Convexity of $f(v) = E[(sum_i X_i v_i)^k]$ subj. to $sum_i v_i^2=1$
$begingroup$
I'm interested in the function
$$f(v) = Eleft[big(sum_i X_i v_ibig)^{2k}right],$$
where $v$ is conditioned to $|v|_2=1$, $kge1$ is integer, and $X_i$ are iid random variables.
For which random distributions is $f$ maximized when all the weight of $v$ is concentrated on a single coordinate? That is, when $v=[1,0dots,0]$.
Clearly when $X_isimmathcal N(0,1)$ we get $Eleft[big(sum_i X_i v_ibig)^{2k}right]=C_k|v|^{2k}_2=C_k$ for some constant $C_k$ depending on $k$. Hence for the normal distribution the `layout' of $v$ doesn't matter. On the other hand, if $X_i$ is an $alpha$-stable distribution, we'd have $Eleft[big(sum_i X_i v_ibig)^{2k}right]=C_k|v|^{2k}_alpha$, which is clearly maximized at $v=[1,0dots,0]$ when $alpha<2$.
It seems to me that $f(v)$ will be convex for any random distribution that has Gaussian moments or larger. I have checked it for the exponential distribution and chi-squared as well, but I don't know how to prove this in general.
Is $f(v)$ convex for any distribution with $Eleft[X^{2k}right]ge (2k-1)!!$ (rhs being the $2k$th moment of a standard normal distribution.)
real-analysis probability statistics probability-distributions convex-analysis
edited Dec 21 '18 at 19:30
Henning Makholm
240k17305541
asked Dec 21 '18 at 18:20
Thomas AhleThomas Ahle
1,5121320
$endgroup$
|
show 2 more comments
$begingroup$
I'm interested in the function
$$f(v) = Eleft[big(sum_i X_i v_ibig)^{2k}right],$$
where $v$ is conditioned to $|v|_2=1$, $kge1$ is integer, and $X_i$ are iid random variables.
For which random distributions is $f$ maximized when all the weight of $v$ is concentrated on a single coordinate? That is, when $v=[1,0dots,0]$.
Clearly when $X_isimmathcal N(0,1)$ we get $Eleft[big(sum_i X_i v_ibig)^{2k}right]=C_k|v|^{2k}_2=C_k$ for some constant $C_k$ depending on $k$. Hence for the normal distribution the `layout' of $v$ doesn't matter. On the other hand, if $X_i$ is an $alpha$-stable distribution, we'd have $Eleft[big(sum_i X_i v_ibig)^{2k}right]=C_k|v|^{2k}_alpha$, which is clearly maximized at $v=[1,0dots,0]$ when $alpha<2$.
It seems to me that $f(v)$ will be convex for any random distribution that has Gaussian moments or larger. I have checked it for the exponential distribution and chi-squared as well, but I don't know how to prove this in general.
Is $f(v)$ convex for any distribution with $Eleft[X^{2k}right]ge (2k-1)!!$ (rhs being the $2k$th moment of a standard normal distribution.)
real-analysis probability statistics probability-distributions convex-analysis
edited Dec 21 '18 at 19:30
Henning Makholm
240k17305541
asked Dec 21 '18 at 18:20
Thomas AhleThomas Ahle
1,5121320
$endgroup$
2
$begingroup$
$f(v_1,...,v_n)$ is convex (for any random variables for which the expectations are finite) over $mathbb{R}^n$, but not over the domain ${v:||v||=1}$. For example take the convex function $g(x_1,x_2)=(x_1+x_2)^{2k}$. Then $g(1,0)=1 < g(1/sqrt{2},1/sqrt{2})=sqrt{2}^{2k}$.
$endgroup$
– Michael
Dec 21 '18 at 18:37
1
$begingroup$
In other words, convexity seems to have nothing to do with the problem. It would also help to clarify how many terms your sum is over (i.e., the dimension of $v$ is what?), and if the square is inside or outside the expectation (I assume inside). That is, does $E(sum X_i)^2$ mean $E[(sum X_i)^2]$ or $E[sum X_i]^2$?
