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 1d revised $\DeclareMathOperator{cov}{Cov}\cov \, (A,B)\geq 0$ and $\cov(B,C)\geq 0\Rightarrow \cov(A,C)\geq 0$? added 9 characters in body 1d revised Radius of convergence of the series $\sum\limits_{n=1}^\infty (-1)^n\frac{1\cdot3\cdot5\cdots(2n-1)}{3\cdot6\cdot9\cdots(3n)}x^n$ edited title 1d comment Radius of convergence of the series $\sum\limits_{n=1}^\infty (-1)^n\frac{1\cdot3\cdot5\cdots(2n-1)}{3\cdot6\cdot9\cdots(3n)}x^n$ The ratio test is actually (slightly) easier to perform without the rewriting. 1d comment radius of convergence of $\sum_{n=1}^\infty n!^2x^{n^2}$ That, or, forgetting one has a power series, the ratio test on $c_n=n!^2x^{n^2}$. 1d answered How do we plot nonlinear differential equations 1d comment How do we plot nonlinear differential equations How the plotting method should depend on it being out of phase or nonlinear escapes me, I must admit. For streamplots, simply use streamplot[{0.01*7*x - 0.0001*4*x*y,-0.01*8*y + 0.0001*5*x*y},{x, -0.1, 600}, {y, -0.1, 600}]. 1d comment Men and Women enter a supermarket according to independent poisson process (stochastic process) Alternatively, each new customer is a man with probability $m/(m+w)$, independently of the other arrivals, where $m=2$ is the rate of men's arrivals and $w=4$ is the rate of women's arrivals. Hence the probability that the $k$ next arrivals after any given arrival are men's arrivals is $(m/(m+w))^k=1/3^k$. 1d comment How do we plot nonlinear differential equations Lotka-Volterra predator-prey solutions are out of phase. What is your question? 1d comment Probability of extinction in branching process @teddy1203109238 There is a general theorem about this in your notes. 1d comment Is this backwards process a Markov chain? How do I know? I made the computation. For the same reason, I know that nothing but the definition is involved. 1d revised Given $\mathbb{E}[X|Y] = Y$ a.s. and $\mathbb{E}[Y|X] = X$ a.s. show $X = Y$ a.s. added 50 characters in body; edited tags 1d comment Given $\mathbb{E}[X|Y] = Y$ a.s. and $\mathbb{E}[Y|X] = X$ a.s. show $X = Y$ a.s. This is the correct proof, +1. 1d comment How to calculate this integral? What @heropup said. The right thing to do is to write down a solution yourself, to post it as an answer and even, after a while, possibly to accept it. 2d comment How to calculate this integral? Sorry? This shows that the centered normal distribution with a well chosen variance is stationary, no? 2d comment Given $\mathbb{E}[X|Y] = Y$ a.s. and $\mathbb{E}[Y|X] = X$ a.s. show $X = Y$ a.s. @ki3i Indeed this is the idea of the $L^1$ proof. Note however that $\mathbb E[(X-Y){\bf 1}_{Y\leqslant c}{\bf 1}_{X> c}]>0$ is not guaranteed, only $\mathbb E[(X-Y){\bf 1}_{Y\leqslant c}{\bf 1}_{X> c}]\geqslant0$. And later on, the step [$\mathbb E[(X-Y){\bf 1}_{Y\leqslant c}{\bf 1}_{X\leqslant c}]<0$ for every $c$ implies \$X