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4

You know those numbers on houses that tell you the address? Like, say, 1205 Main Street, Sometown USA. If you run a store that sells those numbers, you should probably stock more 1s and 2s than 8s and 9s. Building contractor supply stores probably know about Benford's law this way. Hardware stores (like ACE, TruValue, etc.) probably don't. In fact, the next ...


3

Of course it is a probability density function, since $f(x) > 0 ~\forall x \in (-\infty, \infty)$ $\int_{-\infty}^{\infty}f(x) = 1$ This is actually the Cauchy distribution. To verify the second fact, you may note: $$\int_{-\infty}^{\infty}f(x) = \int_{-\infty}^{\infty} \frac{1}{\pi (1 + x^2)} = \frac{1}{\pi} \tan^{-1} \bigg[ {x} ...


3

$P(Z\ge 4.03)+P(Z\le -4.03)=P(|Z|\ge 4.03)=P(Z^2\ge 4.03^2)=P(\chi ^2_1\ge 4.03^2)$ But $P(Z\ge 4.03)=P(Z\le -4.03)$ so $P(\chi ^2_1\ge 4.03^2)=2P(Z\ge 4.03).$ There's the mysterious $1/2.$ Also $P(Z>4.03)=P(Z\ge 4.03)$ because $Z$ is a continuous rv and so $P(Z=4.03)=0.$ And $ P(Z\ge 4.03)\lt 0.0003 $ does not mean they are comparing it to 0.0003, I ...


2

I like to mention the following voting system: let each voter rank all candidates in order of preference (a total preorder will do, i.e., voters are allowed to rank two candidates equally). Now tabulate, for each pair $(i,j)$ of candidates, the number of voters who placed $i$ before $j$ minus the number who placed $j$ before $i$. This gives an ...


2

We have not been told about the physical properties of these dice. We will assume, without justification, that the dice are "fair" (all $10$ sides are equally likely). The numbers $1$ to (weird) $100$ are equally likely. Whatever Alicia picks, the probability Beti's dice roll matches it is $\frac{1}{100}$. As for both getting (say) $49$, the probability is ...


2

Let's look at a prefect game. In each turn, we will flip one of the cards and after we've seen it, we will flip the other one, so the second (fourth, sixth...) choice of cards depend on the choice of the first (third, fifth...) card. Since there is 1 in 39 cards to complete the pair once we've selected the first card, there is a chance of $\frac{1}{39}$ of ...


2

The reason it is giving half the value is that it should give half the value. If the standard deviation is twice the size, then the graph of the probability function is twice the "width". But the area under the probability function is always 1, so all the heights must be divided by 2 in order to maintain the area at 1. So probability density functions' ...


1

I will construe this to mean that if $X\sim N(\mu,\sigma^2)$ then $(X-\mu)/\sigma\sim N(0,1)$. That $X\sim N(\mu,\sigma^2)$ we may construe to mean $$ \Pr(X\in A)=\int_A \frac{1}{\sqrt{2\pi}} \exp\left( \frac{-1}{2}\cdot \left(\frac{x-\mu}{\sigma}\right)^2 \right)\, \frac{dx}{\sigma} $$ for every Borel-measurable set $A\subseteq\mathbb R$. Now $$ ...


1

Let $C=$ Number of Clashes and $p_0 = P(C=0)$. In general, your distribution is a discrete probability mass function. However, since you only care about a single value $(C=0)$ the rest of the distribution is somewhat irrelevant. Therefore, one simple approach is to develop a conservative confidence interval for $p_0$ by inverting the hypothesis test for a ...


1

Your approach is right, as you correctly determined the cumulative distribution to find the median and then identified the median class. The upper limit, however, is not $17$ but $17.5$, because you have to average the higher boundary of the class (which is $17$) and the lower boundary of the successive class (which is $18$). This can also be interpreted as ...


1

Regression models are not created to do what you want. For example, if you had each of those 4 variables in the model like this: $$y=b0+b1*x1+b2*x2+b3*x3+b4*x4$$ then we can increase the predicted y to infinity by increasing any x variable that has a positive coefficient. Furthermore, we cannot fit the model in one range of observed x values and then use x ...


1

First, if you use $f(x)=x^p$, then the $f$-mean is the same as the $p$-mean. Second, "the best measure of central tendency" depends on the problem one is trying to solve much more than on the data set (although these aspects are of course related). Third, a very important measure of central tendency - median - is not included in either generalization.


1

The two ways you propose are not equivalent in an OLS regression. You should stick to the version where every fruit has a dummy. A generic OLS model is of the form $$y_i=\alpha+\beta_1 orange_i+\beta_2 apple_i+X\gamma+\epsilon_i.$$ So you need 2 dummies $orange,apple$, and if both are zero the constant $\alpha$ is the unconditional mean of pears. in general, ...


1

Let $F_Y(y)$ be the CDF of $Y = F(X)$. Then, for any $y \in [0,1]$ we have: $F_Y(y) = \Pr[Y \le y] = \Pr[F(X) \le y] = \Pr[X \le F^{-1}(y)] = F(F^{-1}(y)) = y$. What distribution has this CDF?


