Bound on third moment when the variable is bounded and has finite expectation
If I have a random variable $X$ such that $|X| leq 1$ and $mathbb{E}(X) = c < infty$, I want to find the tightest upper bound for $mathbb{E}(X^3)$. What would be the most optimal bound to use?
probability-theory
asked Nov 30 at 19:23
Hanae Sakuraba
82
add a comment |
If I have a random variable $X$ such that $|X| leq 1$ and $mathbb{E}(X) = c < infty$, I want to find the tightest upper bound for $mathbb{E}(X^3)$. What would be the most optimal bound to use?
probability-theory
asked Nov 30 at 19:23
Hanae Sakuraba
82
Let $P(X=1)=c$ and $P(X=0)=1-c$. Then $E(X)=E(X^3)=c$. I suspect this is optimal. In general you need to know the distribution of $X$.
– herb steinberg
Nov 30 at 21:15
add a comment |
If I have a random variable $X$ such that $|X| leq 1$ and $mathbb{E}(X) = c < infty$, I want to find the tightest upper bound for $mathbb{E}(X^3)$. What would be the most optimal bound to use?
probability-theory
asked Nov 30 at 19:23
Hanae Sakuraba
82
If I have a random variable $X$ such that $|X| leq 1$ and $mathbb{E}(X) = c < infty$, I want to find the tightest upper bound for $mathbb{E}(X^3)$. What would be the most optimal bound to use?
probability-theory
probability-theory
asked Nov 30 at 19:23
Hanae Sakuraba
82
asked Nov 30 at 19:23
Hanae Sakuraba
82
asked Nov 30 at 19:23
Hanae Sakuraba
82
asked Nov 30 at 19:23
Hanae Sakuraba
82
asked Nov 30 at 19:23
Hanae Sakuraba
82
82
Let $P(X=1)=c$ and $P(X=0)=1-c$. Then $E(X)=E(X^3)=c$. I suspect this is optimal. In general you need to know the distribution of $X$.
– herb steinberg
Nov 30 at 21:15
add a comment |
Let $P(X=1)=c$ and $P(X=0)=1-c$. Then $E(X)=E(X^3)=c$. I suspect this is optimal. In general you need to know the distribution of $X$.
– herb steinberg
Nov 30 at 21:15
Let $P(X=1)=c$ and $P(X=0)=1-c$. Then $E(X)=E(X^3)=c$. I suspect this is optimal. In general you need to know the distribution of $X$.
– herb steinberg
Nov 30 at 21:15
Let $P(X=1)=c$ and $P(X=0)=1-c$. Then $E(X)=E(X^3)=c$. I suspect this is optimal. In general you need to know the distribution of $X$.
– herb steinberg
Nov 30 at 21:15
add a comment |
1 Answer
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Let $f(x)=x^3$ and let $Ssubset mathbb R^2$ be the convex hull of the points ${(x,f(x)): xin[-1,1]}$. The set $S$ is the set of all possible $(EX,EX^3)$ values for random variables $X$ with $P(|X|le1)=1.$ The intersection of $S$ with the vertical line cut out by $x=c$ is the interval from $(c,L(c))$ to $(c,U(c))$, where $L$ and $U$ are the lower convex and upper concave envelopes of $f$ on $[-1,1]$. Then the desired upper bound is $U(c)$.
More explicitly: Draw a line through $(1,1)$, tangent at $(-1/2, -1/8)$ to the graph of $f$. The graph of $U$ follows that of $f$ for $xin[-1,-1/2]$ and that of the tangent line for $xin[-1/2,1]$. The formula for $U$ is $$U(x)=begin{cases}-x^3&xle -1/2\ -frac 1 8 +frac 3 4 (x+1/2)& xgt-frac 1 2 end{cases}.$$
If $c=0$, for example, $U(0)=1/4$, which is attained by a random variable $X$ taking the value $X=1$ with probability $1/3$ and the value $X=-1/2$ with probability $2/3$. Then $$EX^3=frac 2 3 (-frac 1 8 )+frac 1
3(1)=frac 1 4 .$$
answered Nov 30 at 22:00
kimchi lover
9,63131128
Why does this property of belonging to the convex hull holds?
