New answers tagged

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Recall that if $U\sim\mathsf U(0,1)$ and the distribution function $F_Y$ of $Y$ is absolutely continuous, then the random variable $F_Y^{-1}(U)$ has the same distribution as $Y$. In particular, if $F_Y(y) = (1-e^{-y})\mathsf 1_{(0,\infty)}(y)$ then $$F(Y) = U \iff Y = -\log(1-U). $$ Set $Y_k = -\log(1-X_k)-1$, then $Y_k\stackrel d= Y-1$ where $Y\sim\...


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If $X_i\equiv\mathcal{U}(0,1)\to Y_i=1-X_i\equiv\mathcal{U}(0,1)$. Now, note that $\mathbb{P}(\{-\sum \ln Y_i\geq n\})= \mathbb{P}(\{- \ln \prod Y_i\geq n\})= \mathbb{P}(\{ \prod Y_i\geq e^{-n}\})$ You see the way to finish?


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The two word answer is "polar coordinates". In more detail, let $f:S^{n-1}\to\Bbb R$ be a continuous function. Then $$ \eqalign{ \Bbb E[f(X)] &=\int_{\Bbb R^n}f(x_1/z,\ldots,x_n/z)(2\pi)^{-n/2}e^{-z^2/2}\,dx_1\cdots dx_n\cr &=(2\pi)^{-n/2}\int_0^\infty\left[\int_{S^{n-1}} f(u)\,\sigma_{n-1}(du)\right]e^{-r^2/2}r^{n-1}\,dr\cr &=c_n\int_{S^{n-1}} ...


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If $F$ is a CDF and it must be shown that $F$ is the CDF corresponding with the uniform distribution on $[0,1]$ then it is enough to prove that $F(y)=y$ for every $y\in(0,1)$. This because the values of $F$ on elements not in $(0,1)$ are determined by the following rules for a CDF: If $y\geq1$ then $1\geq F(y)\geq F(z)=z$ for every $z\in(0,1)$ and ...


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The pdf of $Y$ is obtained by taking the joint pdf of $(X,Y)$ and marginalizing $X$ out. That is: $$f_Y(y)=\int_{-\infty}^\infty f_{X,Y}(x,y) dx.$$ The joint pdf of $(X,Y)$ is the product of the conditional pdf $f_{Y|X}(y|x)$ and the pdf of $X$, $f_X$. (If this seems weird to you, it is basically analogous to the familiar identity $P(A \cap B)=P(A \mid B) ...


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The second method of solution that you wrote does not work, because $$\Pr[Y < X + 20] \ne \frac{X + 20 - 2000}{200}$$ if $2200 \ge X > 2200 - 20$, since you would get a probability exceeding $1$. Similarly, $$\Pr[Y < X - 20] \ne \frac{x - 20 - 2000}{200}$$ if $2000 \le X < 2000 + 20$, as this would give you a probability less than $0$. Instead, ...


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What prevents you from using Kullback-Leibler divergence (KL divergence) as a measure of distance from the uniform distribution? I do agree with you on the fact that KL divergence is not a true measure of "distance" because it does not satisfy (a) symmetry, and (b) triangle inequality. Nonetheless, it can serve as a criterion for measuring how far/close a ...


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Total variation distance, also known as statistical distance, is a good metric (very stringent). (Note that up to a factor $2$, it's equivalent to $\ell_1$ distance between the vectors of probabilities.) It also has a nice interpretation in terms of closeness of events. $\ell_2$ will be much more forgiving towards small differences, and put the emphasis on ...


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The distribution of $ X_i$ and the normal distribution are both symmetric about the mean, which is an advantage if the Central Limit theorem (CLT) is applied. And the required condition of $n>30$ (rule of thumb) is fullfilled. We can use both distributions and see if the approximation is good enough. Using Irwin–Hall distribution $$1-P\left( \sum_{i=...


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The sum of iid uniform random variables is an Irwin-Hall distributed random variable: $$P(\sum_{i=1}^{48} X_i > 20) = 1 - P(\sum_{i=1}^{48} X_i \le 20)$$ $$= 1 - \frac{1}{48!} \sum_{k=0}^{20} (-1)^k \binom{48}{k}(20-k)^{48}$$


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In this answer, I made the (natural) assumption that the 100 random variable are not only identically distributed, but also independent. Hint: look at the cumulative distribution function of $X\stackrel{\rm def}{=} \max_{1\leq k\leq n} X_k$ ($n=100$ here), and use independence of the $X_k$'s to factor. (Spoiler below.) Spoiler #1 Spoiler #2


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You're correct. $$E[X^3] = E[X] = 0$$ and hence $$\operatorname{Cov}(X,X^2) = 0$$ They are uncorrelated but not independent: Observe that $$P(X \in (\frac{-1}{2},\frac{1}{2}), X^2 \in (\frac{-1}{2},\frac{1}{2})) \ne P(X \in (\frac{-1}{2},\frac{1}{2})) P(X^2 \in (\frac{-1}{2},\frac{1}{2}))$$ This is a classic example of a counterexample to the converse of ...


