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5

Maximising the standard deviation or variance for a given mean is equivalent to maximising the sum of squares of the values for a given sum of the values. Meanwhile if $b \ge a$ and $\delta \gt 0$, $$(a-\delta)^2 +(b+\delta)^2 $$ $$= a^2-2a\delta+\delta^2+b^2+2b\delta + \delta^2 $$ $$= a^2+b^2 +2(b-a)\delta+2\delta^2 $$ $$\gt a^2+b^2$$ so the sum of ...


3

First, you might as well assume that the mean is zero, so that $a < 0 < b$; that just simplifies things. (Otherwise: let $a' = a - \mu; b' = b - \mu$ and you can convert your problem for $[a, b], \mu$ to mine for $[a, b], 0$ by subtracting $\mu$ from all the sample values, etc.) The SD is $$ s = \sum x_i^2 $$ and you want to maximize this subject to ...


3

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


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^{...


2

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)...


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.


2

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 ...


2

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 ...


2

As commented, the $z$-score value is strange. Since we want to find the $1\%$ of newborns who exceed a certain weight, we can conduct a $99\%$ one-sided confidence interval to determine this. I used a $z$-score of $2.327$ ("guesstimate") for the calculation.


2

If you look at the normal table you will find that $\Pr(Z\gt 2.33)\approx 0.01$, or equivalently $\Pr(Z\le 2.33)\approx 0.99$. (The right number is between $2.32$ and $2.33$.) It looks as if there was in error in the table look up.


2

There is no real sense in formulating an expected error because you should say an expected error relative to some distribution, but you don't know what that distribution is. Nonetheless, the "standard" statistical answer here is as follows. The number of $1$s is distributed as Binomial($m,p$) where $p=1/n$ would be the unbiased case. This has mean $mp$ and ...


2

(since I can't add a comment I will repost a answer I posted here) There is a simple proof that requires only linear algebra. First notice that if you take $\beta=(\beta_0,\beta_1,\ldots,\beta_{n−1})^T$ and the matrix $X=[\mathbb 1,x_1,x_2,\ldots,x_{n−1}]$, where $\mathbb 1=(1,1,\ldots,1)^T$ and $x_i=(x_{i1},\ldots,x_{i(n−1)})^T$, you can write the model ...


1

It is best to use a boxplot to find outliers. The problem with using the sample mean $\bar X$ and the sample SD $S$ is that an outlier seriously affects the values of $\bar X$ and $S$. By contrast, the boxplot uses the median and the interquartile range to detect outliers. These measures of location and dispersion, respectively, are not much affected by ...


1

The rule using the IQR is often used with boxplots. Using R statistical software, we obtain the following: x = c(200, 330, 675, 999, 1200, 3000, 25000) IQR(x) ## 1597.5 sd(x) ## 9093.544 boxplot(x, horizontal=T) The vertical line at 200 is the smallest data value above the lower 'fence'; the lower end of the box is at 502.5; the heavy line inside ...


1

The size of a test is, loosely speaking, the maximum Type I error probability. As the Wikipedia article put it, this is the probability of falsely rejecting the null. Your misconception stems from a misinterpretation of $\alpha$: here, we wish to control $\alpha$ to be small, not large. A size $\alpha = 0.1$ test has at most a $10\%$ chance of ...


1

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 ...


1

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 ...


1

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

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, $X^2=1$)


1

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 = $\sqrt{120\cdot0.3\cdot 0.7}=\sqrt{25.20}=5.02$.


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 + ...


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 ...


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 ...


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 $\...



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