– Daniel
Nov 30 at 22:37
@Daniel You mean, that $S$ has all possible $(EX,EX^3)$ pairs in it? Intuitively, because the $E$ operator averages, and the convex hull of a set is the set of all possible averages. A high-brow version of this idea is the basis of en.wikipedia.org/wiki/Choquet_theory .
– kimchi lover
Nov 30 at 22:42
I never saw this idea, but it seems interesting. I can see easily that $X$ being a constant with probability 1 implies $(c, f(c))$ is in the possible pairs, and thus that any point in the convex hull is possible by considering the random variable $X$ that is assumes value $c_1$ with probability $alpha$ and $c_2$ with probability $1 - alpha$. However I do not see how to ensure that a general random variable is also in $S$. edit I see it now, just consider a sequence of discrete random variables converging to the general one... Thanks!
– Daniel
Nov 30 at 22:57
@Daniel The same reasoning should make you believe this for "simple" random variables, taking on finitely many values only. There are 2 parts to my claim: each $(EX,EX^3)$ is a point in $S$, and that each point in $S$ arises this way. Which gives you more trouble?
– kimchi lover
Nov 30 at 23:06
A truly clever idea. I have never thought in such way. Thanks!
– Hanae Sakuraba
Nov 30 at 23:10
|
show 3 more comments
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Let $f(x)=x^3$ and let $Ssubset mathbb R^2$ be the convex hull of the points ${(x,f(x)): xin[-1,1]}$. The set $S$ is the set of all possible $(EX,EX^3)$ values for random variables $X$ with $P(|X|le1)=1.$ The intersection of $S$ with the vertical line cut out by $x=c$ is the interval from $(c,L(c))$ to $(c,U(c))$, where $L$ and $U$ are the lower convex and upper concave envelopes of $f$ on $[-1,1]$. Then the desired upper bound is $U(c)$.
More explicitly: Draw a line through $(1,1)$, tangent at $(-1/2, -1/8)$ to the graph of $f$. The graph of $U$ follows that of $f$ for $xin[-1,-1/2]$ and that of the tangent line for $xin[-1/2,1]$. The formula for $U$ is $$U(x)=begin{cases}-x^3&xle -1/2\ -frac 1 8 +frac 3 4 (x+1/2)& xgt-frac 1 2 end{cases}.$$
If $c=0$, for example, $U(0)=1/4$, which is attained by a random variable $X$ taking the value $X=1$ with probability $1/3$ and the value $X=-1/2$ with probability $2/3$. Then $$EX^3=frac 2 3 (-frac 1 8 )+frac 1
3(1)=frac 1 4 .$$
answered Nov 30 at 22:00
kimchi lover
9,63131128
Why does this property of belonging to the convex hull holds?
– Daniel
Nov 30 at 22:37
@Daniel You mean, that $S$ has all possible $(EX,EX^3)$ pairs in it? Intuitively, because the $E$ operator averages, and the convex hull of a set is the set of all possible averages. A high-brow version of this idea is the basis of en.wikipedia.org/wiki/Choquet_theory .
– kimchi lover
Nov 30 at 22:42
I never saw this idea, but it seems interesting. I can see easily that $X$ being a constant with probability 1 implies $(c, f(c))$ is in the possible pairs, and thus that any point in the convex hull is possible by considering the random variable $X$ that is assumes value $c_1$ with probability $alpha$ and $c_2$ with probability $1 - alpha$. However I do not see how to ensure that a general random variable is also in $S$. edit I see it now, just consider a sequence of discrete random variables converging to the general one... Thanks!
