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The probability that one particular ball is never drawn in $n$ tries is: $$P(A_n) = P(B_n)= P(C_n) = {(\tfrac 2 3)}^{n}$$ The probability that two particular balls are never drawn in $n$ tries is: $$P(A_n\cap B_n) = P(B_n\cap C_n)= P(A_n\cap C_n) = {(\tfrac 1 3)}^{n}$$ The probability that three particular balls are never drawn in $n$ tries is: $$P(A_n\cap ... 3 Notice that your even is a zero-one event. Define the event A_n:=\{X_n<c\} for some c>0. Then$$\sum_{n\geq 0} P(A_n)=\sum_{n\geq 0} c/n=\infty.$$So by the reverse Borel Cantelli lemma, A_n occurs infinitely often with probability 1. This implies that P(\omega: X_n(\omega)\rightarrow\infty)<1, and since it's a zero-one event, its ... 2 For x>1, we have$$\mathbb P(\xi_n\leqslant x) = \mathbb P\left(\bigcap_{i=1}^n \left\{\eta_i\leqslant x\right\}\right)=\prod_{i=1}^n\mathbb P\left(\eta_i\leqslant x \right) = \left(1 - x^{-\alpha}\right)^n. Hence \begin{align} \mathbb P(\zeta_n\leqslant x) &= \mathbb P\left(\xi_n n^{-\frac1\alpha}\leqslant x\right)\\ &= \mathbb ... 2 The case n\leq 2 is trivial. Assume n > 2: Following your notation the probabaility that you don't draw a certain ball until n is (2/3)^n. P(A_n \cap B_n \cap C_n)=0 because you have to draw some balls. A_n \cap B_n is the event that you draw n green balls in a row. Did that help? 2 Certainly n\geq3 and say we're interested in \mathcal P(X=n). There are 3^n possible outcomes. Now, if we needed n drawings to get the three colors, that means that with n-1 drawings we had only two colors, say, colors B and W. There are 2^{n-1} of drawing only B's and W's. We take out the cases all B's and all W's, so we have 2^{n-1}-2 ... 2 If Y_1,\dots,Y_n are iid and \sigma:\{1,\dots,n\}\to\{1,\dots,n\} is a permutation then:F_{Y_{\sigma(1)},\dots Y_{\sigma(n)}}(y_1,\dots,y_n)=F(y_1)\times\cdots\times F(y_n)=F_{Y_1,\dots,Y_n}(y_1,\dots,y_n)$$where F is the common CDF of the Y_i. So the random vectors \langle Y_{\sigma(1)},\dots,Y_{\sigma(n)}\rangle and \langle ... 2 It's a good start to try to solve it in a probabilistic way: notice that the Poisson random variable has the reproducibility property, that is, if X_{k} \sim \text{Poisson}(1), k = 1, 2, \ldots, n independently, then$$S_n = \sum_{k = 1}^n X_{k} \sim \text{Poisson}(n),$$whose distribution function F_{S_n} satisfies:$$F_{S_n}(n) = P[S_n \leq n] = ...

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$$E[Y_1] = \int_0^1 2y_1^2dy_1 = \frac23$$ $$E[Y_2] = \int_0^1 2y_2^2dy_2 = \frac23$$

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The answer to your first question is yes: if $c \ne 0$ and $c \in \mathbb R$, then $cX \sim \operatorname{Normal}(c\mu, |c|\sigma)$ if $X$ is normal with mean $\mu$ and standard deviation $\sigma$. This is because the normal distribution belongs to a location-scale family: its PDF is $$f_X(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-(x-\mu)^2/(2\sigma^2)}, \quad ... 1 The notation is used to say that a random variable follows a specific distribution, in this case the Chi distribution with parameter (n-1). 1 Yes. In elementary probability, the definition of independence of random variables is that their pdfs (if they exist) split up ie$$f_{X_1, \dots, X_n} (x_1, \dots, x_n) = \prod_{i=1}^{n} f_{X_i}(x_i) \tag{$*$} $$Some books use cdfs instead ie$$F_{X_1, \dots, X_n} (x_1, \dots, x_n) = \prod_{i=1}^{n} F_{X_i}(x_i) \tag{$**$} $$(*) and (**) can be ... 1 The initial sentence on the linked page is wrong. The author is being sloppy with language. The pdf in the first line is the joint pdf of the two random variables, not the pdf of the sum. And it isn't even quite that. The function$$ \lambda^2 e^{-\lambda(x_1+x_2)} \quad\text{for }x_1,x_2>0 $$is the probability density with respect to the measure ... 1 Correct. The presence of a density function implies that the distribution is absolutely continuous with respect to Lebesgue measure, and in that case the density (i.e., the Radon-Nikodym derivative of the distribution) is only defined up to sets of Lebesgue measure zero. 1 Due to the memoryless property of the exponential distribution$$P(X>30 \mid X>10)=P(X>20)=1-F(20)$$where F(x)=1-e^{-λx} for x>0. In your expression$$\frac{P(X>30)}{P(X>10)}=\frac{1-F(30)}{1-F(10)}=\frac{1-(1-e^{-30/20})}{1-(1-e^{-10/20})}=\frac{e^{-30/20}}{e^{-10/20}}1 We have four times involved so define an r.v. for each. The range for each of the times is 1 minute and it is simpler to use minutes as our units. We define the random variables as: \begin{align} X &= \text{Day 1 Start Time - 5:31pm (in minutes)} \\ Y &= \text{Day 2 Start Time - 5:31pm (in minutes)} \\ Z &= \text{Day 1 End Time - 5:46pm (in ... 1 (1.) Let s be defined as S_n = x_1+\dotsb+x_n. Then the distribution of s = S_n isP(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.$$As for an approximation, this is not the one you are looking for, but it might help. If ... 1 This is the "coupon collector's problem". As the Wikipdia article explains, the expected value is k times the kth harmonic number:$$ k\left( 1 + \frac 1 2 + \frac 1 3 + \cdots + \frac 1 k \right). 1 You want to know something about \sum_iX_i. However, X_i has a different distribution for different values of i, so this is difficult. Estimating \sum_i X_i+Y_i is easier, since X_i+Y_i has a Bin(n,p) distribution for all i. So Y_i is just an auxiliary variable to show that X_i is dominated by a Bin(n,p) variable. I would say that ... 1 Without loss of generality, let's assume \theta = 0 and \sigma = 1 for simplicity (for the general case, consider the transformation Z = \frac{X - \theta}{\sigma}). Rigorously, we have to show E[|g(X)X|] < \infty at beginning. Indeed, denote \frac{1}{\sqrt{2\pi}} by c, it follows that \begin{align} & E[|g(X)X|] \\ = & ... 1 It is a standard result in measure theory that for nonnegative functions f\int_A f d \mu = \lim_{n \to \infty} n \mu(f^{-1}([n,\infty)) + \sum_{k=1}^{n^2-1} \frac{k}{n} \mu(f^{-1}([k/n,(k+1)/n)).$$(The details of the partitioning are not so important; the important matter is that the mesh size goes to zero and the upper bound goes to infinity.) For ... 1 Suppose X is normal with mean \mu and standard deviation \sigma. Then Z=\frac{X-\mu}{\sigma} is normal with mean 0 and standard deviation 1, and X=\sigma Z + \mu. Then$$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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For both questions, the answer boils down to the triangle inequality. It is more straight forward for the second: if $x\in B(2re_i,r)$, then $d(x,0)\leq d(x,2re_i)+d(2re_i,0)<r+2r=3r$. For the first question, we first need to calculate the distance between the centers of two circles. Since $d(2re_i,2re_j)=2rd(e_i,e_j)=2r\sqrt{2}$, a point in our ...

