$for which the probability that$X$is at least$Bin(n,\theta)$is greater than$\frac{\alpha}{2}$. Similarly, the second set consists of all$\theta$for which the probability that$X$is at most$Bin(n,\theta)$is greater than$\frac{\alpha}{2}$. Since we have the intersection of these two sets, the ... 2 You need $$\frac{d}{d\sigma}\left(-n\log\sigma- \frac 1 2 \sigma^{-2} \sum_{i=1}^n x_i^2\right).$$ (I've written$x_i$where you have$x_i-\mu$since in this case it is known that$\mu=0$.) The derivative is $$\frac{-n}{\sigma} +\sigma^{-3} \sum_{i=1}^n x_i^2.$$ That differs from what you have. $$\frac{d}{d\sigma} \sigma^{-2} = -2\sigma^{-3}.$$ 2 This question is a Bernoulli trial (more specifically a binomial distribution)(an action is performed and only one of the two possible outcomes is happens at once). Bernoulli's trials are a foundation and will be heavily used later in your class and I highly recommend you attempt this question by yourself. Here are a few tips to get you started: (1) Define ... 1 Parallel resistance is $$R_p = \frac{R_1 R_2}{R_1+R_2}$$ Uncertainty can be calculated by using differentials and a linear model. $$dR_p = \frac{\partial R_p}{\partial R_1}dR_1 + \frac{\partial R_p}{\partial R_2} dR_2= \frac{R_2^2}{(R_1+R_2)^2}dR_1+\frac{R_1^2}{(R_1+R_2)^2}dR_2$$ Hence, the total max ... 1 The term $$\frac{\bar x}{nS_{xx}}$$ is independent of$i$and can be taken out of the summation. Thus $$\sum_{i}\frac{\bar{x}(x_i-\bar{x})}{nS_{xx}}=\frac{\bar x}{nS_{xx}}\sum_{i}\left(x_i-\bar{x}\right)=$$ where the latter term$\sum_i \left(x_i-\bar{x}\right)$is known to be equal to zero. But, for the sake of completeness $$\sum_i ... 1 When the variance is unknown, you might want to estimate the variance \sigma^2 , or to estimate the standard deviation \sigma and the square it. Sometimes the result is the same, sometimes it's not. For Maximum Likelihood it happens to be the same. But, anyway, it's good to state it clearly firsthand: which is the parameter I'm estimating? If it's ... 1 For your first question, suppose X and Y are independent random variables. The statement is that for any Borel measurable functions f and g, f(X) and g(Y) are independent. In fact, independence of X and Y is equivalent to (and in some formulations is defined as) the events X \in A and Y \in B being independent for all Borel subsets A ... 1 For E_2, two die can sum to 5 with the rolls (4,1),(1,4),(2,3) and (3,2). This is \frac{4}{36} = \frac{1}{9}. Now imagine you have to make bet on the probability of E_1 occurring and you win the bet if E_1 does occur. If you know that E_2 occurs, there is a \frac{1}{2} probability that E_1 will occur. Otherwise, the probability is less ... 1 If you knew the mean value of your distribution, the variance should be divided by the number of samples n. On the other hand if you extract the mean value from your data, you are fixing a relation on your n samples (their sum is n\bar X) so you are left with the equivalent of n-1 samples. 1 Any weighted average of the (X_i - \mu)^2's is an unbiased estimator. This is why you "should use \mu instead of \bar{x} and divide by n, if the true mean is known". Unfortunately, \mu is usually unknown. Of all the procedures that try to correct this problem by replacing \mu by a function of the X_i, the one that takes the replacement of ... 1 You might think about orthogonal projection of the response vector$$ y=\begin{bmatrix} y_1 \\ \vdots \\ y_n \end{bmatrix} $$onto the column space of the "design" matrix$$ X = \begin{bmatrix} 1 & \cos\theta_1 & \sin\theta_1 \\ \vdots & \vdots & \vdots \\ 1 & \cos\theta_n & \sin\theta_n \end{bmatrix}. $$The projection, which is ... 1 The number of unacceptable chocolate bars in a sample of 1000 is a random variable X which has the binomial distribution with parameters n=1000 and p=0.10, in symbols$$X \sim \mathrm{Bin}(n=1000,\,p=0.10)$$To calculate the required probabilities we will use the normal approximation to the binomial distribution, that is$$X \sim ... 1 The problem is with$E \left[ \left(\sum X_i \right)^2 \right]$Let$Y=\sum X_i.$We know$E[Y]= n \mu.$We also know$\mathrm{Var} (Y)= n \sigma^2.$That means$E[Y^2]=n \sigma^2 + n^2 \mu^2.$Then plugging into your equation we find$\mathrm{Var} (\bar X ) = {1 \over n^2} \left[ n \sigma^2 + n^2 \mu^2 \right] - \mu^2 = {\sigma^2 \over n}$1 I suggest adapting this code I've taken from the link I gave in my first comment. >> A = rand(700,1); >> MAX = max(A); >> STD = std(A); >> MAX = max(A); >> MIN = min(A); >> STEP = (MAX - MIN) / 1000; >> PDF = normpdf(MIN:STEP:MAX, M, S); >> plot(MIN:STEP:MAX, PDF); In your case the distribution isn't random, ... 1 Your initial calculation shows there are$7^4-6^4=1105$combinations with largest number$6$. This is not divisible by$7^3=343$, not even by$7$. The error is that combinations with more than one$6$get double counted in your approach. The logical value for$K$is$4$, which selects which die is the$6$.$4 \cdot 343=1372$But you are counting the ... 1 I would recommend finding the lat/lon differences in total or principal. Then with that you can get an average and variance. Then you can normalize and just use a standard z table. go here: http://stattrek.com/probability-distributions/chi-square.aspx and here: http://www.statlect.com/probability_exercises.htm 1 The identity to prove is that, for every function$f$such that$f(0)=0$, $$E(f(X))=qE(f(X+1)).$$ To show this, one computes$E(f(X))$and$E(f(X+1))$, using the definition of the distribution of$X$, and one watches the simplification occur... A more general formula, valid when$f(0)\ne0$and possibly preferable from a probabilistic point of view, is$$... 1 Note that choosing a card from a well-shuffuled stack of infinitely many cards is mathematically the same as drawing a card from a stack of$52$, recording the card you get, and replacing your card into the stack. That is, we note that the probability of drawing any given card is$1/52\$, regardless of how many or which cards have already been chosen.

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The utility of the game values in Combinatorial Game Theory (and in particular, of the Surreal numbers as they show up in that context) is not really related to how easy it is to win, but merely related to the value in sums with other games. For an extreme example, both {99|} (value 100) and {0|} (value 1) are games in which Left wins by making their only ...

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