Conditional Expectation Inequality for bounded momentConditional expectation w.r.t. random variable and w.r.t. $sigma$-algebra, equivalenceEstimator for conditional expectation w.r.t an eventVariance of Conditional Expectation From First PrinciplesConditional expectation with respect to sigma algebraInequality for Poisson point processIs my understanding of a multivariate discrete conditional expectation calculation correct?How is the conditional expectation calculated?show that Conditional Expectation inequality existsConditional expectation counterexamplesIntuition behind the expectation of a conditional expectation
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Conditional Expectation Inequality for bounded moment
Conditional expectation w.r.t. random variable and w.r.t. $sigma$-algebra, equivalenceEstimator for conditional expectation w.r.t an eventVariance of Conditional Expectation From First PrinciplesConditional expectation with respect to sigma algebraInequality for Poisson point processIs my understanding of a multivariate discrete conditional expectation calculation correct?How is the conditional expectation calculated?show that Conditional Expectation inequality existsConditional expectation counterexamplesIntuition behind the expectation of a conditional expectation
$begingroup$
I was going through a proof in Agnostic Estimation of Mean and Variance (Lemma 3.12, page 21) and encountered the following:
- Let $X$ be a random variable with $mathbfE[X] = mu$ and $mathbfE[(X-mu)^2] = sigma^2$
- Let $A$ be an event that occurs with probability $1-epsilon$ and let $A^c$ be the complement (i.e. $P(A^c)=epsilon$)
- Let $d Omega$ be the probability measure
Using $Ebig[(Y - E[Y])^4big] geq big(Ebig[(Y-E[Y])^2big]big)^2$ for a random variable $Y$, and $P(A^c)=epsilon$, we have
$$
frac1epsilon bigg( int_A^c (X - mu)^2 dOmega bigg)^2 leq int_A^c (X - mu)^4 dOmega
$$
What I do not understand is where the $frac1epsilon$ is coming from when just using the proposed inequality with $Y = X mid A^c$?
And how to show the inequality still holds?
probability probability-theory inequality conditional-expectation expected-value
asked Mar 21 at 19:50
eeekeeek
12
$endgroup$
add a comment |
$begingroup$
I was going through a proof in Agnostic Estimation of Mean and Variance (Lemma 3.12, page 21) and encountered the following:
- Let $X$ be a random variable with $mathbfE[X] = mu$ and $mathbfE[(X-mu)^2] = sigma^2$
- Let $A$ be an event that occurs with probability $1-epsilon$ and let $A^c$ be the complement (i.e. $P(A^c)=epsilon$)
- Let $d Omega$ be the probability measure
Using $Ebig[(Y - E[Y])^4big] geq big(Ebig[(Y-E[Y])^2big]big)^2$ for a random variable $Y$, and $P(A^c)=epsilon$, we have
$$
frac1epsilon bigg( int_A^c (X - mu)^2 dOmega bigg)^2 leq int_A^c (X - mu)^4 dOmega
$$
What I do not understand is where the $frac1epsilon$ is coming from when just using the proposed inequality with $Y = X mid A^c$?
And how to show the inequality still holds?
probability probability-theory inequality conditional-expectation expected-value
asked Mar 21 at 19:50
eeekeeek
12
$endgroup$
add a comment |
$begingroup$
I was going through a proof in Agnostic Estimation of Mean and Variance (Lemma 3.12, page 21) and encountered the following:
- Let $X$ be a random variable with $mathbfE[X] = mu$ and $mathbfE[(X-mu)^2] = sigma^2$
- Let $A$ be an event that occurs with probability $1-epsilon$ and let $A^c$ be the complement (i.e. $P(A^c)=epsilon$)
- Let $d Omega$ be the probability measure
Using $Ebig[(Y - E[Y])^4big] geq big(Ebig[(Y-E[Y])^2big]big)^2$ for a random variable $Y$, and $P(A^c)=epsilon$, we have
$$
frac1epsilon bigg( int_A^c (X - mu)^2 dOmega bigg)^2 leq int_A^c (X - mu)^4 dOmega
$$
What I do not understand is where the $frac1epsilon$ is coming from when just using the proposed inequality with $Y = X mid A^c$?
And how to show the inequality still holds?
probability probability-theory inequality conditional-expectation expected-value
asked Mar 21 at 19:50
eeekeeek
12
$endgroup$
I was going through a proof in Agnostic Estimation of Mean and Variance (Lemma 3.12, page 21) and encountered the following:
- Let $X$ be a random variable with $mathbfE[X] = mu$ and $mathbfE[(X-mu)^2] = sigma^2$
- Let $A$ be an event that occurs with probability $1-epsilon$ and let $A^c$ be the complement (i.e. $P(A^c)=epsilon$)
- Let $d Omega$ be the probability measure
Using $Ebig[(Y - E[Y])^4big] geq big(Ebig[(Y-E[Y])^2big]big)^2$ for a random variable $Y$, and $P(A^c)=epsilon$, we have
$$
frac1epsilon bigg( int_A^c (X - mu)^2 dOmega bigg)^2 leq int_A^c (X - mu)^4 dOmega
$$
What I do not understand is where the $frac1epsilon$ is coming from when just using the proposed inequality with $Y = X mid A^c$?
