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The easy way to disprove it is to show a counterexample. I will modify your approach slightly to use positive integers. So let's say you can randomly generate 1's and 2's but want to randomly generate 1's, 2's, and 3's. Under the proposed method, we would generate, say, a 1 if the sum of 3 randomly generated numbers is 0 mod 3, a 2 if the sum is 1 mod 3, ...

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You choose $n$ numbers. Success occurs when you choose at least one is a five (event $F$) and at least one even number (event $E$). Failure occurs when neither or only one event happens. You wish to find: $\mathsf P(E\cap F)$ You should the law of complements, the principle of inclusion and exclusion, and that: \begin{align} \mathsf P(E^c) &= \frac ... 2 Not sure this is the most efficient, but: The probability it is not divisible by 5 is (1 - 1/9)^{n}. The probability it is not divisible by 2 is (1 - 4/9)^{n}. The probability it is not divisible by 2 and 5 is (1 - 5/9)^{n}. The quantity you seek is 1 - ((1 - 1/9)^{n} +(1 - 4/9)^{n} -(1 - 5/9)^{n})= 1 - (8/9)^n -(5/9)^n + (4/9)^n. 3 Let X and Y be independent random variable with distribution uniform on [0,1], and let W=\max(X,Y). For 0\le w\le 1, we haveF_W(w)=\Pr(W\le w)=\Pr\left((X\le w)\cap(Y\le w)\right)=w^2$$For the density function f_W(w) of W, differentiate F_W(w). We get 2w on the interval (0,1), precisely as the simulation suggests. 3 Corrigendum$$\mathbb{var}[\overline{x}]=\mathbb{var}\left[ \frac{1}{N}\sum_{i=1}^Nx_i \right]\stackrel{\text{iid}}{=} \frac{1}{N^2}\sum_{i=1}^N\mathbb{var}\left[x_i \right]=N \cdot \frac{m_{c_2}}{N^2}=\frac{m_{c_2}}{N}$$Cov({x_i},{x_j}) = E\left[ {{x_i}{x_j}} \right] - E\left[ {{x_i}} \right]E\left[ {{x_j}} \right] = 0 Since \{x_1,\ldots, x_N\} iid : ... 1$$ 𝔼[\frac{1}{N^2}\sum_{i = 1}^n\sum_{j \neq i}x_ix_j] = \frac{1}{N^2}\sum_{i = 1}^n\sum_{j \neq i}𝔼[x_i]𝔼[x_j] $$And this is not equal to zero. 0 UUID v 1 will produce collision if the tiestamps are the same, so this depends on generation rate and timer resolution. The node-id should be the MAC of a built-in NIC, otherwise it should come from a special address block requested fro IEEE or based on a good random number (http://tools.ietf.org/html/rfc4122#section-4.5). The way I read that spec however is ... 1 Consider a distribution with$$P(X=m\delta)=1-\deltaP\left(X=m\tfrac{1-\delta+\delta^2}{\delta}\right)=\delta$$then E[X]=m but P(X\gt a ) can be made arbitrarily small by choosing \delta \lt \frac{a}{m} and if necessarily smaller. So to get something useful, you would need some constraint on X such as a given maximum or a given variance. 0 let's for simplicity generate points within an angle ALPHA of the north pole P = (1 - cos(ALPHA))/2 U = random(0,1) a random value in the range (0,1) PHI = acos(1-2*P*U) THETA = random (-PI,PI) PHI is the angle from the north pole 0 Looks an awful like a version of a jump-diffusion process, especially since its brownian between oscillations. 0 Hint: In the random walk, X_i is step #i (to right if +1, left if -1). Since \alpha > 1/2, this is a biased random walk. What does the condition on \sum_{k=1}^n Z_k say about \sum_{k=1}^n X_k (which is where the random walk ends up after n steps)? 0 The proposition you provide is correct if each of the X_i can only take on non-negative values. (I'm sure there are other more complicated conditions that would also make it correct, but for general distributions, the inequality does not hold.) A counterexample with n=2: Let X_1 and X_2 each be uniformly disributed on -3,-2, and choose k = ... 0 It is true if your random variables are nonnegative:$$ X_1+X_2+\cdots+X_n\leq k\implies X_1\leq k\;\&\;X_2\leq k\;\&\;\ldots\;\&\;X_n\leq k. $$So$$ \Pr(X_1+X_2+\cdots+X_n\leq k)\leq\Pr(X_1\leq k\;\&\;X_2\leq k\;\&\;\ldots\;\&\;X_n\leq k)=\prod_{i=1}^n\Pr(X_i\leq k).  The equality above uses independence.

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The notation $F_X$ is absurd since $X$ is a random process, not a random variable. For every $t$ and $x$ positive, $[X(t)\leqslant x]=[A^t\leqslant x]=[A\leqslant x^{1/t}]$ hence $F_{X(t)}(x)=F_A(x^{1/t})=\min\left\{1,\frac18x^{3/t}\right\}$. Equivalently, $F_{X(t)}(x)=0$ for every $x\leqslant0$, $F_{X(t)}(x)=\frac18x^{3/t}$ for every $0\leqslant ... 0 each bus has$(1-\frac{1}{100})^{100}\sim\frac{1}{e}$to be left empty. By linearity of expectaions the answer is$100(1-\frac{1}{100})^{100}\sim\frac{100}{e}\$.

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