$endgroup$
– Michael
Dec 21 '18 at 18:48
$begingroup$
@Michael Thank you, I've added more brackets to my expectations. Regarding convexity, I do indeed mean convex over the domain. Is there a better way to state this?
$endgroup$
– Thomas Ahle
Dec 21 '18 at 19:01
1
$begingroup$
A convex function must have a convex domain. There is no function that is convex over the domain ${v: ||v||=1}$.
$endgroup$
– Michael
Dec 21 '18 at 19:02
1
$begingroup$
You likely want the $X_i$ to have mean zero since if $E[X_i]=m$ we get $f(1,0,0,..,0)=E[X_1^{2k}]$ and $f(1/sqrt{n}, ..., 1/sqrt{n})geq (msqrt{n})^{2k}$, which goes to infinity with $n$ whenever $mneq 0$. It would also help to know whether or not $n=k$.
$endgroup$
– Michael
Dec 21 '18 at 19:06
|
show 2 more comments
$begingroup$
I'm interested in the function
$$f(v) = Eleft[big(sum_i X_i v_ibig)^{2k}right],$$
where $v$ is conditioned to $|v|_2=1$, $kge1$ is integer, and $X_i$ are iid random variables.
For which random distributions is $f$ maximized when all the weight of $v$ is concentrated on a single coordinate? That is, when $v=[1,0dots,0]$.
Clearly when $X_isimmathcal N(0,1)$ we get $Eleft[big(sum_i X_i v_ibig)^{2k}right]=C_k|v|^{2k}_2=C_k$ for some constant $C_k$ depending on $k$. Hence for the normal distribution the `layout' of $v$ doesn't matter. On the other hand, if $X_i$ is an $alpha$-stable distribution, we'd have $Eleft[big(sum_i X_i v_ibig)^{2k}right]=C_k|v|^{2k}_alpha$, which is clearly maximized at $v=[1,0dots,0]$ when $alpha<2$.
It seems to me that $f(v)$ will be convex for any random distribution that has Gaussian moments or larger. I have checked it for the exponential distribution and chi-squared as well, but I don't know how to prove this in general.
Is $f(v)$ convex for any distribution with $Eleft[X^{2k}right]ge (2k-1)!!$ (rhs being the $2k$th moment of a standard normal distribution.)
real-analysis probability statistics probability-distributions convex-analysis
edited Dec 21 '18 at 19:30
Henning Makholm
240k17305541
asked Dec 21 '18 at 18:20
Thomas AhleThomas Ahle
1,5121320
$endgroup$
I'm interested in the function
$$f(v) = Eleft[big(sum_i X_i v_ibig)^{2k}right],$$
where $v$ is conditioned to $|v|_2=1$, $kge1$ is integer, and $X_i$ are iid random variables.
For which random distributions is $f$ maximized when all the weight of $v$ is concentrated on a single coordinate? That is, when $v=[1,0dots,0]$.
Clearly when $X_isimmathcal N(0,1)$ we get $Eleft[big(sum_i X_i v_ibig)^{2k}right]=C_k|v|^{2k}_2=C_k$ for some constant $C_k$ depending on $k$. Hence for the normal distribution the `layout' of $v$ doesn't matter. On the other hand, if $X_i$ is an $alpha$-stable distribution, we'd have $Eleft[big(sum_i X_i v_ibig)^{2k}right]=C_k|v|^{2k}_alpha$, which is clearly maximized at $v=[1,0dots,0]$ when $alpha<2$.
It seems to me that $f(v)$ will be convex for any random distribution that has Gaussian moments or larger. I have checked it for the exponential distribution and chi-squared as well, but I don't know how to prove this in general.
