Mixing
Holder's Inequality with Indicator Functions probability
$begingroup$
Suppose I have two events $A$ and $B$. If $A$ and $B$ are independent, I can say that
$$mathbb{E}[mathbb{1}_{A}mathbb{1}_{B}] = mathbb{E}[mathbb{1}_A] mathbb{E}[mathbb{1}_B] = Pr(A) Pr(B)$$
where $mathbb{1}_{A}$ is the indicator function of $A$ and likewise for $B$. If $A$ and $B$ are not independent, then may I apply Holder's Inequality to the indicator functions? That is, may I say that
$$ mathbb{E}[|mathbb{1}_{A}mathbb{1}_{B}|] leq mathbb{E}[|mathbb{1}_A|^p]^{frac{1}{p}}mathbb{E}[|mathbb{1}_B|^q]^{frac{1}{q}}$$
for $p,q in [1,infty)$ such that $p^{-1} + q^{-1} = 1$? This would imply that (for integer $p,q$)
$$mathbb{E}[mathbb{1}_{A}mathbb{1}_{B}]leq Pr(A)^{frac{1}{p}}Pr(B)^{frac{1}{q}}$$
With regard to other, similar questions:
This question deals with expectations conditioned on events $A$ and $B$ as does this question. Many questions deal with the more general real analysis version of Holder's Inequality. I simply want to know whether I am correctly applying Holder's Inequality in the context of probability to (possibly correlated) indicator functions.
probability expected-value holder-inequality
asked Jan 28 at 8:03
sharkbaitsharkbait
254
$endgroup$
add a comment |
$begingroup$
Suppose I have two events $A$ and $B$. If $A$ and $B$ are independent, I can say that
$$mathbb{E}[mathbb{1}_{A}mathbb{1}_{B}] = mathbb{E}[mathbb{1}_A] mathbb{E}[mathbb{1}_B] = Pr(A) Pr(B)$$
where $mathbb{1}_{A}$ is the indicator function of $A$ and likewise for $B$. If $A$ and $B$ are not independent, then may I apply Holder's Inequality to the indicator functions? That is, may I say that
$$ mathbb{E}[|mathbb{1}_{A}mathbb{1}_{B}|] leq mathbb{E}[|mathbb{1}_A|^p]^{frac{1}{p}}mathbb{E}[|mathbb{1}_B|^q]^{frac{1}{q}}$$
for $p,q in [1,infty)$ such that $p^{-1} + q^{-1} = 1$? This would imply that (for integer $p,q$)
$$mathbb{E}[mathbb{1}_{A}mathbb{1}_{B}]leq Pr(A)^{frac{1}{p}}Pr(B)^{frac{1}{q}}$$
With regard to other, similar questions:
This question deals with expectations conditioned on events $A$ and $B$ as does this question. Many questions deal with the more general real analysis version of Holder's Inequality. I simply want to know whether I am correctly applying Holder's Inequality in the context of probability to (possibly correlated) indicator functions.
probability expected-value holder-inequality
asked Jan 28 at 8:03
sharkbaitsharkbait
254
$endgroup$
1
$begingroup$
What you have done is fine for $1<p<infty$. For $p=1$ the inequality simply says $E|I_AI_B|leq P(A)$ if $P(B)neq 0$ and $E|I_AI_B|=0$ if $P(B)=0$ which is of no interest.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 8:07
add a comment |
$begingroup$
Suppose I have two events $A$ and $B$. If $A$ and $B$ are independent, I can say that
$$mathbb{E}[mathbb{1}_{A}mathbb{1}_{B}] = mathbb{E}[mathbb{1}_A] mathbb{E}[mathbb{1}_B] = Pr(A) Pr(B)$$
where $mathbb{1}_{A}$ is the indicator function of $A$ and likewise for $B$. If $A$ and $B$ are not independent, then may I apply Holder's Inequality to the indicator functions? That is, may I say that
$$ mathbb{E}[|mathbb{1}_{A}mathbb{1}_{B}|] leq mathbb{E}[|mathbb{1}_A|^p]^{frac{1}{p}}mathbb{E}[|mathbb{1}_B|^q]^{frac{1}{q}}$$
for $p,q in [1,infty)$ such that $p^{-1} + q^{-1} = 1$? This would imply that (for integer $p,q$)
$$mathbb{E}[mathbb{1}_{A}mathbb{1}_{B}]leq Pr(A)^{frac{1}{p}}Pr(B)^{frac{1}{q}}$$
With regard to other, similar questions:
This question deals with expectations conditioned on events $A$ and $B$ as does this question. Many questions deal with the more general real analysis version of Holder's Inequality. I simply want to know whether I am correctly applying Holder's Inequality in the context of probability to (possibly correlated) indicator functions.
