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seen Aug 7 '13 at 5:49

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! .
Jan
12
accepted Considering $\int_0^\infty2^{-x}(1+x)^ndx$