Is $f(v)$ convex for any distribution with $Eleft[X^{2k}right]ge (2k-1)!!$ (rhs being the $2k$th moment of a standard normal distribution.)
real-analysis probability statistics probability-distributions convex-analysis
real-analysis probability statistics probability-distributions convex-analysis
edited Dec 21 '18 at 19:30
Henning Makholm
240k17305541
asked Dec 21 '18 at 18:20
Thomas AhleThomas Ahle
1,5121320
edited Dec 21 '18 at 19:30
Henning Makholm
240k17305541
asked Dec 21 '18 at 18:20
Thomas AhleThomas Ahle
1,5121320
edited Dec 21 '18 at 19:30
Henning Makholm
240k17305541
edited Dec 21 '18 at 19:30
Henning Makholm
240k17305541
edited Dec 21 '18 at 19:30
Henning Makholm
240k17305541
240k17305541
asked Dec 21 '18 at 18:20
Thomas AhleThomas Ahle
1,5121320
asked Dec 21 '18 at 18:20
Thomas AhleThomas Ahle
1,5121320
asked Dec 21 '18 at 18:20
Thomas AhleThomas Ahle
1,5121320
1,5121320
2
$begingroup$
$f(v_1,...,v_n)$ is convex (for any random variables for which the expectations are finite) over $mathbb{R}^n$, but not over the domain ${v:||v||=1}$. For example take the convex function $g(x_1,x_2)=(x_1+x_2)^{2k}$. Then $g(1,0)=1 < g(1/sqrt{2},1/sqrt{2})=sqrt{2}^{2k}$.
$endgroup$
– Michael
Dec 21 '18 at 18:37
1
$begingroup$
In other words, convexity seems to have nothing to do with the problem. It would also help to clarify how many terms your sum is over (i.e., the dimension of $v$ is what?), and if the square is inside or outside the expectation (I assume inside). That is, does $E(sum X_i)^2$ mean $E[(sum X_i)^2]$ or $E[sum X_i]^2$?
$endgroup$
– Michael
Dec 21 '18 at 18:48
$begingroup$
@Michael Thank you, I've added more brackets to my expectations. Regarding convexity, I do indeed mean convex over the domain. Is there a better way to state this?
$endgroup$
– Thomas Ahle
Dec 21 '18 at 19:01
1
$begingroup$
A convex function must have a convex domain. There is no function that is convex over the domain ${v: ||v||=1}$.
$endgroup$
– Michael
Dec 21 '18 at 19:02
1
$begingroup$
You likely want the $X_i$ to have mean zero since if $E[X_i]=m$ we get $f(1,0,0,..,0)=E[X_1^{2k}]$ and $f(1/sqrt{n}, ..., 1/sqrt{n})geq (msqrt{n})^{2k}$, which goes to infinity with $n$ whenever $mneq 0$. It would also help to know whether or not $n=k$.
$endgroup$
– Michael
Dec 21 '18 at 19:06
|
show 2 more comments
2
$begingroup$
$f(v_1,...,v_n)$ is convex (for any random variables for which the expectations are finite) over $mathbb{R}^n$, but not over the domain ${v:||v||=1}$. For example take the convex function $g(x_1,x_2)=(x_1+x_2)^{2k}$. Then $g(1,0)=1 < g(1/sqrt{2},1/sqrt{2})=sqrt{2}^{2k}$.
$endgroup$
– Michael
Dec 21 '18 at 18:37
1
$begingroup$
In other words, convexity seems to have nothing to do with the problem. It would also help to clarify how many terms your sum is over (i.e., the dimension of $v$ is what?), and if the square is inside or outside the expectation (I assume inside). That is, does $E(sum X_i)^2$ mean $E[(sum X_i)^2]$ or $E[sum X_i]^2$?
$endgroup$
– Michael
Dec 21 '18 at 18:48
$begingroup$
@Michael Thank you, I've added more brackets to my expectations. Regarding convexity, I do indeed mean convex over the domain. Is there a better way to state this?
$endgroup$
– Thomas Ahle
Dec 21 '18 at 19:01
1
$begingroup$
A convex function must have a convex domain. There is no function that is convex over the domain ${v: ||v||=1}$.
$endgroup$
– Michael
Dec 21 '18 at 19:02
1
$begingroup$
You likely want the $X_i$ to have mean zero since if $E[X_i]=m$ we get $f(1,0,0,..,0)=E[X_1^{2k}]$ and $f(1/sqrt{n}, ..., 1/sqrt{n})geq (msqrt{n})^{2k}$, which goes to infinity with $n$ whenever $mneq 0$. It would also help to know whether or not $n=k$.