– Daniel
Nov 30 at 22:57
@Daniel The same reasoning should make you believe this for "simple" random variables, taking on finitely many values only. There are 2 parts to my claim: each $(EX,EX^3)$ is a point in $S$, and that each point in $S$ arises this way. Which gives you more trouble?
– kimchi lover
Nov 30 at 23:06
A truly clever idea. I have never thought in such way. Thanks!
– Hanae Sakuraba
Nov 30 at 23:10
|
show 3 more comments
Let $f(x)=x^3$ and let $Ssubset mathbb R^2$ be the convex hull of the points ${(x,f(x)): xin[-1,1]}$. The set $S$ is the set of all possible $(EX,EX^3)$ values for random variables $X$ with $P(|X|le1)=1.$ The intersection of $S$ with the vertical line cut out by $x=c$ is the interval from $(c,L(c))$ to $(c,U(c))$, where $L$ and $U$ are the lower convex and upper concave envelopes of $f$ on $[-1,1]$. Then the desired upper bound is $U(c)$.
More explicitly: Draw a line through $(1,1)$, tangent at $(-1/2, -1/8)$ to the graph of $f$. The graph of $U$ follows that of $f$ for $xin[-1,-1/2]$ and that of the tangent line for $xin[-1/2,1]$. The formula for $U$ is $$U(x)=begin{cases}-x^3&xle -1/2\ -frac 1 8 +frac 3 4 (x+1/2)& xgt-frac 1 2 end{cases}.$$
If $c=0$, for example, $U(0)=1/4$, which is attained by a random variable $X$ taking the value $X=1$ with probability $1/3$ and the value $X=-1/2$ with probability $2/3$. Then $$EX^3=frac 2 3 (-frac 1 8 )+frac 1
3(1)=frac 1 4 .$$
answered Nov 30 at 22:00
kimchi lover
9,63131128
Why does this property of belonging to the convex hull holds?
– Daniel
Nov 30 at 22:37
@Daniel You mean, that $S$ has all possible $(EX,EX^3)$ pairs in it? Intuitively, because the $E$ operator averages, and the convex hull of a set is the set of all possible averages. A high-brow version of this idea is the basis of en.wikipedia.org/wiki/Choquet_theory .
– kimchi lover
Nov 30 at 22:42
I never saw this idea, but it seems interesting. I can see easily that $X$ being a constant with probability 1 implies $(c, f(c))$ is in the possible pairs, and thus that any point in the convex hull is possible by considering the random variable $X$ that is assumes value $c_1$ with probability $alpha$ and $c_2$ with probability $1 - alpha$. However I do not see how to ensure that a general random variable is also in $S$. edit I see it now, just consider a sequence of discrete random variables converging to the general one... Thanks!
– Daniel
Nov 30 at 22:57
@Daniel The same reasoning should make you believe this for "simple" random variables, taking on finitely many values only. There are 2 parts to my claim: each $(EX,EX^3)$ is a point in $S$, and that each point in $S$ arises this way. Which gives you more trouble?
– kimchi lover
Nov 30 at 23:06
A truly clever idea. I have never thought in such way. Thanks!
– Hanae Sakuraba
Nov 30 at 23:10
|
show 3 more comments
Let $f(x)=x^3$ and let $Ssubset mathbb R^2$ be the convex hull of the points ${(x,f(x)): xin[-1,1]}$. The set $S$ is the set of all possible $(EX,EX^3)$ values for random variables $X$ with $P(|X|le1)=1.$ The intersection of $S$ with the vertical line cut out by $x=c$ is the interval from $(c,L(c))$ to $(c,U(c))$, where $L$ and $U$ are the lower convex and upper concave envelopes of $f$ on $[-1,1]$. Then the desired upper bound is $U(c)$.
