Zhou Heng
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 Feb 17 asked Ask for a question about independence Feb 17 accepted A question about independence of bivariate random variables Feb 16 asked A question about independence of bivariate random variables Feb 16 accepted A question about independence wrt joint random variable Feb 16 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. Feb 16 comment A question about independence wrt joint random variable Thank you. So 1) is not true. What about 2)? Feb 16 asked A question about independence wrt joint random variable Feb 14 accepted A question about conditional independence Feb 14 asked A question about conditional independence Jan 22 comment Ask for a proof of an upper bound Thank you!_____ Jan 22 accepted Ask for a proof of an upper bound Jan 22 comment Ask for a proof of an upper bound wonderful!______ Jan 22 asked Ask for a proof of an upper bound Jan 20 comment limit of probability Using the same method Mercy gave, we can prove that $P(A_n\cup B_n)\to 1$. Jan 20 awarded Commentator Jan 20 comment limit of probability Thank you!_____ Jan 20 accepted limit of probability Jan 20 asked limit of probability Jan 16 awarded Tumbleweed Jan 12 comment Considering $\int_0^\infty2^{-x}(1+x)^ndx$ Cool! Thank you! .