# Tag Info

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To be concrete, assume every $X_i$ is a real valued random variable. Then a first observation is that you confuse $X_1(\omega)$, say, which is a number, and $X_1$, which is a function. For example, the random vector you consider is $X=(X_1,\ldots,X_n)$ (with or without transpose), which is a function from $\Omega$ to $\mathbb R^n$, certainly not ...

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Your random variable $Z$ of course is distributed as $\chi^2_n$. The multvariate normal density $N_n(\mathbf 0, I_n)$ is $z\in\R^n\mapsto\text{constant}\cdot \exp(-\|z\|^2)$. This can be factored as one factor that depends on $z$ only through $\|z\|$ and another that depends on $z$ only through $z/\|z\|$ (the latter factor in this case is $1$). From that ...

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I can see the probability distribution for $E_k$ given $N$ is of a Gamma random variate with scale parameter $k_BT$ and shape parameter $N-1$. You can use any of the standard Statistical packages to generate $E_k$ for a given $N>2$ and a constant $k_BT$. For example, function rgamma generates Gamma random variates for given Gamma parameters in R that uses ...

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Let $F(f(x))=f(k)$ denote the Fourier transform of $f(x)$. Let $(f*g)(t)=\int_{-\infty}^{\infty}f(t-x)g(x)dx$ denote the convolution of f and g. To rewrite the expression in terms of characteristic functions we apply Fourier transform on both sides. We have $f(k)=bF((f*f)(by))=f^2(k/b)$ where we have used the properties $F((f*g)(x))=f(k)g(k)$ and ...

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Let $g:=f(X_1,\dots,X_n)$. If $B\subset\mathbb R$ is a Borel set, then $g^{-1}(B)=\{\omega,(X_1(\omega),\dots,X_n(\omega))\in f^{-1}(B)\}$. Since $f^{-1}(B)$ is a Borel subset of $\mathbb R^n$, $g^{-1}(B)\in\mathcal S$. Indeed, take $\mathcal B$ the class of Borel subsets $S$ of $\mathbb R^n$ such that $\{\omega,(X_1(\omega),\dots,X_n(\omega))\in S\}\in ... 0 Yes, we can. Note that the$X_i$are random variable also on the measure space$(\Omega, \mathcal S, P)$(as they are trivially$\mathcal S$-measurable). Now apply 1.3.3. 0 Note that $$E[X_k]=\frac25,\qquad E[X_kX_\ell]=\frac15,\qquad N=\sum_{k=1}^3X_k,$$ hence $$E[X_kN]=\frac25+\frac15+\frac15=\frac45,\qquad E[N]=3\cdot\frac25=\frac65,$$ and $$\mathrm{Cov}(X_k,N)=\frac45-\frac25\cdot\frac65=\frac8{25}.$$ 0 Yes. The generators produce three numbers; order them as$x\ge y\ge z $and define $$t=\frac{x-y}{x-z}$$ The quantity$t$is uniformly distributed on$[0,1]$. Sketch of proof. The probability density of the triple$(x,y,z)$is$6$times the Lebesgue measure restricted to the tetrahedron with vertices$(0,0,0)$,$(1,0,0)$,$(1,1,0)$,$(1,1,1)$. The ... 0$X(t)$is a time-varying stochastic process here. So, for fixed$t$,$X(t)$is a random variable and is defined as:$X(t)=A\mathbb{cos}(2\pi ft+\theta)$, where$\theta$is Uniformly distributed over [$-\pi,\pi]$If$\theta$follows$U[-\pi,\pi]$, then for fixed$t$,$Y(t)=2\pi ft+\theta$follows$U[2\pi ft-\pi,2\pi ft+\pi]$so that, ... 1 It is not clear what you mean by$N(2)$. The number of arrivals in$2$minutes indeed has Poisson distribution with parameter$2\lambda$, where$\lambda$is the parameter of the exponential. What$\lambda$we would get from your calculation can only be clear if the calculation is shown. If$X$is the Poisson with parameter$2\lambda$, and you put$\Pr(X\ge ...

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I feel as though you can solve this just by sticking to the exponential distribution. You could solve for $\lambda$ with the following equation: $$\frac{1}{2} = \int_{0}^2 \lambda e^{-\lambda t}dt$$ and then once you have that, you can solve for the expected time with $$E = \lambda\int_0^{\infty} t e^{-\lambda t} dt$$ using the definition of expected ...