$endgroup$
– Michael
Dec 21 '18 at 19:06
2
2
$begingroup$
$f(v_1,...,v_n)$ is convex (for any random variables for which the expectations are finite) over $mathbb{R}^n$, but not over the domain ${v:||v||=1}$. For example take the convex function $g(x_1,x_2)=(x_1+x_2)^{2k}$. Then $g(1,0)=1 < g(1/sqrt{2},1/sqrt{2})=sqrt{2}^{2k}$.
$endgroup$
– Michael
Dec 21 '18 at 18:37
$begingroup$
$f(v_1,...,v_n)$ is convex (for any random variables for which the expectations are finite) over $mathbb{R}^n$, but not over the domain ${v:||v||=1}$. For example take the convex function $g(x_1,x_2)=(x_1+x_2)^{2k}$. Then $g(1,0)=1 < g(1/sqrt{2},1/sqrt{2})=sqrt{2}^{2k}$.
$endgroup$
– Michael
Dec 21 '18 at 18:37
1
1
$begingroup$
In other words, convexity seems to have nothing to do with the problem. It would also help to clarify how many terms your sum is over (i.e., the dimension of $v$ is what?), and if the square is inside or outside the expectation (I assume inside). That is, does $E(sum X_i)^2$ mean $E[(sum X_i)^2]$ or $E[sum X_i]^2$?
$endgroup$
– Michael
Dec 21 '18 at 18:48
$begingroup$
In other words, convexity seems to have nothing to do with the problem. It would also help to clarify how many terms your sum is over (i.e., the dimension of $v$ is what?), and if the square is inside or outside the expectation (I assume inside). That is, does $E(sum X_i)^2$ mean $E[(sum X_i)^2]$ or $E[sum X_i]^2$?
$endgroup$
– Michael
Dec 21 '18 at 18:48
$begingroup$
@Michael Thank you, I've added more brackets to my expectations. Regarding convexity, I do indeed mean convex over the domain. Is there a better way to state this?
$endgroup$
– Thomas Ahle
Dec 21 '18 at 19:01
$begingroup$
@Michael Thank you, I've added more brackets to my expectations. Regarding convexity, I do indeed mean convex over the domain. Is there a better way to state this?
$endgroup$
– Thomas Ahle
Dec 21 '18 at 19:01
1
1
$begingroup$
A convex function must have a convex domain. There is no function that is convex over the domain ${v: ||v||=1}$.
$endgroup$
– Michael
Dec 21 '18 at 19:02
$begingroup$
A convex function must have a convex domain. There is no function that is convex over the domain ${v: ||v||=1}$.
$endgroup$
– Michael
Dec 21 '18 at 19:02
1
1
$begingroup$
You likely want the $X_i$ to have mean zero since if $E[X_i]=m$ we get $f(1,0,0,..,0)=E[X_1^{2k}]$ and $f(1/sqrt{n}, ..., 1/sqrt{n})geq (msqrt{n})^{2k}$, which goes to infinity with $n$ whenever $mneq 0$. It would also help to know whether or not $n=k$.
$endgroup$
– Michael
Dec 21 '18 at 19:06
$begingroup$
You likely want the $X_i$ to have mean zero since if $E[X_i]=m$ we get $f(1,0,0,..,0)=E[X_1^{2k}]$ and $f(1/sqrt{n}, ..., 1/sqrt{n})geq (msqrt{n})^{2k}$, which goes to infinity with $n$ whenever $mneq 0$. It would also help to know whether or not $n=k$.
$endgroup$
– Michael
Dec 21 '18 at 19:06
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
In fact, we can show that $$
g(v) = Eleft[(sum_{i} v_i X_i)^{2k}right]^{frac{1}{2k}}$$ is subadditive, homogeneous of degree $1$ over $vinmathbb{R}^n$ (this implies convexity.) Subadditivity comes from Minkowski's inequality which is implying that
$$begin{eqnarray}
g(v+w) &=& Eleft[(sum_{i} v_i X_i +sum_{i} w_i X_i)^{2k}right]^{frac{1}{2k}}\&leq& Eleft[(sum_{i} v_i X_i)^{2k}right]^{frac{1}{2k}} + Eleft[(sum_{i} w_i X_i)^{2k}right]^{frac{1}{2k}}\&=&g(v) + g(w).
