Conditional expected value of order statistic
$begingroup$
Let $theta_1,dots,theta_N$ be a collection of independent RVs, ditributed uniformly on $[0,1]$. Further let $theta^{(r)}$ be the $r$th order statistic where
$theta^{(1)}leqdots theta^{(r)}leqdots theta^{(N)}$.
Is the following reasoning correct?
$mathbb{E}[theta_i vert theta_igeq theta^{(m)}]
= mathbb{P}(theta_i =theta^{(m)})mathbb{E}[theta_i vert theta_i=theta^{(m)}]
+ dots
+mathbb{P}(theta_i =theta^{(N)})mathbb{E}[theta_i vert theta_i=theta^{(N)}]\
=frac{1}{N}sum_{l=m}^{N}mathbb{E}[theta^{(l)}]
=frac{1}{N}sum_{l=m}^{N}frac{l}{N+1}$
Thanks in advance!
conditional-expectation order-statistics
asked Dec 7 '18 at 11:02
mcksmcks
13
$endgroup$
add a comment |
$begingroup$
Let $theta_1,dots,theta_N$ be a collection of independent RVs, ditributed uniformly on $[0,1]$. Further let $theta^{(r)}$ be the $r$th order statistic where
$theta^{(1)}leqdots theta^{(r)}leqdots theta^{(N)}$.
Is the following reasoning correct?
$mathbb{E}[theta_i vert theta_igeq theta^{(m)}]
= mathbb{P}(theta_i =theta^{(m)})mathbb{E}[theta_i vert theta_i=theta^{(m)}]
+ dots
+mathbb{P}(theta_i =theta^{(N)})mathbb{E}[theta_i vert theta_i=theta^{(N)}]\
=frac{1}{N}sum_{l=m}^{N}mathbb{E}[theta^{(l)}]
=frac{1}{N}sum_{l=m}^{N}frac{l}{N+1}$
Thanks in advance!
conditional-expectation order-statistics
asked Dec 7 '18 at 11:02
mcksmcks
13
$endgroup$
add a comment |
$begingroup$
Let $theta_1,dots,theta_N$ be a collection of independent RVs, ditributed uniformly on $[0,1]$. Further let $theta^{(r)}$ be the $r$th order statistic where
$theta^{(1)}leqdots theta^{(r)}leqdots theta^{(N)}$.
Is the following reasoning correct?
$mathbb{E}[theta_i vert theta_igeq theta^{(m)}]
= mathbb{P}(theta_i =theta^{(m)})mathbb{E}[theta_i vert theta_i=theta^{(m)}]
+ dots
+mathbb{P}(theta_i =theta^{(N)})mathbb{E}[theta_i vert theta_i=theta^{(N)}]\
=frac{1}{N}sum_{l=m}^{N}mathbb{E}[theta^{(l)}]
=frac{1}{N}sum_{l=m}^{N}frac{l}{N+1}$
Thanks in advance!
conditional-expectation order-statistics
asked Dec 7 '18 at 11:02
mcksmcks
13
$endgroup$
Let $theta_1,dots,theta_N$ be a collection of independent RVs, ditributed uniformly on $[0,1]$. Further let $theta^{(r)}$ be the $r$th order statistic where
$theta^{(1)}leqdots theta^{(r)}leqdots theta^{(N)}$.
Is the following reasoning correct?
$mathbb{E}[theta_i vert theta_igeq theta^{(m)}]
= mathbb{P}(theta_i =theta^{(m)})mathbb{E}[theta_i vert theta_i=theta^{(m)}]
+ dots
+mathbb{P}(theta_i =theta^{(N)})mathbb{E}[theta_i vert theta_i=theta^{(N)}]\
=frac{1}{N}sum_{l=m}^{N}mathbb{E}[theta^{(l)}]
=frac{1}{N}sum_{l=m}^{N}frac{l}{N+1}$
Thanks in advance!
conditional-expectation order-statistics
conditional-expectation order-statistics
asked Dec 7 '18 at 11:02
mcksmcks
13
asked Dec 7 '18 at 11:02
mcksmcks
13
edited Dec 9 '18 at 10:08
mcks
asked Dec 7 '18 at 11:02
mcksmcks
13
asked Dec 7 '18 at 11:02
mcksmcks
13
asked Dec 7 '18 at 11:02
mcksmcks
13
13
add a comment |
add a comment |
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