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visits member for 3 years, 5 months
seen Jun 21 at 13:34

Feb
19
comment Why does this covariance matrix have additional symmetry along the anti-diagonals?
@Michael: Please check the timestamps of the edits and comments. Cheers.
Feb
16
awarded  Nice Answer
Feb
13
awarded  Yearling
Jan
1
comment Inequalities of the quantile function
possible duplicate of Quantile function properties
Dec
29
comment Triple Euler sum result $\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$
Please post the symbolic input you entered.
Dec
22
comment $\int_{t=-\infty}^x (G(t)-F(t))\mbox{d}t\geq 0\forall x$ and $\frac{\mbox{d}F(t)}{\mbox{d}G(t)}$ increasing $\Longrightarrow G(x)\geq F(x)\forall x$?
Are $1-$ and $2-$ supposed to be item identifiers in a list? If so, the way you have it typeset makes it very easy to confuse it with something very different!
Dec
22
comment Distribution for ratio of dependent quadratic forms.
(As a side note, since $\mathbf x_1^T \mathbf x_1^{\perp} = \mathbf x_1^T \mathbf Q^T \mathbf Q x_1^{\perp}$ for any orthogonal $\mathbf Q$ and by standard properties of the multivariate normal, without loss of generality we can assume $\mathbf A$ to be diagonal.)
Dec
22
comment Distribution for ratio of dependent quadratic forms.
Is there a way you can make your question a little more precise. Employing a literal interpretation, this is not even true in the simplest of cases. Take $\boldsymbol{\mu} = 0$ and $\boldsymbol{\Sigma} = \mathbf{A} = \mathbf{I}$ and choose $\mathbf{x}_1 = (x_{0,1}, 0, \ldots, 0)$. Then $s$ cannot be $F$ distributed since $s \geq 1$ with probability 1.
Dec
7
comment Probability question with false negative and false positive
@OldJohn: You may (claim to) be old, but you sure are fast! Beat me by a whisker. :-)
Nov
28
comment show that if $X\ge 0$ , $E(X)\le \sum_{n=0}^{\infty}P(X>n)$.
Crosspost: stats.stackexchange.com/q/77922/2970
Nov
25
revised Famous black mathematicians
Minor grammatical fixes.
Nov
25
comment Famous black mathematicians
(+1) ...and Blackwell's Theorem, one of the fundamental contributions to renewal theory which extends the elementary renewal theorem. Among (many) other contributions...
Nov
25
comment Famous black mathematicians
@000: I find your remark a little bit ignorant (in the literal, not offensive, sense of the word). While "fame" is a rather relative and slippery term, David Blackwell certainly fits the bill, in my mind. Fundamental contributions in probability, stochastic processes and statistics, with his name attached to among the most famous theorems in mathematical statistics and renewal theory. He also "dabbled" in game theory.
Nov
19
comment How to prove that $2\sqrt{a^{ea}b^{eb}}\ge a^{eb}+b^{ea}$ for $a > 0, b > 0$?
Or, just assume, without loss of generality, that $b \geq a$ and divide. :-)
Nov
19
revised How to prove that $2\sqrt{a^{ea}b^{eb}}\ge a^{eb}+b^{ea}$ for $a > 0, b > 0$?
General cleanup.
Nov
19
comment How to prove that $2\sqrt{a^{ea}b^{eb}}\ge a^{eb}+b^{ea}$ for $a > 0, b > 0$?
It may also be worth pointing out that if we replace the RHS of the OP's first equation with the corresponding geometric mean, namely $2 a^{eb/2} b^{ea/2}$, the result becomes (almost trivially) true. Thus, the analogous GM version of the second inequality in the OP's post is true. (The HM version is not.)
Nov
19
comment How to prove that $2\sqrt{a^{ea}b^{eb}}\ge a^{eb}+b^{ea}$ for $a > 0, b > 0$?
(+1) Indeed, even the most crude analysis of the second term in your display equation shows that $a = 2^{1 / 2 e} \cdot 8 \leq 9.09$ is already enough.
Nov
18
comment how to prove $\int_{0}^{a}B(t)dt\sim N(0,\frac{a^3}{3})$
Please do not crosspost identical questions simultaneously. Please see this meta response for more information.
Nov
17
comment If $|X_{n}| \leq Y$ almost surely, show that $\sup_{n}|X_{n}|\leq Y$ almost surely as well.
(+1) Because it seems to me that for students first encountering the subject, what the standard shorthand notation hides is often as much (or more of) an impediment to learning the material than any difficulty with the concepts themselves.
Nov
17
comment If $|X_{n}| \leq Y$ almost surely, show that $\sup_{n}|X_{n}|\leq Y$ almost surely as well.
I detect a study group in the making.