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Jun
29
revised What distribution has $X^n$ if $X$ is normal distributed?
added 10 characters in body
Jun
29
revised What is the significance of multiplying 2 Gaussian PDFs?
added 78 characters in body
Jun
29
revised What distribution has $X^n$ if $X$ is normal distributed?
added 97 characters in body; edited tags
Jun
29
revised What distribution has $X^n$ if $X$ is normal distributed?
rolled back to a previous revision
Jun
29
comment Probability that after 10,000 steps (+-1) you'll end up at the origin. How to use Central Limit Theorem?
Sorry but this is cheating... Including $1/\sqrt{n}$ in $a$ and $b$ yields a very small interval, which is natural since one chooses carefully $a$ and $b$ to catch one value of $S_n$. Yes this kind of mathematical nonsense is everywhere, textbooks, applications to the life sciences, others (ever heard of CLTs applied to samples of size 10 in reputable medicine journals?) -- and this is not a reason to reproduce them without thinking.
Jun
28
comment Probability that after 10,000 steps (+-1) you'll end up at the origin. How to use Central Limit Theorem?
.../... Finally, note that a sure sign that something is amiss with this approach is that it would yield the same asymptotics for $$P(S_{2n}=0)\qquad\text{and}\qquad P(S_{2n+1}=0),$$ which behave rather differently when $n\to\infty$.
Jun
28
comment Probability that after 10,000 steps (+-1) you'll end up at the origin. How to use Central Limit Theorem?
This was already mentioned à propos another answer, now deleted, but here we go again: CLT cannot yield this kind of information. What CLT says is that $$P(a\sqrt{n}\leqslant S_n\leqslant b\sqrt{n})$$ converges when $n\to\infty$, for every $(a,b)$ fixed, but it says nothing about the asymptotics of $$P(a_n\sqrt{n}\leqslant S_n\leqslant b_n\sqrt{n})$$ when $n\to\infty$, for $(a_n,b_n)$ depending on $n$ -- and yet this is what one needs here, with $a_n$ and $b_n$ roughly $\pm1/\sqrt{n}$. .../...
Jun
28
comment Probability that all colors are chosen
Seems tailored to apply the principle of inclusion-exclusion. Did you try it?
Jun
28
comment Does this series converge, and if so to what value?: $\sum_{n=0}^\infty \left\{\frac{1}{(n+1)^2} \right\}\ln(2n+1)$
If you can read French, the convergence is folklore: fr.wikipedia.org/wiki/S%C3%A9rie_de_Bertrand Re the value of the sum, I see no reason to expect it should be a simple combination of some known constants.
Jun
28
comment Product Sigma Algebra, Explicitly
Yes $\mathcal F_1\otimes\mathcal F_2=2^{\Omega_1}\otimes2^{\Omega_2}=2^{\Omega_1\times\Omega_2}$. (Not the most exciting example of sigma-algebras, but still...)
Jun
28
comment Jacobian Transformation p.d.f
Seems perfect to me. +1.
Jun
28
comment What distribution has $X^n$ if $X$ is normal distributed?
I mean that "this yields a full answer for every odd exponent n. And now for the even exponents..."
Jun
28
comment Jacobian Transformation p.d.f
The densities of $U$ and $V$ are correct now (but it is not clear to me what the notation $f_{X,Y}^{-1}(u,v)$ is referring to).
Jun
28
revised Jacobian Transformation p.d.f
deleted 96 characters in body
Jun
28
comment Question on nonhomogeneous ODE
@NonStandard Fully rigorous, thank you (note that the fact that the word "remark" appears in an answer does not mean that the answer is a remark). Which step do you have trouble to follow?
Jun
28
comment Trapping region for $\ddot x + (x^2-2) \dot x + x + \sin(x) =0$
@NonStandard No work to show since this is a WA diagram, as mentioned at the beginning of the post. Also mentioned is that this was posted waiting for a full answer by the OP.
Jun
28
comment Solving $2^x - 3^x + 6^x =0$.
@NonStandard The reason why this is community. Re the OP, you might be missing the target here since the OP seemed to consider the explanation clarified things.
Jun
28
comment What distribution has $X^n$ if $X$ is normal distributed?
Yet this yields a full answer for every odd exponent n. And now for the even exponents...
Jun
28
comment For which values of $c_1, c_2$ and $c_3$ is (1, 2, -2) a local minimum
It seems you showed a necessary condition is that $$\frac{c_1+c_3}{4} = \frac{c_2+c_3}{8}$$
Jun
28
comment $X_1$, $X_2$ i.i.d., prove that $E(X_1\mid X_1+X_2) = E(X_2\mid X_1+X_2)$
Hint: $E(h(U+V)U)$ depends only on the joint distribution of $(U,V)$. Apply this to $(U,V)=(X_1,X_2)$ and $(U,V)=(X_2,X_1)$.