Need help proving an expected value inequality involving a sign function
I am trying to prove the following inequality:
$E(f(a,b)) le e^{E(-a cdot b)}$, where $E$ denotes expectation, and
$f(a,b) = {^{1 | a ne sign(b)}_{0 | a = sign(b)}$.
Additional info is that $a,b$ are random variables such that $a = {-1, +1}$, and $|b| le 1$.
From this, I was able to conclude the following:
Because $|b|$ is always less than 1 and $a = pm 1$, $-1 le -acdot b le 1$, and thus $-1 le E(-acdot b) le 1$.
From definition of $f(a,b)$,
$0 le E(f(a,b))le 1$
Furthermore, I was able to verify the inequality for boundary cases.
If $a$ never match the sign of $b$, then $E(f(a,b)) = 1$.
For $-a cdot b$, if $b > 0$, then $a = -1$, and thus $-a cdot b = -(-1 cdot b) = b > 0$.
If $b < 0$, then $a = 1$, and thus $-a cdot b = -(1 cdot b) = -b > 0$. Thus, for no matches, $0 < E(-acdot b) le 1$, so $1 < e^{E(-acdot b)} le e$, which makes it greater than or equal to 1.
If $a$ always match the sign of $b$, then $E(f(a,b)) = 0$.
For matches the sign of $a$ will always be the same as $b$, so $a cdot b$ will always be positive, and thus $-1 le -a cdot b < 0$, so $e^{-1} le e^{E(-acdot b)} < 1$.
What I am not sure about is how to prove that for any number of matches, this inequality holds. Help would be greatly appreciated!
expected-value
asked Nov 24 at 20:27
William Deng
32
add a comment |
I am trying to prove the following inequality:
$E(f(a,b)) le e^{E(-a cdot b)}$, where $E$ denotes expectation, and
$f(a,b) = {^{1 | a ne sign(b)}_{0 | a = sign(b)}$.
Additional info is that $a,b$ are random variables such that $a = {-1, +1}$, and $|b| le 1$.
From this, I was able to conclude the following:
Because $|b|$ is always less than 1 and $a = pm 1$, $-1 le -acdot b le 1$, and thus $-1 le E(-acdot b) le 1$.
From definition of $f(a,b)$,
$0 le E(f(a,b))le 1$
Furthermore, I was able to verify the inequality for boundary cases.
If $a$ never match the sign of $b$, then $E(f(a,b)) = 1$.
For $-a cdot b$, if $b > 0$, then $a = -1$, and thus $-a cdot b = -(-1 cdot b) = b > 0$.
If $b < 0$, then $a = 1$, and thus $-a cdot b = -(1 cdot b) = -b > 0$. Thus, for no matches, $0 < E(-acdot b) le 1$, so $1 < e^{E(-acdot b)} le e$, which makes it greater than or equal to 1.
If $a$ always match the sign of $b$, then $E(f(a,b)) = 0$.
For matches the sign of $a$ will always be the same as $b$, so $a cdot b$ will always be positive, and thus $-1 le -a cdot b < 0$, so $e^{-1} le e^{E(-acdot b)} < 1$.
What I am not sure about is how to prove that for any number of matches, this inequality holds. Help would be greatly appreciated!
expected-value
asked Nov 24 at 20:27
William Deng
32
add a comment |
I am trying to prove the following inequality:
$E(f(a,b)) le e^{E(-a cdot b)}$, where $E$ denotes expectation, and
$f(a,b) = {^{1 | a ne sign(b)}_{0 | a = sign(b)}$.
Additional info is that $a,b$ are random variables such that $a = {-1, +1}$, and $|b| le 1$.
From this, I was able to conclude the following:
Because $|b|$ is always less than 1 and $a = pm 1$, $-1 le -acdot b le 1$, and thus $-1 le E(-acdot b) le 1$.
From definition of $f(a,b)$,
$0 le E(f(a,b))le 1$
Furthermore, I was able to verify the inequality for boundary cases.
If $a$ never match the sign of $b$, then $E(f(a,b)) = 1$.
For $-a cdot b$, if $b > 0$, then $a = -1$, and thus $-a cdot b = -(-1 cdot b) = b > 0$.
If $b < 0$, then $a = 1$, and thus $-a cdot b = -(1 cdot b) = -b > 0$. Thus, for no matches, $0 < E(-acdot b) le 1$, so $1 < e^{E(-acdot b)} le e$, which makes it greater than or equal to 1.
If $a$ always match the sign of $b$, then $E(f(a,b)) = 0$.
For matches the sign of $a$ will always be the same as $b$, so $a cdot b$ will always be positive, and thus $-1 le -a cdot b < 0$, so $e^{-1} le e^{E(-acdot b)} < 1$.
What I am not sure about is how to prove that for any number of matches, this inequality holds. Help would be greatly appreciated!
expected-value
asked Nov 24 at 20:27
William Deng
32
I am trying to prove the following inequality:
$E(f(a,b)) le e^{E(-a cdot b)}$, where $E$ denotes expectation, and
$f(a,b) = {^{1 | a ne sign(b)}_{0 | a = sign(b)}$.
Additional info is that $a,b$ are random variables such that $a = {-1, +1}$, and $|b| le 1$.
From this, I was able to conclude the following:
Because $|b|$ is always less than 1 and $a = pm 1$, $-1 le -acdot b le 1$, and thus $-1 le E(-acdot b) le 1$.
From definition of $f(a,b)$,
$0 le E(f(a,b))le 1$
Furthermore, I was able to verify the inequality for boundary cases.
If $a$ never match the sign of $b$, then $E(f(a,b)) = 1$.
For $-a cdot b$, if $b > 0$, then $a = -1$, and thus $-a cdot b = -(-1 cdot b) = b > 0$.
If $b < 0$, then $a = 1$, and thus $-a cdot b = -(1 cdot b) = -b > 0$. Thus, for no matches, $0 < E(-acdot b) le 1$, so $1 < e^{E(-acdot b)} le e$, which makes it greater than or equal to 1.
If $a$ always match the sign of $b$, then $E(f(a,b)) = 0$.
For matches the sign of $a$ will always be the same as $b$, so $a cdot b$ will always be positive, and thus $-1 le -a cdot b < 0$, so $e^{-1} le e^{E(-acdot b)} < 1$.
What I am not sure about is how to prove that for any number of matches, this inequality holds. Help would be greatly appreciated!
expected-value
expected-value
asked Nov 24 at 20:27
William Deng
32
asked Nov 24 at 20:27
William Deng
32
asked Nov 24 at 20:27
William Deng
32
asked Nov 24 at 20:27
William Deng
32
asked Nov 24 at 20:27
William Deng
32
32
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