# Tag Info

0

It seems to me the answer should be about $\frac{1}{8}$. You're correct that the valid potential lengths are 14:01 - 15:59 and 15:01 - 16:59 for day 1 and day 2, respectively. Just to make the numbers nicer let's recenter to 0, so that the times are 0:00 - 1:58 for day 1 and 1:00 - 2:58 for day 2. Note the possible time in seconds for day 1's length is ...

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The notation is used to say that a random variable follows a specific distribution, in this case the Chi distribution with parameter $(n-1)$.

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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: \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 minutes)} \\ W &= ...

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(1.) Let $s$ be defined as $S_n = x_1+\dotsb+x_n$. Then the distribution of $s = S_n$ is $$P(S_n = k) = \Sigma^{(k)}\prod_{i=1}^np_i^{y_i}(1-p_i)^{1-y_i},$$ 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 ...

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So you set $W=X+Y$ and $V=X-Y$. Solving for $X$ and $Y$ yields $$X=\frac{W+V}{2}\text{ and }Y=W-\frac{W+V}{2}$$ Now we can use the transformation of variable formula to get the joint distribution of $(W,V)$, i.e., $$f_{W,V}(w,v)=f_X(x)f_Y(y)\times|J|$$ where $|J|$ is the Jacobian. I'm too lazy to do the work, but that should get you started.

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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 $$E[X \mid X \in [a,b]]=E[\sigma Z + \mu \mid X \in [a,b]] \\ = \mu + \sigma E[Z \mid X \in [a,b]] \\ = \mu + \sigma E \left [Z \left | Z \in \left [ ... 0 Suppose for simplicity that you have a standard normal X with pdf f. One of the main properties of f is that it satisfies f'=-xf which implies \int_a^b xf\,dx=-\int_a^b df=f(a)-f(b). It follows that$$E(X|X\in[a,b])=\frac {f(a)-f(b)}{\Phi(b)-\Phi(a)}$$Set b=\infty (\Phi(\infty)=1, f(\infty)=0) if you want a one-sided bound. 0 The "mean" of a continuous probability distribution, P(x), is, by definition, the integral of xP(x). To restrict a normal distribution, y= Ae^{\frac{(x- \mu)^2}{\sigma^2}}, between x= a and x= b, with a< b, we have to divide by the probability x is between a and b, the integral of P(x) between a and b. Here, mean is \frac{\int_a^b xe^{\frac{(x- ... 0 I think you just need to take the values/points (which are above 10MW), sum them up and divide the sum by their count. That's all, no? Whether the distribution is normal or not, that's not relevant here, I think. 0 Since I already typed it for the other question: The marginal density of Y is f_Y(y) = 2y\mathsf 1_{(0,1)}(y) and the conditional density of X\mid Y=y is f_{X\mid Y=y}(x) = \frac1y\mathsf 1_{(0,y)}(x). Therefore the joint density is given by the product$$f_{X,Y}(x,y) = f_Y(y)f_{X\mid Y=y}(x) = 2\cdot\mathsf 1_{(0,1)}(y) \mathsf 1_{(0,y)}(x). $$To ... 0 Let \mu=E(X). Then cov(X)=E(XX^T)-\mu\mu^T and cov(AX)=E((AX)(AX)^T)-(A\mu)(A\mu)^T=A(E(XX^T)-\mu\mu^T)A^T=Acov(X)A^T. 2 For x>1, we have$$\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 ... 1 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 ... 1 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 ... 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. 0 Let \{N(t):t\geqslant0\} be the arrival process of the busses, with arrival times \{S_n\}. What we're concerned with here is the asymptotic distribution of the forward recurrence time, i.e.A(x):=\lim_{t\to\infty} \mathbb P(t-S_{N(t)}\leqslant x). $$For intuition, if the current time is t, then the time that the last bus arrived is S_{N(t)}, ... 0 First note that p_n=\frac1n - \frac1{n+1} hence p_n = \frac1{n(n+1)} is the probability mass function of a random variable X. Now if |s|<1, we have$$\mathbb E\left[|s^X|\right] = \sum_{n=1}^\infty \frac{|s|^n}{n(n+1)}\leqslant\sum_{n=1}^\infty \frac1{n(n+1)}=1, $$and so by absolute convergence of the above series, the generating function of X ... 0 In this answer, it is shown that$$ \begin{align} e^{-n}\sum_{k=0}^n\frac{n^k}{k!} &=\frac12+\frac{2/3}{\sqrt{2\pi n}}+O\left(\frac1n\right) \end{align} $$2 It's a good start to try to solve it in a probabilistic way: notice that the Poisson random variable has the reproducibility property, that is, if X_{k} \sim \text{Poisson}(1), k = 1, 2, \ldots, n independently, then$$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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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 $$f_X(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-(x-\mu)^2/(2\sigma^2)}, \quad ... 0 Z = \alpha X + \beta X^2. We have to be careful because this is not a one-to-one function of X. Let's suppose e.g. that \beta > 0. Then the minimum possible value of \alpha x + \beta x^2 is -a^2/(4b). For z > -a^2/(4b),$$\eqalign{\mathbb P(Z \le z) &= \mathbb P\left( \dfrac{-a - \sqrt{a^2 + 4 b z}}{2b} \le X \le \dfrac{-a + ...

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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 ...

