In this statement “If a random variable Y is symmetrically distributed around its mean $\mu_Y$, that is, $Y=\mu_Y+U$, where U is symmetrically distributed aronud zero- then the odd moments of Y are powers of $\mu_Y$, i.e. $E[(Y)^j]= (\mu_Y)^j $ for any odd integer j, assuming the moment exists”. I think this statement is true for […]

Hi I’m stuck on this question: Recall that $X$ is said to have a lognormal distribution with parameters $\mu$ and $\sigma^2$ if $\log(X)$ is normal with mean $\mu$ and variance $\sigma^2$. Suppose $X$ is such a lognormal random variable. Find $\mathrm{E}[X]$. Find $\mathrm{Var}(X)$. I know that the approach is to find the moment generating function […]

We have geometric mean of pairwise arithmetic means on the left, which obeys the following inequality: $$\frac{x+y+z}{3} \geq \color{blue}{ \frac{\sqrt[3]{(x+y)(y+z)(z+x)}}{2} } \geq \sqrt[3]{xyz}$$ And on the right we have root-mean-square of geometric means, obeying the same inequality: $$\frac{x+y+z}{3} \geq \color{blue}{\sqrt{\frac{xy+yz+zx}{3}} } \geq \sqrt[3]{xyz}$$ This time I checked with Wolfram Alpha first, and apparently, the inequality […]

This question already has an answer here: Proving the AM-GM inequality for 2 numbers $\sqrt{xy}\le\frac{x+y}2$ 5 answers How to prove that $\frac{a+b}{2} \geq \sqrt{ab}$ for $a,b>0$? [duplicate] 3 answers

For $x,y$ positive real numbers, and $p\neq 0$ real, define the Hölder $p$-mean $$M_p(x,y) := \left(\frac{x^p+y^p}{2}\right)^{1/p}$$ whereas $$M_0(x,y) := \sqrt{xy}$$ is the limit of $M_p(x,y)$ when $p\to 0$, i.e., the geometric mean. Furthermore, when $p,q$ are two real numbers, define $L_{p,q}(x,y)$ as the obvious fixed point satisfying the equation $$L_{p,q}(x,y) = L_{p,q}(M_p(x,y),M_q(x,y))$$ —meaning that $L_{p,q}(x,y)$ […]

Take: $$a_0=x,~~~~b_0=y$$ $$a_{n+1}=\frac{a_n+\sqrt{a_nb_n}}{2},~~~~b_{n+1}=\frac{b_n+\sqrt{a_nb_n}}{2}$$ Then we obtain as a limit the logarithmic mean of $x,y$: $$\lim_{n \to \infty} a_n=\lim_{n \to \infty} b_n=\frac{x-y}{\ln x-\ln y}$$ I don’t know how to prove this. But I do know that numerically it fits really well. In fact, the best approximation is obtained if we take geometric mean of $a_n,b_n$: $$x=5,~~~~y=3$$ […]

In answering calculate the mean and variance of the highest number drawn on lottery based on the lowest number drawn, I couldn’t find the mean and variance of the order statistics of a discrete uniform sample without replacement anywhere, so I figured I’d derive them here for future reference. Draw $k$ distinct numbers uniformly from […]

This question already has an answer here: If $\sigma_n=\frac{s_1+s_2+\cdots+s_n}{n}$ then $\operatorname{{lim sup}}\sigma_n \leq \operatorname{lim sup} s_n$ 2 answers

We know that if we iterate arithmetic and harmonic means of two numbers, we get their geometric mean. So, basically if we need to compute the square root of $x$: $$\sqrt{x}=\sqrt{1 \cdot x}=AHM(1,x)$$ $$a_0=1,~~~~b_0=x$$ $$a_{n+1}=\frac{a_n+b_n}{2},~~~~~b_{n+1}=\frac{2a_nb_n}{a_n+b_n}=\frac{a_nb_n}{a_{n+1}}$$ As far as I know, this expression will converge for any real positive $x$. See this and this question for […]

I want to create an ordered sequence of various ‘three-number means‘ with as many different elements in it as possible. So far I’ve got $12$ ($8$ unusual ones are highlighted): $$\sqrt{\frac{x^2+y^2+z^2}{3}} \geq \color{blue}{ \frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}}} \geq $$ $$\geq \color{blue}{\frac{\sqrt{(x+y)^2+(y+z)^2+(z+x)^2}}{2 \sqrt{3}} } \geq \frac{x+y+z}{3} \geq $$ $$ \geq \color{blue}{ \frac{\sqrt[3]{(x+y)(y+z)(z+x)}}{2} } \geq \color{blue}{\sqrt{\frac{xy+yz+zx}{3}}} \geq $$ $$\geq […]

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