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Calculate the chance of $0,1,2$ duds from the binomial distribution You get 3 from a pack with no duds, 0 from the ones with one dud, -3 from the ones with two, and -6 from the rest.

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RNGs are subjected to a wide range of tests. The simplest, having a reasonable balance of $0$'s and $1$'s would reject your first example. You can make this precise by computing the chance (over a much longer run than you show) that the imbalance is a random fluctuation. If you insist that the chance of a random fluctuation is less than $10^{-6}$, for ...

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Hint: Let $Y$ be the random time taken for the power conditioning chip (PC) to fail and $X$ be the random time taken for the signal processing (SP) chip to fail. How do you denote the following event?$$\mathbb{P}[\{\text{Time required for SP chip to fail} > \text{Time required for PC chip to fail}\}]$$ If you figure out the above, you have your answer.

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Note that the Taylor series of $-\log(1-x)$ is $$\sum_{k=1}^{\infty}\frac{x^k}{k}=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\cdots$$ Therefore your expression becomes \begin{align*} \sum_{k=1}^{\infty}\frac1k\left(\frac58\right)^{k-1}\frac38&=\frac38\cdot\frac85\sum_{k=1}^{\infty}\frac1k\left(\frac58\right)^k \\ ... 1 Hint:\ln(1-x)= -\sum_{n=1}^\infty\frac{x^n}{n}$$Therefore$$-\frac{\ln(1-x)}{x}= \sum_{n=1}^\infty\frac{x^{n-1}}{n}.$$0 Denote by r the outcome "numerical value that comes up if we roll once a standard six-faced fair die" and by I_i \equiv I\{r=i\} the indicator function that takes the value 1 when r=i and zero in all other cases. The \{r=i\}'s are the elementary events. Now, we can express the value of r as$$r = \sum_{i=1}^6iI(r=i)$$We want the expected ... 2 Let B\in\mathcal{B}_{\mathbb{R}^{2}}. P_{\left(\hat{X},\hat{Y}\right)}\left(B\right) is an abbreviation of P_{\left(X,Y\right)}\left\{ \left(\hat{X},\hat{Y}\right)\in B\right\} . Here P_{\left(X,Y\right)} is the probability measure belonging to the space on wich \left(\hat{X},\hat{Y}\right) is defined. P_{\left(X,Y\right)}\left\{ ... 0 Hint:$$\frac{1_{\{X > n\}}}{X} \leq \frac{1_{\{X > n\}}}{n}$$\forall n \geq 1 and a slight variation of above for the second problem. Then you just need to show$$\lim_{n \to \infty}P(X > n) = 0$$This follows from something you are given. Can you do it now? 0 Presumably what is meant is X = \mu_1 + Y where Y is Gaussian with mean 0 and variance \sigma_1^2, and independent of \mu_1. 1 Hint: How can you find$$ f(X)=\int f(X, \mu_1)d\mu_1 $$from the information given? 0 If anyone's interested, after some thought, it looks like Brownian motion fits that bill. If w_i is a set of random numbers like in a random walk, e.g. 1,1,-1,1,-1,-1, etc. a random walk sequence can be constructed like so: x_i = \sum_{j=0}^i w_j Brownian motion just puts an exponential damping on the sum so that terms back beyond some specified time ... 0 We have \mathbb P\{X\geqslant c\}\in\{0,1\} for each c. Define$$c_0:=\inf\{c\mid \mathbb P\{X\geqslant c\}=0\}$$and show that X=c_0 almost everywhere. 0 Just so that the answer is here and clear. Label the non-spades 1\cdots39. Let X_i=1 iff card i is drawn before the fifth spade (and X_i=0 iff card i is drawn after the fifth spade). Then X=\sum\limits_{i=1}^{39}X_i= the number of non-spade cards drawn before the fifth spade. Let Y=5+X. Then Y= the number of cards drawn in order to ... 2 Let h(x)=x^3 and f be the PDF of U. Then we aim at finding the mean and variance of Y=g(h(U)). For the mean we have$$ {\rm E}[Y]=\int_{-\infty}^\infty g(h(x))f(x)\,\mathrm dx=\int_{-1}^1g(x^3)f(x)\,\mathrm dx since f(x)=0 outside of [-1,1]. Now, note that x\mapsto g(x^3) is an odd function (since it's a composition of two odd functions) and ... 0 I'm an electrical engineer and conceptually, commutation (which is what this is called) is analogous to sliding one waveform across the other. The "fixed" PDF is a square wave 1 unit high between 0 and 1. The "sliding" PDF is also 1 unit high and 1 unit wide and as it slides up from the left, the values are 0 until it just starts to overlap at x=-1, then ... 3 For y\geqslant0, let A_y=[−y+1≤X^2≤y+1] then; If 0\leqslant y\leqslant1, then A_y=[-\sqrt{y+1}\leqslant X\leqslant-\sqrt{1-y}]\cup[\sqrt{1-y}\leqslant X\leqslant\sqrt{y+1}]. If y\geqslant1, then A_y=[-\sqrt{y+1}\leqslant X\leqslant\sqrt{y+1}]. This allows to deduce F_Y in terms of F_X. 0 First find the CDF for a given value of N. Since the max of the U_k is less than x iff all the U_k are less than x and since they are independent you get the CDF given N is x^N now you take expectation of this expression with respect to N, using the fact N is a geometric distribution. Call V=\max U_i we have so far F_{V|N}(x|N)=\Pr[V\le x|N]=x^N ... 0 Note that \max_i U_i \le t holds exactly iff U_i \le t for all i. Hence \begin{align*} \def\P{\mathbb P} \P[\max_i U_i \le t] &= \P\left[\bigcap_{i\le n} \{U_i \le t\}\right]\\ &= \prod_{i=1}^n \P[U_i \le t] \end{align*} Now, as U_i is uniformly [0,1]-distributed, \P[U_i \le t] = t for all t \in [0,1], that is \P[\max_i U_i \le ...