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$\textbf{hint}$ If you require a method of solving the last integral, then use differentiation under the integral sign. $$\int_a^b x^2\mathrm{e}^{-x^2}dx = -\partial_{\alpha}\int_a^b \mathrm{e}^{-\alpha x^2}dx$$ where you evaluate the derivative w.r.t $\alpha$ at $\alpha=1$ and use the limits (you should see something you recognize)

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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 ... 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 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 ... 0 I believe that the distribution of your sample mean will tend towards a Gaussian distribution regardless of the underlying distribution (i.e. central limit theorem). The variance of the mean will be$\frac{\sigma^2}{N}$, where$\sigma^2$is the variance of the underlying distribution, and$N$is the number of data points used to calculate the mean. By the ... 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 ... 0 Another way to do it, assuming some knowledge of elementary probabilistic properties. You proved that$f(y_{1},y_{2})=f_{1}(y_{1})f_{2}(y_{2})$, hence $$\mathbb{P}[Y_{1}\in A_{1}, Y_{2}\in A_{2}]=\mathbb{P}[Y_{1}\in A_{1}]\cdot\mathbb{P}[Y_{2}\in A_{2}]$$ Which is the definition of the independence of$Y_{1}$and$Y_{2}$, which directly implies ... 2 $$E[Y_1] = \int_0^1 2y_1^2dy_1 = \frac23$$ $$E[Y_2] = \int_0^1 2y_2^2dy_2 = \frac23$$ 0 I suspect you need to know more about$F(X_1,X_2)$. Suppose$X_1$and$X_1^\prime$are independently$N(0,1)$and$X_2=X_1$while$X_2^\prime = X_1^\prime$. Then$\Pr[(X_1 - X_1')(X_2 - X_2') > 0] = 1$. Suppose$X_1$and$X_1^\prime$are independently$N(0,1)$while$X_2=-X_1$and$X_2^\prime = -X_1^\prime$. Then$\Pr[(X_1 - X_1')(X_2 - X_2') > ...

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Hint: If $I:=\left\{ \langle D,W,L\rangle\in\mathbb{Z}_{\geq0}\mid D+W+L=13\right\}$ then: $$3^{13}=\sum_{\langle D,W,L\rangle\in I}\frac{13!}{D!W!L!}$$ However, for the answer you need a more complicated index set $I$.

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The exponential distribution is part of the gamma distribution family, if $X$ has ${\Gamma(\alpha,\beta)}$ distribution: $$f_X(x)=\frac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x}$$ $\alpha$ is the shape parameter (more important) and $\beta$ is the scale parameter. The exponential distribution with parameter $\lambda$ is $\Gamma(1,\lambda)$, and ...

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|] \\ = & ...

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This is the "coupon collector's problem". As the Wikipdia article explains, the expected value is $k$ times the $k$th harmonic number: $$k\left( 1 + \frac 1 2 + \frac 1 3 + \cdots + \frac 1 k \right).$$

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Ok. I got it. The Lemma is true. Given a sequence of probability measures $P_n$, we consider $P_n(x)$ for every $x \in \mathcal{X}$. Using Weierstrass Bolzano theorem, noting that $P_n(x)$ is a bounded sequence of real numbers, there exists a convergent subsequence. We do this iteratively: WLOG, assume $\mathcal{X} = \{1,2,...,N\}$. Step $1$: Find a ...

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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 $0\lt ... 0 For the employer to know the order in which the computers were checked, he must have been noting the first computer as version$A$and the first computer with a different version than$A$as$B$and then forgetting what version$A$and$B$correspond to. An example of a sample path of him noting: A, A, B, A, B, B, A, ... When he has$N$of the same system ... 1 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 ... 1 $$\frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{\infty}x\mathrm{e}^{-\frac{(x-\mu)^2}{2\sigma^2}}dx$$ 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 ... 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 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? 4 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: $$P(A_n\cap ... 1 It is a standard result in measure theory that for nonnegative functions f$$\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)).$$(The details of the partitioning are not so important; the important matter is that the mesh size goes to zero and the upper bound goes to infinity.) For ... 1 Yes. In elementary probability, the definition of independence of random variables is that their pdfs (if they exist) split up ie$$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{$**$} $$(*) and (**) can be ... 1 In general, for smooth g(X) you can do a Taylor expansion around the mean \mu=E(X):$$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 $$where m_i is the i-th centered moment. In our case m_2=m_3 =\lambda, ... 1 Around \lambda=0 you have already written the series expansion so$$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)$$Around \lambda=\infty, \epsilon_\lambda=\frac {P_\lambda-\lambda}\lambda is tightly concentrated (approximately normal) around 0 with ... 2$$E[g(\theta)|Y=y]\neq E[g(\theta)|Y=y']\implies E[g(\theta)|X=x,Y=y]\neq E[g(\theta)|X=x,Y=y']?$$The right side (, which holds for all x) is more restrictive/stringent than the left side, and thus the proposition should not be true in general under no additional assumptions. Of course, a "trivial" interpretation of the inequality suggests that the ... 0 The constraints are -y \leq x \leq y, 0 < y < \infty. That first constraint is equivalent to$$ |x| \leq y $$which, when combined with the second constraint, naturally yields$$ |x| \leq y < \infty$\$ Whence the integration limits.

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As A.S.'s comment indicates, both distributions relate to the same kind of process (a Poisson process), but they govern different aspects: The Poisson distribution governs how many events happen in a given period of time, and the exponential distribution governs how much time elapses between consecutive events. By way of analogy, suppose that we have a ...

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