4,844 reputation
31732
bio website none
location Not from around here
age
visits member for 3 years, 10 months
seen Nov 30 at 2:44

Dec
16
awarded  Enlightened
Dec
16
awarded  Nice Answer
Sep
30
awarded  Explainer
Jun
18
comment Does affine equivariance implies shape unbiasedness?
Hi @user: The edit is helpful. It doesn't appear to address my remarks on the equivariance definition, though. Cheers.
Jun
18
comment Does affine equivariance implies shape unbiasedness?
Can you clear up the statement of affine equivariance? Currently none of the operations are well defined (matrix multiplication where dimensions don't match and addition of matrix and vector). What, if any, distributional assumptions on $\mathbf X $ are you making? (Independent rows?)
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.