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Hint: Expand the right hand side and try to minimize the function with respect to $a$. Then check the sign of the $2$nd derivative to make sure you've minimized the function accurately. Adding on to it: Write out the following and see how the sides of the expression compare. $$\mathbb{E}[(X-a)^2]=\mathbb{E}[(X-\mathbb{E}[X]+\mathbb{E}[X]-a)^2]=?$$In the ...

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Hint: The probabilities are much simpler than the ones proposed. Let $Y$ be the number of trials until we get success. Then $Y=1,2,3,4,5$, each with probability $\frac{1}{5}$.

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It can be verified in Maple as follows: $$with(Statistics): X := RandomVariable(Exponential(a)): Y := RandomVariable(Gamma(b, L)): Z := X/Y; CDF(Z, t);$$ produces $$\cases{0&t\leq 0\cr - \left( bt+a \right) ^{-L}{a}^{L}+1&0<t\cr}.$$

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Simulate them. Write a quick computer program (Mathematica or Matlab (with Statistics toolbox) or octave) to simulate all three random variables, sample each 1000 times, and see if the associated empirical distributions are similar.

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By definition, two stochastic processes $(X_t)_{t \geq 0}$ and $(Y_t)_{t \geq 0}$ are independent iff their corresponding canonical filtrations are independent, i.e. $$\mathcal{F}^X := \sigma(X_s; s \geq 0) \quad \text{and} \quad \sigma(Y_s; s \geq 0) =: \mathcal{F}^Y .$$ are independent. From $$\mathcal{F}^X = \sigma \bigg( \underbrace{\bigcup_{n \in ... 1 I would not agree that this is correct. In particular, if X ~ N(\mu, \sigma^2) with pdf f(x), and cdf F(x), then the effect of the quality control mechanism is to doubly truncate the distribution, so that post-quality control, the distribution of nuts is say g(x):$$g(x) = \frac{f(x)}{F(b)-F(a)}$$where b=445 is the upper bound, and a=420 is ... 1 It is probably not worthwhile to try to locate a computational error, since the analysis is incorrect from the beginning. Let X be the number of hard questions chosen. We are taking 6 questions from the 10, without replacement. There are \binom{10}{6} equally likely ways to do this. The random variable X can take on values 0, 1, 2, or 3. ... 1 Recall that for every nondecreasing sequence (A_n) of events of union A=\bigcup\limits_nA_n, one has$$P[A]=\lim\limits_{n\to\infty}P[A_n]. $$Apply this to A_n=[X\leqslant x,Y\leqslant y_n] where the sequence (y_n) is nondecreasing and with limit +\infty, then A=[X\leqslant x] hence$$P[X\leqslant x]=\lim\limits_{n\to\infty}P[X\leqslant ...

