l*********t
发帖数: 89 | 1 What does a distribution with a maximal variance look like which is only
defined between 0 and 1? Give the proof. | t*******e
发帖数: 172 | 2 I can not help present the following idea.
It is easy to find two point distribution with probability 1/2 give us the
variance 1/4.
Then, we wanna prove it is an upper bound, I am not sure weather this method
was mentioned in this webpage.
Let us do the Monte Carlo Simulation, given a distribution p(x), we take n
samples from it. Then the variance is estimated by
(n(X_1^2+...+X_n^2)-(X_1+..+X_2)^2)/n^2, which is no more than 1/4, let n
tend to infinity, we know the variance no more than 1/4. This proves the
upper bound is 1/4. | l*********t
发帖数: 89 | 3 Cool idea. Thank you!
method
【在 t*******e 的大作中提到】
: I can not help present the following idea.
: It is easy to find two point distribution with probability 1/2 give us the
: variance 1/4.
: Then, we wanna prove it is an upper bound, I am not sure weather this method
: was mentioned in this webpage.
: Let us do the Monte Carlo Simulation, given a distribution p(x), we take n
: samples from it. Then the variance is estimated by
: (n(X_1^2+...+X_n^2)-(X_1+..+X_2)^2)/n^2, which is no more than 1/4, let n
: tend to infinity, we know the variance no more than 1/4. This proves the
: upper bound is 1/4.
| xf
发帖数: 68 | 4 Just start with definition
For any rv X and real number a, Var(X)<=E(X-a)^2.
Set a=1/2. when X in[0,1]
then Var(X)<=E(1/2)^2=1/4 | c******s
发帖数: 270 | 5 Var(x) = E[x^2] - E[x]^2
<= E[x] - E[x]^2
= 1/4 - (1/2 - E[x])^2
<= 1/4
when E[x^2] = E[x] =1/2, it reaches the higher bound |
|