1

$$ Prob(Y\leq x)=P(F(X)\leq x)=P(X\leq F^{-1}(x))=x \\ $$ The last equality is from the definition of the quantile function.


1

Your problem has nothing to do with "the notion of a set". It's just that you don't have an algorithm to determine whether a particular individual is a member of a certain set. This is a not uncommon situation, whether in the real world or in the world of mathematics. In mathematics, the situation can be worse: most sets of integers have no description at ...


1

Just integrate over -infinity to positive infinity, $$\int_{-\infty}^{\infty}f(x) dx = 1$$ $$=\int_{-\infty}^{\infty}\dfrac1{π(1+x^2)}dx$$ $$=\dfrac{1}{\pi}\int_{-\infty}^{\infty}\dfrac1{(1+x^2)}dx$$ We split this in two limits to make it easier to handle, $$=\dfrac{1}{\pi}\int_{-\infty}^0\dfrac1{(1+x^2)}dx + ...


1

Let $D_i$ be the result of die roll $i$. Note that each die face has $\frac{1}{6}$ chance of occurring, so lets focus on the statistical properties of $\frac{(O_i-E_i)^2}{E_i}$. For a given value of $N=$ number of die rolls, let $Y^i_N=\sum\limits_{j=1}^N \mathbb{I}_{i}(D_i)$, where $\mathbb{I}_{i}(D_i)=1$ iff $D_i=i$ and is $0$ otherwise. Therefore, ...


1

It is important to realize that in continuous probability, the probability of a simple event happening is zero. Take your rain example: the probability of it raining exactly 2 inches is essentially zero. This is because there are an infinite number of possible observations (1.9999...., 2.00000...1..., etc.). As such, for continuous probability, we refer to ...


1

If the rate is $r$ per unit of time then the parameter is $\lambda = rT$ so the likelihood function is $$(rT)^n \frac{e^{-rT}}{n!}$$ If you take the derivative of this with respect to $r$ and set this equal to $0$ to solve to find the maximum likelihood estimate of $r$, you do not get $T/n$. This is in fact obvious from dimensional analysis. But you do get ...


1

See this section on Wikipedia The values in the table come from evaluating the integral $$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^x e^{-t^2/2}\mathop{dt}.$$ It is hard to evaluate this integral, which is why we rely on tables instead.


1

P is the number of regressors, without counting the constant term. If you are running simple linear regression with a single independent variable, then $P=1$.


1

If your sample size remains constant, you will have $k>n$ where $i=1,...,n$ so your matrix will not be invertible and an infinite (continuum) betas will be able to minimize the square errors.


1

If I understood correctly you need to find the mean and the variance of $\hat \sigma^2$? Not dealing with the fact why did you use this estimate, one can do the following: $$\mathbb{E}\left[\hat \sigma^2\right]=\mathbb{E}\left[\frac{1}{N} \sum_{n=0}^{N-1} x_n^2\right]=\frac{1}{N} \sum_{n=0}^{N-1}\mathbb{E}\left[ x_n^2\right]=\mu_2'.$$ where $\mu_2'$ is the ...


1

I think to resolve $\theta$ by gradient will be hard way (or impossible??). Because it different with linear classfication, it will not has close form. So i suggest you can use other method example Newton's method. BTW, do you find $\theta$ using above way?


1

Instead of fitting the function, fit the first (positive) differences $d_i=s_{i-1}-s_{i}$ Edits based on OP comments While addressing OP's comments, I happened upon an easier approach that I liked better: Since the sequence of $s_i$ are decreasing, let's model each $s_i$ as the asymptote $\theta$ plus a positive term $\epsilon_i$ such that ...


1

If I differentiate $\mathbb V(\overline{y}_2')$ it gives: $\frac{S_2^2\cdot \rho^2 \cdot (n^2-2nu + \rho^2u^2)}{(n^2-u^2\rho ^2)^2}$ To get the fraction equal to zero, $n^2-2nu + \rho^2u^2$ has to be zero, if $S_2^2\cdot \rho^2 \neq 0$ . To solve this equation, you can use the the general formula to solve an quadratic equation: $x_{1,2}=\frac{-b \pm ...


1

The definition of variance for a random variable $X$, $var(X) = E[(X - \mu)^2]$, since $\mu = 0$, it's obvious that $var(X) = E[X^2]$.


1

You seem to be missing some information when describing your question, and this missing information is critical to the reason behind your question. Suppose I observe one individual from country $A$. Their income is a single normally distributed random variable with mean $\mu_A = 18000$ and standard deviation $\sigma_A = 6000$. If I observe two randomly ...


1

$h(t,b)=1_{\{t<b\}}$ is the indicator function of $t<b$. I believe, where $t=\max (x_1,...x_n)$ To see why this is the case: the pdf of uniform distributions on $(a,b)$ is given by $$f(x)=\frac{1}{b-a}1_{x\in [a,b]} $$ This is exactly what you said, but using indicator function makes it clearer. $1_{x\in [a,b]}$ takes value 1 when $x\in [a,b]$ and ...



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