n******r
发帖数: 3 | 1
It is not that hard as you thought, but you may need Cauchy-Schwarz
inequality.
a <= E|X| = E{|X|*I[X=sa]} [where I is indicator function]
<= sa + sqrt(E{|X|^2}*E{I[X>=sa]}) [use C-S inequality and I^2=I]
= sa + sqrt(1*P(X>=sa))
So P(X>=sa) >= (1-s)^2*a^2. # |
g*******g
发帖数: 45 | 2 I found this problem in a text book. It keeps confusing me.. |
w*******e
发帖数: 14 | 3
2B+1W imples P=2/3 but
P=2/3 not implies originally 2B+1W.
actually, P(balck)=1(choose from the third ball whic is black)*1/3+
1/2(choose from the two random balls)*2/3=2/3
【在 g*******g 的大作中提到】
: I found this problem in a text book. It keeps confusing me..
|
H****h
发帖数: 1037 | 4 解释太少。从第一步开始就不一定成立，即所谓P(BB)=P(BW)=P(WB)=P(WW)=1/4。
【在 g*******g 的大作中提到】
: I found this problem in a text book. It keeps confusing me..
|
