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4

Strictly speaking, at any given point the pdf doesn't mean anything. You can change the value of the pdf at that particular point to be whatever you want, and the distribution of the random variable is the same. Still, "generically", $f(x)$ is essentially the probability to find the random variable in the interval $(x-\epsilon,x+\epsilon)$, divided by $2 ...


3

The calculation on the right-hand side is correct; just the notation is bad, because as you say the right-hand factors in both terms are conditional probabilities. A better way to write this would be $$ P(\text{C})=P(\text{W})+P(\text{Bl})= ...


3

Basic approach. Let $a_1, a_2, \ldots, a_{31}$ be the people in question. $a_1$ starts the rumor, and tells a randomly selected person. Then that second person tells a randomly selected person. And so on. (a) The first time the rumor is transmitted, of course, $a_1$ cannot tell $a_1$ themselves. Thereafter, though, $a_1$ is avoided each time with ...


3

Sub $u=x+y$, $v=x-y$ with Jacobian $J=1/2$: $$\int_0^{\infty} dx \, \int_0^{\infty} dy \frac{f(x+y)}{x+y} = \frac12 \int_0^{\infty} du \int_{-u}^u dv \frac{f(u)}{u} = \int_0^{\infty} du \, f(u) = 1$$


3

Oussama Boussif, in a truly nice use of Pascals's inversion formula, showed that $\delta(k) = 2k $ is a solution. However, the OP said that the solution must have $\delta(k) \in \{1, 2, ...\} $. To satisfy this, use $\sum_{k=0}^{\theta} (-1)^k \binom{\theta}{k} = 0 $. Therefore, for any real $a$, $\delta(k) =2k+a(-1)^k $ is a solution. Choosing $a = 1$ ...


3

Let's introduce the following Pascal inversion formula: $$ {b}_{n}=\sum_{k=0}^{n}{n\choose k}a_k\\ a_n={(-1)}^{p}\sum_{k=0}^{n}(-1)^k {n\choose k}a_k $$ So: $$ \sum_{k=0}^{\theta}{\theta\choose k}\delta{(k)} = {2}^{\theta}\theta $$ Using the latter, we get: $$ \delta{(\theta)}={(-1)}^{\theta}\sum_{k=0}^{\theta}(-1)^k {\theta\choose k}k2^k $$ Let's ...


2

The random variable $V$ is log-normally distributed if and only if $\ln V$ is a normal random variable. Since the logarithm is a monotonous function, the median of $\ln V$ is the logarithm of the median of $V$. Equivalently, the median of $V$ is the exponential of the median of $\ln V$. (1) The variable $\ln V$ has a normal distribution, which is ...


2

$${ n \choose r+1} = \frac{n-r}{n + 1} {n \choose r}$$


2

Your answer to (i) is slightly wrong as with replacement you can require more than $n$ selections so it should be $\displaystyle \sum_{k=1}^\infty k\frac{(n-1)^{k-1}}{n^{k}}$ but this can be simplified to $n$. An easier approach is to solve $E[X]=1 + \dfrac{n-1}{n}E[X]$, since this is a geometric distribution. Your answer to (ii) is correct but this can be ...


2

The PDF of $M_n$ is $$ \frac1{\beta(\tfrac{n+1}2,\tfrac{n+1}2)} x_+^{(n-1)/2} (1-x)_+^{(n-1)/2} $$ where $x_+$ is defined to be $x$ if $x > 0$, and zero otherwise. So the PDF of $(M_n - EM_n)/\sqrt{\text{var}(M_n)}$ is $$ \frac{1}{\sqrt{4(n+2)} \beta(\tfrac{n+1}2,\tfrac{n+1}2)} \left(\frac12 + \frac x{\sqrt{4(n+2)}}\right)_+^{(n-1)/2} \left(\frac12 - ...


2

My guess is that whoever wrote the sign was trying to write the statistical formula for the sample (excess) kurtosis, but mangled it; so, I would guess the intended codeword is probably just "kurtosis".


2

You have the first answer correct.   The probability of selecting urn A and then silver from Urn A is indeed: $\tfrac 1 4$. $$\mathsf P(A\cap S) = \mathsf P(A)\;\mathsf P(S\mid A)$$ For the second question: Hint:   Each individual coin had an equal probability of being chosen.   The one chosen was gold.   So how many gold coins were ...


2

The velocity of a moving particle is the rate of change of the displacement at that time. The probability density of a continuous real valued random variable is the rate of change of the cumulative probability at that point.


2

$\sum_iX_i=n\bar X$, so $$\begin{align*} \sum_i(X_i-\bar X)^2&=\sum_iX_i^2-2\bar X\sum_iX_i+\sum_i\bar X^2\\ &=\sum_iX_i^2-2\bar X(n\bar X)+n\bar X^2\\ &=\sum_iX_i^2-n\bar X^2\;. \end{align*}$$ Moreover, $$\begin{align*} \sum_i(X_i-\bar X)(Y_i-\bar Y)&=\sum_iX_iY_i-\bar X\sum_iY_i-\bar Y\sum_iX_i+\sum_i\bar X\bar Y\\ ...