And how to show the inequality still holds?
probability probability-theory inequality conditional-expectation expected-value
probability probability-theory inequality conditional-expectation expected-value
asked Mar 21 at 19:50
eeekeeek
12
asked Mar 21 at 19:50
eeekeeek
12
edited Mar 25 at 13:32
eeek
asked Mar 21 at 19:50
eeekeeek
12
asked Mar 21 at 19:50
eeekeeek
12
asked Mar 21 at 19:50
eeekeeek
12
12
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
So this is actually trivial but the notation was confusing me.
Because $dOmega$ is the probability measure for the entire space, $int_A^c(X-mu)^2 dOmega=P(A^c)Var(X mid A^c)$ so we'd have to normalize it.
$$
beginaligned
&bigg( int_A^c (X - mu)^2 d Omega bigg)^2 leq int_A^c (X - mu)^4 dOmega
\
&implies bigg( frac1P(A^c) int_A^c (X-mu)^2 dOmegabigg)^2 leq frac1P(A^c) int_A^c (X - mu)^4 dOmega\
&implies frac1P(A^c)bigg( int_A^c (X-mu)^2 dOmegabigg)^2 leq int_A^c (X - mu)^4 dOmega
endaligned
$$
answered Mar 25 at 22:56
eeekeeek
12
$endgroup$
add a comment |
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$begingroup$
So this is actually trivial but the notation was confusing me.
Because $dOmega$ is the probability measure for the entire space, $int_A^c(X-mu)^2 dOmega=P(A^c)Var(X mid A^c)$ so we'd have to normalize it.
$$
beginaligned
&bigg( int_A^c (X - mu)^2 d Omega bigg)^2 leq int_A^c (X - mu)^4 dOmega
\
&implies bigg( frac1P(A^c) int_A^c (X-mu)^2 dOmegabigg)^2 leq frac1P(A^c) int_A^c (X - mu)^4 dOmega\
&implies frac1P(A^c)bigg( int_A^c (X-mu)^2 dOmegabigg)^2 leq int_A^c (X - mu)^4 dOmega
endaligned
$$
answered Mar 25 at 22:56
eeekeeek
12
$endgroup$
add a comment |
$begingroup$
So this is actually trivial but the notation was confusing me.
Because $dOmega$ is the probability measure for the entire space, $int_A^c(X-mu)^2 dOmega=P(A^c)Var(X mid A^c)$ so we'd have to normalize it.
$$
beginaligned
&bigg( int_A^c (X - mu)^2 d Omega bigg)^2 leq int_A^c (X - mu)^4 dOmega
\
&implies bigg( frac1P(A^c) int_A^c (X-mu)^2 dOmegabigg)^2 leq frac1P(A^c) int_A^c (X - mu)^4 dOmega\
&implies frac1P(A^c)bigg( int_A^c (X-mu)^2 dOmegabigg)^2 leq int_A^c (X - mu)^4 dOmega
endaligned
$$
answered Mar 25 at 22:56
eeekeeek
12
$endgroup$
add a comment |
$begingroup$
So this is actually trivial but the notation was confusing me.
Because $dOmega$ is the probability measure for the entire space, $int_A^c(X-mu)^2 dOmega=P(A^c)Var(X mid A^c)$ so we'd have to normalize it.
$$
beginaligned
&bigg( int_A^c (X - mu)^2 d Omega bigg)^2 leq int_A^c (X - mu)^4 dOmega
\
&implies bigg( frac1P(A^c) int_A^c (X-mu)^2 dOmegabigg)^2 leq frac1P(A^c) int_A^c (X - mu)^4 dOmega\
&implies frac1P(A^c)bigg( int_A^c (X-mu)^2 dOmegabigg)^2 leq int_A^c (X - mu)^4 dOmega
endaligned
$$
answered Mar 25 at 22:56
eeekeeek
12
$endgroup$
So this is actually trivial but the notation was confusing me.
Because $dOmega$ is the probability measure for the entire space, $int_A^c(X-mu)^2 dOmega=P(A^c)Var(X mid A^c)$ so we'd have to normalize it.
$$
beginaligned
&bigg( int_A^c (X - mu)^2 d Omega bigg)^2 leq int_A^c (X - mu)^4 dOmega
\
&implies bigg( frac1P(A^c) int_A^c (X-mu)^2 dOmegabigg)^2 leq frac1P(A^c) int_A^c (X - mu)^4 dOmega\
&implies frac1P(A^c)bigg( int_A^c (X-mu)^2 dOmegabigg)^2 leq int_A^c (X - mu)^4 dOmega
endaligned
$$
answered Mar 25 at 22:56
eeekeeek
12
answered Mar 25 at 22:56
eeekeeek
12
answered Mar 25 at 22:56
eeekeeek
12
answered Mar 25 at 22:56
eeekeeek
12
12
add a comment |
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