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$$\frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{\infty}x\mathrm{e}^{-\frac{(x-\mu)^2}{2\sigma^2}}dx$$ let $z=\frac{x-\mu}{\sigma}$ then we get $$\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(\sigma z+\mu)\mathrm{e}^{-z^2/2}dz$$ where you should use $$\dfrac{d}{dz}\mathrm{e}^{-z^2/2}=-z\mathrm{e}^{-z^2/2}$$ to help. But, you should really know how to ...

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This question seems to belong to StackOverflow's R-tag In R we do not generally care about whether a variable is integer or floating if they are used for calculations as an operation such as 2L/3.0 will return a floating variable anyways. For example an integer is also a floating variable: is.numeric(2L) returns TRUE while is.integer(2) returns FALSE. For ...

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Denote $\frac{1}{\sqrt{2\pi}}$ by $c$. If $X$ and $Y$ are independent, then \begin{align} & F_{X/Y}(z) \\ = & P\left[\frac{X}{Y} \leq z\right] \\ = & \int_{-\infty}^\infty P\left[\frac{X}{y} \leq z\right]f_Y(y)dy \\ = & \int_{-\infty}^0 P[X \geq y z]ce^{-y^2/2}dy + \int_0^\infty P[X \leq y z]ce^{-y^2/2}dy \\ = & \int_{-\infty}^0 ...

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If all you want is the joint density function, you don't need to integrate: you have $f_{U,V}(u,v)=u$, which I think is right. You do still need the ranges to complete the function definition though. It's clear that with $U=X+Y,$ we have $0\lt U\lt 2$. We need to split this, however, as follows: Case $0\lt U\leq 1$: It's easy to see that we need $0\lt ... 1 When$X$is finite, you can treat$\mathcal P$as a subset of$\Bbb R^X$. Total variation is the norm there, and all the norms over finite-dimensional spaces are equivalent, so the induced metric is equivalent to the Eucledian one. Embedded$\mathcal P$is bounded and closed in Eucledian metric, hence compact. 1 Around$\lambda=0$you have already written the series expansion so $$E(\sqrt{ P_\lambda})=(1-\lambda+O(\lambda^2))(\lambda+\frac {\sqrt 2}2\lambda^2+O(\lambda^3))=\lambda-(1-\frac {\sqrt 2}2)\lambda^2+O(\lambda^3)$$ Around$\lambda=\infty$,$\epsilon_\lambda=\frac {P_\lambda-\lambda}\lambda$is tightly concentrated (approximately normal) around$0$with ... 1 In general, for smooth$g(X)$you can do a Taylor expansion around the mean$\mu=E(X)$: $$g(X)=g(\mu) + g'(\mu)(X-\mu)+ \frac{g^{''}(\mu)}{2!}(X-\mu)^2+ \frac{g^{'''}(\mu)}{3!}(X-\mu)^3+\cdots$$ So $$E[g(X)]=g(\mu) + \frac{g^{''}(\mu)}{2!}m_2+ \frac{g^{'''}(\mu)}{3!} m_3+\cdots$$ where$m_i$is the$i$-th centered moment. In our case$m_2=m_3 =\lambda\$, ...

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That's about representation of probability spaces. The original definition is: for random experiment we need to have some sample probability space, and observations are measurable maps from that space to the range space. These maps induce the joint probability measure. Now, usually you know what is the range space, and what is the joint distribution, so ...

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