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Apr
25
comment Prove $\{s_n\}$ converges if $\{a_n = s_n + 2s_{n+1}\}$ converges.
And now, duplicates, near-duplicates, functional duplicates of the question abound.
Apr
25
comment Random variables set representation in the sample space
A unique one.
Apr
25
comment Prove $\{s_n\}$ converges if $\{a_n = s_n + 2s_{n+1}\}$ converges.
@Golbez The mystery is solved, please see my previous comment.
Apr
25
comment Prove $\{s_n\}$ converges if $\{a_n = s_n + 2s_{n+1}\}$ converges.
@Umakant You know what? The conditions $a_n = s_n + 2s_{n\color{red}{-1}}$ and $a_n = s_n + 2s_{n\color{red}{+1}}$ are not equivalent.
Apr
25
comment Random variables set representation in the sample space
Obviously. To steal a quote from a famous philosopher of the late 20th century... what else?
Apr
25
comment Conditional expectation, sigma algebra
?? Each $A_i$ is in $\Sigma_Y$ hence each $\mathbf 1_{A_i}$ is $\Sigma_Y$-measurable hence $Z=E(X\mid Y)$ is $\Sigma_Y$-measurable, an assertion which is strictly equivalent to the assertion that $\Sigma_Z\subseteq\Sigma_Y$.
Apr
25
comment Random variables set representation in the sample space
Well, $P(Y=f(X))=P(A)$ where $A=\{\omega\in\Omega\mid Y(\omega)=f(X(\omega))\}$. Is this your question?
Apr
25
comment How do I show that $0<a_n^2<a_n$ If $\sum _{n=1}^\infty a_n$ is convergent?
What exactly makes that "$a_n$ must decrease to $0$ at some point"? This seems obviously not guaranteed.
Apr
25
comment Prove $\{s_n\}$ converges if $\{a_n = s_n + 2s_{n+1}\}$ converges.
@BerrickFillmore Indeed not the same problem, since one is correct and the conclusion of the other is wrong.
Apr
25
comment Conditional expectation, sigma algebra
$$E(X|Y)=\sum_iE(X|Y=y_i)\,\mathbf 1_{A_i}\qquad A_i=\{Y=y_i\}\in\Sigma_Y$$
Apr
25
comment Lotka-Volterra Problem From Arnold's Ordinary Differential Equations
The thing is that the system is "almost separable" (to use a non canonical term). To wit, if $(x(t),y(t))$ solves Lotka-Volterra then $dx=x(k-ay)dt$ and $dy=y(bx-\ell)dt$ hence $(bx-\ell)dx/x=(k-ay)dy/y$ (Houston, separation completed...), that is, $p(x(t))-p(x(0))=q(y(0))-q(y(t))$ where $(p,q)$ are the functions $p(z)=\int(bz-\ell)dz/z$ and $q(z)=\int(az-k)dz/z$, QED. The technique would apply to every system $\dot x=A(x)B(y)$, $\dot y=C(x)D(y)$.
Apr
24
revised Lotka-Volterra Problem From Arnold's Ordinary Differential Equations
added 49 characters in body; edited tags
Apr
24
comment Lotka-Volterra Problem From Arnold's Ordinary Differential Equations
Again: the transform $(x,y)\to(u,v)$ is not necessary to show that the $(x,y)$ solution curves are the level sets of the function $p(x)+q(y)$. The solutions of the $(u,v)$ differential system are indeed confined to the level sets of the function $p(u)+q(v)$ but to parts of these level sets since each $(u,v)$ solution can at most span one quarter of such a level set because no solution can cross the lines $u=\ell/b$ and $v=k/a$. And finally, I fail to see where Arnol'd uses $(u,v)$ to show that $p(x)+q(y)$ is invariant along every solution $(x,y)$ of the original Lotka-Volterra system.
Apr
24
comment Show $\mathbb{E}(X \mid Y,Z) = \mathbb{E}(X \mid Y)$ if $Z$ is independent of $X$ and $Y$
Right. Then my advice would be to reread my first comment much more carefully.
Apr
24
comment Phase Diagram Bifurcations and much more.
Did you try to plot ONE phase diagram of this, for starters? Or do you fail to know what is a phase diagram? Actually, what is your question exactly?
Apr
24
comment Define $(a_n)_{n\in \mathbb{N}}$ by $a_n:=\prod_{k=2}^n 1-\frac{1}{k^{1+c}}$. Does $\lim_{n\to\infty} a_n=0$ hold?
Hands-on, purely real, approach: for every $x$ small enough, $$e^{-2x}\leqslant1-x\leqslant e^{-x}.$$
Apr
24
comment Show $\mathbb{E}(X \mid Y,Z) = \mathbb{E}(X \mid Y)$ if $Z$ is independent of $X$ and $Y$
Hint: You want to show that $E(E(X|Y)u(Y,Z))=E(Xu(Y,Z))$ for every (bounded measurable) function $u$. Can you show that for a whole class of functions $u$ this identity indeed holds?
Apr
24
comment Summation of $3^k$ from $2$ to $72$
How? Why? A general result or a special case? So much mystery...
Apr
24
comment Homogeneous differential equation - cannot manipulate equation
And now that you know "why", @Doc, what comes next?
Apr
24
comment Lotka-Volterra Problem From Arnold's Ordinary Differential Equations
.../... while every $(x,y)$-curve does. 3.2. Second, the Ansatz $(x,y)\to(u,v)$ is simply not used to find the $(x,y)$ solution curves. 4. Finally, Arnol'd very much proves something, namely, that the $(x,y)$ solution curves are exactly the level sets of the functional $bx+ay-\ell\log x-k\log y$ (see the quantity $p(x)+q(y)$ in the book).