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No, the required integral is $$\begin{align}\mathsf P(-2\ln(X)\leq t) =&~ \mathsf P(X\geq\mathsf e^{-t/2})\\[1ex] =&~ \mathbf 1_{\exp(-t/2)\in[0;1]}\int_{\exp(-t/2)}^1\operatorname d x \\[1ex]=&~ (1-\mathsf e^{-t/2})~\mathbf 1_{t\in[0;\infty)}\end{align}$$


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The PDF of $T$ is $$P(T)=P(X)\left|\frac{dX}{dT}\right|$$ $$=1\times\left|\frac{dX}{dT}\right|$$ $$=\left|\frac{X}{2}\right|$$ $$=\frac{1}{2}e^{-T/2}$$ The CDF is $$P(T\le t)=\int_{0}^{t}\frac{1}{2}e^{-T/2}dT$$ $$=-\left[e^{-T/2}\right]_{0}^t$$ $$=1-e^{-t/2}$$


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Yes. That is what you need to do. Do you know what to do next? Hint:


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That is indeed correct. It might also help to draw a simple picture of the curve on which the point $(X,Y)$ is constrained to lie, and see if that tells you something about what to expect the covariance to be. The easier question is whether their independent: Notice that if you know what $X$ is, that tells you what $Y=X^2$ is.


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Simply note that \begin{align*} P\big(\max (X,Y) >0.5\big) &=1-P\big(\max (X,Y) \le 0.5\big)\\ &=1- P\big((X\le 0.5)\cap (Y\le 0.5)\big)\\ &=1-P\big((X\le 0.5)\big)P\big((Y\le 0.5)\big). \end{align*}


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Uniformly randomly permute $n-k$ white balls and $k$ red balls and take the positions of the red balls as the numbers drawn. Consider the indicator variables $Y_{im}$ that are $1$ if the $m$-th white ball is to the left of the $i$-th red ball and $0$ otherwise. Then $X_{(i)}=\sum_mY_{im}+i$. To find the mean, we need $\def\xp#1{\mathbb E\left[#1\right]}\xp{...


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I don't know where you are getting this "Usually distribution notation is such that you give the name of the distribution, then its mean, and finally the variance" from. As for the uniform distribution, the best way to imagine the uniform distribution is to know where it starts and where it ends. It gives you a quick idea of what the curve looks like. The ...


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Are you from France or somewhere in Europe? In France, you use $]0,1[$ to denote the open unit interval. Some other countries use $(0,1)$. Regardless, $U$ can be regarded as a function taking two inputs and producing a mathematical object. I find $U(0,1)$ much more intuitive than $U(0.5,1/12)$. I would say it is more convenient to be given the bounds of the ...


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Any of the common programming languages has such a function build in, otherwise you can google it and there are some websites that provide for you.


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More generally, let $n\in\Bbb N$ and $\alpha\in[0,\pi]$. Then we can compute the probability that $n$ random points $x_1,\ldots, x_n$ independently and uniformly distributed on a circle are in a common arc subtended by $\alpha$ as follows: All points being in one $\alpha$-arc is equivalent to there being exactly one $k$, $1\le k\le n$, such that the $\alpha$...


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Please look at this work 1 which has investigated the interference from neighbor cells, and hence, investigated the distribution of distances between users. Hope it works for you.


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We can use Chebyshev's inequality on the random variable $\frac{1}{n}\sum_{i=1}^nX_i$, which states the following: $$\mathbb{P}(|Y-\mathbb{E}(Y)|\geq a)\leq\frac{1}{a^2} \text{Var}(Y)$$ Thus $$\mathbb{P}(|\frac{1}{n}\sum_{i=1}^nX_i-\frac{1}{2}|>\frac{1}{2})\leq4 \text{Var}\left(\frac{1}{n}\sum_{i=1}^nX_i\right)=\frac{4}{n^2}n\cdot\frac{1}{12}=\frac{1}{3n}...



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