More explicitly: Draw a line through $(1,1)$, tangent at $(-1/2, -1/8)$ to the graph of $f$. The graph of $U$ follows that of $f$ for $xin[-1,-1/2]$ and that of the tangent line for $xin[-1/2,1]$. The formula for $U$ is $$U(x)=begin{cases}-x^3&xle -1/2\ -frac 1 8 +frac 3 4 (x+1/2)& xgt-frac 1 2 end{cases}.$$
If $c=0$, for example, $U(0)=1/4$, which is attained by a random variable $X$ taking the value $X=1$ with probability $1/3$ and the value $X=-1/2$ with probability $2/3$. Then $$EX^3=frac 2 3 (-frac 1 8 )+frac 1
3(1)=frac 1 4 .$$
answered Nov 30 at 22:00
kimchi lover
9,63131128
Let $f(x)=x^3$ and let $Ssubset mathbb R^2$ be the convex hull of the points ${(x,f(x)): xin[-1,1]}$. The set $S$ is the set of all possible $(EX,EX^3)$ values for random variables $X$ with $P(|X|le1)=1.$ The intersection of $S$ with the vertical line cut out by $x=c$ is the interval from $(c,L(c))$ to $(c,U(c))$, where $L$ and $U$ are the lower convex and upper concave envelopes of $f$ on $[-1,1]$. Then the desired upper bound is $U(c)$.
More explicitly: Draw a line through $(1,1)$, tangent at $(-1/2, -1/8)$ to the graph of $f$. The graph of $U$ follows that of $f$ for $xin[-1,-1/2]$ and that of the tangent line for $xin[-1/2,1]$. The formula for $U$ is $$U(x)=begin{cases}-x^3&xle -1/2\ -frac 1 8 +frac 3 4 (x+1/2)& xgt-frac 1 2 end{cases}.$$
If $c=0$, for example, $U(0)=1/4$, which is attained by a random variable $X$ taking the value $X=1$ with probability $1/3$ and the value $X=-1/2$ with probability $2/3$. Then $$EX^3=frac 2 3 (-frac 1 8 )+frac 1
3(1)=frac 1 4 .$$
answered Nov 30 at 22:00
kimchi lover
9,63131128
edited Nov 30 at 22:33
answered Nov 30 at 22:00
kimchi lover
9,63131128
answered Nov 30 at 22:00
kimchi lover
9,63131128
answered Nov 30 at 22:00
kimchi lover
9,63131128
9,63131128
Why does this property of belonging to the convex hull holds?
– Daniel
Nov 30 at 22:37
@Daniel You mean, that $S$ has all possible $(EX,EX^3)$ pairs in it? Intuitively, because the $E$ operator averages, and the convex hull of a set is the set of all possible averages. A high-brow version of this idea is the basis of en.wikipedia.org/wiki/Choquet_theory .
– kimchi lover
Nov 30 at 22:42
I never saw this idea, but it seems interesting. I can see easily that $X$ being a constant with probability 1 implies $(c, f(c))$ is in the possible pairs, and thus that any point in the convex hull is possible by considering the random variable $X$ that is assumes value $c_1$ with probability $alpha$ and $c_2$ with probability $1 - alpha$. However I do not see how to ensure that a general random variable is also in $S$. edit I see it now, just consider a sequence of discrete random variables converging to the general one... Thanks!
– Daniel
Nov 30 at 22:57
@Daniel The same reasoning should make you believe this for "simple" random variables, taking on finitely many values only. There are 2 parts to my claim: each $(EX,EX^3)$ is a point in $S$, and that each point in $S$ arises this way. Which gives you more trouble?
– kimchi lover
Nov 30 at 23:06
A truly clever idea. I have never thought in such way. Thanks!
– Hanae Sakuraba
Nov 30 at 23:10
|
show 3 more comments
Why does this property of belonging to the convex hull holds?
– Daniel
Nov 30 at 22:37
@Daniel You mean, that $S$ has all possible $(EX,EX^3)$ pairs in it? Intuitively, because the $E$ operator averages, and the convex hull of a set is the set of all possible averages. A high-brow version of this idea is the basis of en.wikipedia.org/wiki/Choquet_theory .