probability expected-value holder-inequality
asked Jan 28 at 8:03
sharkbaitsharkbait
254
$endgroup$
Suppose I have two events $A$ and $B$. If $A$ and $B$ are independent, I can say that
$$mathbb{E}[mathbb{1}_{A}mathbb{1}_{B}] = mathbb{E}[mathbb{1}_A] mathbb{E}[mathbb{1}_B] = Pr(A) Pr(B)$$
where $mathbb{1}_{A}$ is the indicator function of $A$ and likewise for $B$. If $A$ and $B$ are not independent, then may I apply Holder's Inequality to the indicator functions? That is, may I say that
$$ mathbb{E}[|mathbb{1}_{A}mathbb{1}_{B}|] leq mathbb{E}[|mathbb{1}_A|^p]^{frac{1}{p}}mathbb{E}[|mathbb{1}_B|^q]^{frac{1}{q}}$$
for $p,q in [1,infty)$ such that $p^{-1} + q^{-1} = 1$? This would imply that (for integer $p,q$)
$$mathbb{E}[mathbb{1}_{A}mathbb{1}_{B}]leq Pr(A)^{frac{1}{p}}Pr(B)^{frac{1}{q}}$$
With regard to other, similar questions:
This question deals with expectations conditioned on events $A$ and $B$ as does this question. Many questions deal with the more general real analysis version of Holder's Inequality. I simply want to know whether I am correctly applying Holder's Inequality in the context of probability to (possibly correlated) indicator functions.
probability expected-value holder-inequality
probability expected-value holder-inequality
asked Jan 28 at 8:03
sharkbaitsharkbait
254
asked Jan 28 at 8:03
sharkbaitsharkbait
254
asked Jan 28 at 8:03
sharkbaitsharkbait
254
asked Jan 28 at 8:03
sharkbaitsharkbait
254
asked Jan 28 at 8:03
sharkbaitsharkbait
254
254
1
$begingroup$
What you have done is fine for $1<p<infty$. For $p=1$ the inequality simply says $E|I_AI_B|leq P(A)$ if $P(B)neq 0$ and $E|I_AI_B|=0$ if $P(B)=0$ which is of no interest.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 8:07
add a comment |
1
$begingroup$
What you have done is fine for $1<p<infty$. For $p=1$ the inequality simply says $E|I_AI_B|leq P(A)$ if $P(B)neq 0$ and $E|I_AI_B|=0$ if $P(B)=0$ which is of no interest.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 8:07
1
1
$begingroup$
What you have done is fine for $1<p<infty$. For $p=1$ the inequality simply says $E|I_AI_B|leq P(A)$ if $P(B)neq 0$ and $E|I_AI_B|=0$ if $P(B)=0$ which is of no interest.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 8:07
$begingroup$
What you have done is fine for $1<p<infty$. For $p=1$ the inequality simply says $E|I_AI_B|leq P(A)$ if $P(B)neq 0$ and $E|I_AI_B|=0$ if $P(B)=0$ which is of no interest.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 8:07
add a comment |
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$begingroup$
What you have done is fine for $1<p<infty$. For $p=1$ the inequality simply says $E|I_AI_B|leq P(A)$ if $P(B)neq 0$ and $E|I_AI_B|=0$ if $P(B)=0$ which is of no interest.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 8:07