Zhou Heng
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# 28 Comments

 Aug7 comment A problem on a proof in a graph theory textbook I see. "shortest" is the key. Thank you very much! Jul27 comment How to prove connectivity $\leq$ minimum degree? Thanks. Seem we have different definition of connectivity, but yours seems better for this question. May6 comment Relation about Gateaux differentiable and differentiable differentiable is what we meet in undergraduate calculus course (see, e.g., Def. 9.11 in P212 of baby Rudin). Gateaux differentiable (at x) is defined as follows: (1)the directional derivative $f'(x;v)$ exists for all $v\in\mathbb{R}^n$, (2)there exists (unique) $f_G'(x)\in\mathbb{R}^*$ such that $f'(x;v)=$ for all $v\in\mathbb{R}^n$, where $\mathbb{R}^*$ means the dual space of $\mathbb{R}^n$. May5 comment Relation about Gateaux differentiable and differentiable Thank you!_____ May4 comment A question about definition of hyperplane Thank you...... Mar16 comment Ask for a question about independence I answered my question, but I would like to give the credit to you. Mar6 comment Ask for a thing about multivariable concave function Great! So, is sufficiently smooth concave function a special case of Lipschitz continuous function? Mar6 comment Ask for a thing about multivariable concave function I see the mistake in the post. I should add "absolute value of" Mar6 comment Ask for a thing about multivariable concave function why $M$ does not exist for $-x^2$? It is a convex function and the slope of the tangent line on the parabola satisfies the inequality for the fixed $x_0$ and all other $x$. Mar6 comment Ask for a thing about multivariable concave function why not M=1? you know what I mean. And, if we take M=17, the inequality does not hold for all x. Mar2 comment Expression for the size of type class, or multinomial coefficient. It seems that we can use Stirling's approximation $n!\approx\sqrt{2\pi n}(n/e)^n$. But this is only an approximation and a big-O term $O(\ldots(n))$ or something is needed for the equation to hold exactly. I don't know what form of $O(\ldots(n))$ is. Any comment is welcome! ... I have to go to bed now:-( Feb21 comment An optimization problem with regard to permutation function I got it. Thank you! (I mean the former) Feb17 comment Ask for a question about independence @Did: so the independence of $X'$ and $(X,Y)$ can not be derived from the independence of $X$ and $X'$? Note that $Y$ is generated solely from $X$, not like $Y=X+X'$ in the answer where $Y$ is also determined by $X'$. Feb17 comment Ask for a question about independence Yes, I must have omitted some assumptions, but I don't know what they are. Currently the assumptions can be summarized as: $X$ and $X'$ are iid with common probability $p(x)$. $Y$ is generated from $X$ according to the trasition probability $p(y|x)$ (sorry for the typo in the original post; I have corrected it). I really can not think of any other assumptions. Maybe someone who read the proof of Shannon's channel coding theorem before and familiar with random coding and joint typicality can answer this question. Feb17 comment Ask for a question about independence but $X'$ and $Y$ are indeed independent in the proof to show subsequent results. Let's assume $X'$ and $X$ have the same distribution. Feb16 comment A question about independence wrt joint random variable Thank you. I worked it out. We have now $p(x|y_1y_2)=p(x)$. So $p(x|y_1)=\frac{p(xy_1)}{p(y_1)}=\frac{\sum\limits_{y_2}p(xy_1y_2)}{p(y_1)}$$=\f‌​rac{\sum\limits_{y_2}p(x|y_1y_2)p(y_1y_2)}{p(y_1)}=\frac{\sum\limits_{y_2}p(x)p(y‌​_1y_2)}{p(y_1)}=\frac{p(x)\sum\limits_{y_2}p(y_1y_2)}{p(y_1)}=\frac{p(x)p(y_1)}{p‌​(y_1)}=p(x)$. So 2) is true. Thank you again. Feb16 comment A question about independence wrt joint random variable Thank you. So 1) is not true. What about 2)? Jan22 comment Ask for a proof of an upper bound Thank you!_____ Jan22 comment Ask for a proof of an upper bound wonderful!______ Jan20 comment limit of probability Using the same method Mercy gave, we can prove that $P(A_n\cup B_n)\to 1$.