– kimchi lover
Nov 30 at 22:42
I never saw this idea, but it seems interesting. I can see easily that $X$ being a constant with probability 1 implies $(c, f(c))$ is in the possible pairs, and thus that any point in the convex hull is possible by considering the random variable $X$ that is assumes value $c_1$ with probability $alpha$ and $c_2$ with probability $1 - alpha$. However I do not see how to ensure that a general random variable is also in $S$. edit I see it now, just consider a sequence of discrete random variables converging to the general one... Thanks!
– Daniel
Nov 30 at 22:57
@Daniel The same reasoning should make you believe this for "simple" random variables, taking on finitely many values only. There are 2 parts to my claim: each $(EX,EX^3)$ is a point in $S$, and that each point in $S$ arises this way. Which gives you more trouble?
– kimchi lover
Nov 30 at 23:06
A truly clever idea. I have never thought in such way. Thanks!
– Hanae Sakuraba
Nov 30 at 23:10
Why does this property of belonging to the convex hull holds?
– Daniel
Nov 30 at 22:37
Why does this property of belonging to the convex hull holds?
– Daniel
Nov 30 at 22:37
@Daniel You mean, that $S$ has all possible $(EX,EX^3)$ pairs in it? Intuitively, because the $E$ operator averages, and the convex hull of a set is the set of all possible averages. A high-brow version of this idea is the basis of en.wikipedia.org/wiki/Choquet_theory .
– kimchi lover
Nov 30 at 22:42
@Daniel You mean, that $S$ has all possible $(EX,EX^3)$ pairs in it? Intuitively, because the $E$ operator averages, and the convex hull of a set is the set of all possible averages. A high-brow version of this idea is the basis of en.wikipedia.org/wiki/Choquet_theory .
– kimchi lover
Nov 30 at 22:42
I never saw this idea, but it seems interesting. I can see easily that $X$ being a constant with probability 1 implies $(c, f(c))$ is in the possible pairs, and thus that any point in the convex hull is possible by considering the random variable $X$ that is assumes value $c_1$ with probability $alpha$ and $c_2$ with probability $1 - alpha$. However I do not see how to ensure that a general random variable is also in $S$. edit I see it now, just consider a sequence of discrete random variables converging to the general one... Thanks!
– Daniel
Nov 30 at 22:57
I never saw this idea, but it seems interesting. I can see easily that $X$ being a constant with probability 1 implies $(c, f(c))$ is in the possible pairs, and thus that any point in the convex hull is possible by considering the random variable $X$ that is assumes value $c_1$ with probability $alpha$ and $c_2$ with probability $1 - alpha$. However I do not see how to ensure that a general random variable is also in $S$. edit I see it now, just consider a sequence of discrete random variables converging to the general one... Thanks!
– Daniel
Nov 30 at 22:57
@Daniel The same reasoning should make you believe this for "simple" random variables, taking on finitely many values only. There are 2 parts to my claim: each $(EX,EX^3)$ is a point in $S$, and that each point in $S$ arises this way. Which gives you more trouble?
– kimchi lover
Nov 30 at 23:06
@Daniel The same reasoning should make you believe this for "simple" random variables, taking on finitely many values only. There are 2 parts to my claim: each $(EX,EX^3)$ is a point in $S$, and that each point in $S$ arises this way. Which gives you more trouble?
– kimchi lover
Nov 30 at 23:06
A truly clever idea. I have never thought in such way. Thanks!
– Hanae Sakuraba
Nov 30 at 23:10
A truly clever idea. I have never thought in such way. Thanks!
– Hanae Sakuraba
Nov 30 at 23:10
|
show 3 more comments
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Let $P(X=1)=c$ and $P(X=0)=1-c$. Then $E(X)=E(X^3)=c$. I suspect this is optimal. In general you need to know the distribution of $X$.
– herb steinberg
Nov 30 at 21:15