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\begin{align} \Pr(X\le x) & = \Pr(X\le x\ \&\ (Y\le0\text{ or }0<Y\le1\text{ or }1<Y\le2 \text{ or } 2<Y\le3 \text{ or }\ldots)) \\[8pt] & = \Pr(X\le x\ \&\ Y\le0)+\Pr(X\le x\ \&\ 0<Y\le1)+\Pr(X\le x\ \&\ 1<Y\le2)+\cdots \\[8pt] & = \lim_{n\to\infty} \Pr(X\le x\ \&\ Y\le0)+\cdots\cdots+\Pr(X\le x\ \&\ ... 0 Hint: what is the distribution of Y-X ? (1,1) 0 In probability theory f_X(x) denotes PDF (Probability Density Function) and F_X(X) denotes the CDF ( Cumulative Distribution Function ) . If you are given the f_X(x) you can calculate F_X(x) which is equivalent to P(X\le x) ieF_X(x) = \int_{-\infty}^{x}f_X(x)dx$$This is all the background you need to know to deal with this theorem . ... 1 Fifty two cards are labeled with the numbers 1 through 52 . Note that there are 13 spades amongst the cards. For the first spade, We may view these as separating the remaining 39 cards into 14 groups of non-spades - those appearing before the ﬁrst spade, between the ﬁrst and second, etc. Each of these groups is equally likely to appear ﬁrst, so 39/14 ... 0 The proposed strategy is far too complicated. Let D be the region above the curve y=x^2. You want$$\iint_D f(x,y)\,dy\,dx.$$This in a sense finishes things if we do not know f. You can express the integral, if you wish, as an iterated integral, y goes from x^2 to \infty, x then goes from -\infty to \infty. 0 Label the non-spades 1 to 39. Let X_i=1 if non-spade with label i is drawn before the fifth spade, and 0 otherwise. Then the total number Y of cards drawn up to an including the fifth spade is given by$$Y=5+X_1+X_2+\cdots+X_{39}.$$I think you know how to find E(X_i). The number of relevant cards is 14. 1 The main idea is notice that:\Pr[X=Y]=\Pr[\bigcup_{i} (X=i \land Y=i)] =\sum_{i} \Pr[X=i \land Y=i] and use the definition of independence. 0 Obviously this cannot be true, because for k=0 this would yield 0=1. Your comment suggest however that you are interested in the equality:$$ 1-\text{erfc}(x) = \frac{1}{2}\left(\text{erfc}(-x)-\text{erfc}(x)\right) $$This is indeed true, because$$1-\text{erfc}(x) = \text{erf}(x) = \frac{1}{2}\left(\text{erf}(x)-\text{erf}(-x))\right) = ...

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Means follow the rule of linearity . $E[X+Y] = E[X] + E[Y]$ It doesn't matter whether the events are independent or not . So for any linear combination of random variables you can take the mean of the individual random variables and then combine them . You are asking for the intuition for $X+Y$ . These are two events whose outcome can be defined by ...

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In order Player A to score for first time on the $n^{th}$ attempt, he needs to miss the first $n-1$ attempts. The probabilty for miss is $\frac 3{10}$ and for a hit it's $\frac 7{10}$. So the probablity to make his firs basket after $n$ attempts is: $$P_A(n) = \left(\frac{3}{10}\right)^{n-1}\frac7{10}$$ Simularly for Player B we have: P_B(n) = ...

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The chance that both score at very first attempt = 0.7*0.4 at second attempt = (0*3*0.6)*(0.7*0.4) (Both have to fail their first attemts and succeed in second attempt) at nth attempt is (0.3*0.6)^n *(0.7*0.4) (Both have to fail there n attempts and suceed at n+1 th attempt) The combined probabillity hence is 0.28 [1+0.18 + 0.18^2...] ~ 0.341

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Well, if the question is phrased as you state, then they can really only have a joint density, as the other terms dont really fit your example sentence. I can offer you some general instructions for each letter: B) Its just asking you to verify that the double integral of the density over the domain is equal to 1 over the constant, whatever it is, so the ...

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