2

I´m not sure, if all requirements are met. I would do the following transformation: $\begin{align} F_Y(y) & =\mathsf P(X\leq x) & x= -10\ln(0.5\cdot y) \\ & =\mathsf P\big(-10\cdot \ln(0.5\cdot Y)\leq -10\cdot \ln(0.5\cdot y)\big) \\ & = \mathsf P\big(X\leq -10\cdot\ln(0.5\cdot y)\big) \\ & =\Phi \left( \frac{-10\cdot\ln(0.5\cdot ...


1

The pair $(X,Y)$ is uniformly distributed in a triangle. That does not imply that either $X$ or $Y$ is uniformly distributed. Draw the triangle, and you'll see why $X$ is more likely to be in a short interval near $1$ than in an interval of the same length near $0$, so $X$ will not be uniformly distributed. Similarly, $Y$ is more likely to be in a short ...


1

Ha! Nice question, I also had the exact same problem while I was reading Efron's book and it took me some time to figure it out. I guess thus it is by no chance that I have previously posted answers to prove theorems which will help answer your question.. First of all, one important ingredient you forgot to mention in your question is that $\pi_0 + \pi_1 = ...


1

Hint: You can use the fact that $P(A\cup B) = P(A) + P(B) - P(A \cap B)$. Here, $P(A) =$ probability of drawing at least one two, $P(B) =$ probability of drawing at least one three, $P(A\cup B) =$ probability of drawing at least one two or at least one three, and $P(A\cap B) =$ probability of drawing at least one two and at least one three. You say you ...


1

Do you mean?: $\mathsf P(a\leq X\leq b) = \mathsf P(a\,c\leq Y\leq b\,c)$ when $Y=c\,X$ in general, and, for an exponential distribution in particular: $$\int_a^b \lambda e^{-\lambda x}\operatorname d x = \int_{ac}^{bc} \frac \lambda c\cdot e^{-\lambda y/c}\operatorname d y \quad\text{where } X\sim\mathcal{Exp}(\lambda), Y\sim\mathcal{Exp}(\lambda/c)$$ and ...


1

Split it into disjoint events, and add up their probabilities: The probability of choosing white from the first box and then from the second box is: $$\frac{4}{9}\cdot\frac{6}{10}=\frac{24}{90}$$ The probability of choosing black from the first box and then from the second box is: $$\frac{5}{9}\cdot\frac{5}{10}=\frac{25}{90}$$ So the probability ...


1

You need the conditional probability formula $$p(\text{urn B}|\text{gold is selected})=\frac{p(\text{urn B}\cap\text{gold is selected})}{p(\text{gold is selected})}$$ $$= \frac {\frac 12}{\frac 34}=\frac 23$$


1

It looks like statistics; $\bar{x}$ denotes the mean of a sample. You cannot input this on your calculator, or calculate this at all, because the values of the $x_i$ are not given.


1

It's from statistics. $\bar{x}$ is the mean of a sample. It's a quantity you may construct from a sample.


1

Your "curve" only has three points Predict "all negative", in which case you have a false positive rate of $\frac{0}{2}=0$ and a true positive rate of $\frac{0}{3}=0$ Your actual predictions, in which case you have a false positive rate of $\frac{1}{2}=0.5$ and a true positive rate of $\frac{2}{3}=0.666\ldots$ Predict "all positive", in which case you ...


1

Let the result of a game be denoted by $XdY$, meaning that Player $X$ defeats Player $Y$. Then the space of possible results is $$ \{(AdB, AdC), (BdA, BdC), (AdB, CdA, CdB), (BdA, CdB, CdA), (AdB, CdA, BdC, BdA), \ldots\} $$ Partial solution follows. To determine the probability of winning, observe that $C$ wins if and only if the entire series ends ...


1

Take en example: the matrix $$ A=\left[\matrix{1 & 1\\2 & 2}\right]. $$ Applying your method: the first column $$ \left[\matrix{1 \\ 2}\right]=\left[\matrix{1 \\ 1}\right]+\left[\matrix{0 \\ 1}\right], $$ and the second column $$ \left[\matrix{1 \\ 2}\right]=2\left[\matrix{1 \\ 1}\right]-\left[\matrix{1 \\ 0}\right], $$ hence, collecting all "basic" ...


1

The sum of all the $n$ columns is $(0,0,0,\ldots,0)^T$ and so surely the column rank cannot be $n$ as you claim it to be?


1

See here for a review of Slovin's formula, where they say there is no person called Slovin they can atribute the formula to.


1

Yes, you can do this with importance sampling if you just want to correct your estimate of a population statistic (e.g., mean, stdev, etc). However, if you want an actual random sample, then here's one possible approach: Let $c_i$ be the number of projects assigned to person $i$ and $N$ be the total number of people with at least one project. Now, if you ...


1

First question: One observation is not enough because it give you no idea of the variance (or SD) and you need that for a test or for a CI. (If you already know the variance, then you might make a CI, but it would probably be too long to be of any use; see answer to second question below.) Second question: This gets into more complicated territory. It's ...



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