end{eqnarray}$$ Homogeneity can be shown easily:
$$
g(tv) = Eleft[(sum_{i} tv_i X_i)^{2k}right]^{frac{1}{2k}}=t Eleft[(sum_{i} v_i X_i)^{2k}right]^{frac{1}{2k}} = tg(v),quadforall tgeq 0.
$$ This proves $$
g(tv + (1-t)w) leq g(tv) + g((1-t)w) =tg(v) + (1-t)g(w) ,$$ for all $tin [0,1]$ and $v,winmathbb{R}^n$. Since $g$ is a non-negative convex function, and the map $tin [0,infty)mapsto t^{2k}$ is non-decreasing convex, we have $f(v) = g(v)^{2k}$ is also convex.
answered Dec 21 '18 at 19:28
SongSong
11.3k628
$endgroup$
$begingroup$
Convexity came without any requirement on $x$! I guess @Michael was right that convexity is not really the property I need, since my variables live on the sphere. I really appreciate the answer, but do you have any ideas how to show that the sum is maximized when $v=[1, 0,dots]$ as in the question?
$endgroup$
– Thomas Ahle
Dec 22 '18 at 12:22
add a comment |
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1 Answer
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$begingroup$
In fact, we can show that $$
g(v) = Eleft[(sum_{i} v_i X_i)^{2k}right]^{frac{1}{2k}}$$ is subadditive, homogeneous of degree $1$ over $vinmathbb{R}^n$ (this implies convexity.) Subadditivity comes from Minkowski's inequality which is implying that
$$begin{eqnarray}
g(v+w) &=& Eleft[(sum_{i} v_i X_i +sum_{i} w_i X_i)^{2k}right]^{frac{1}{2k}}\&leq& Eleft[(sum_{i} v_i X_i)^{2k}right]^{frac{1}{2k}} + Eleft[(sum_{i} w_i X_i)^{2k}right]^{frac{1}{2k}}\&=&g(v) + g(w).
end{eqnarray}$$ Homogeneity can be shown easily:
$$
g(tv) = Eleft[(sum_{i} tv_i X_i)^{2k}right]^{frac{1}{2k}}=t Eleft[(sum_{i} v_i X_i)^{2k}right]^{frac{1}{2k}} = tg(v),quadforall tgeq 0.
$$ This proves $$
g(tv + (1-t)w) leq g(tv) + g((1-t)w) =tg(v) + (1-t)g(w) ,$$ for all $tin [0,1]$ and $v,winmathbb{R}^n$. Since $g$ is a non-negative convex function, and the map $tin [0,infty)mapsto t^{2k}$ is non-decreasing convex, we have $f(v) = g(v)^{2k}$ is also convex.
answered Dec 21 '18 at 19:28
SongSong
11.3k628
$endgroup$
$begingroup$
Convexity came without any requirement on $x$! I guess @Michael was right that convexity is not really the property I need, since my variables live on the sphere. I really appreciate the answer, but do you have any ideas how to show that the sum is maximized when $v=[1, 0,dots]$ as in the question?
$endgroup$
– Thomas Ahle
Dec 22 '18 at 12:22
add a comment |
$begingroup$
In fact, we can show that $$
g(v) = Eleft[(sum_{i} v_i X_i)^{2k}right]^{frac{1}{2k}}$$ is subadditive, homogeneous of degree $1$ over $vinmathbb{R}^n$ (this implies convexity.) Subadditivity comes from Minkowski's inequality which is implying that
$$begin{eqnarray}
g(v+w) &=& Eleft[(sum_{i} v_i X_i +sum_{i} w_i X_i)^{2k}right]^{frac{1}{2k}}\&leq& Eleft[(sum_{i} v_i X_i)^{2k}right]^{frac{1}{2k}} + Eleft[(sum_{i} w_i X_i)^{2k}right]^{frac{1}{2k}}\&=&g(v) + g(w).
end{eqnarray}$$ Homogeneity can be shown easily:
$$
g(tv) = Eleft[(sum_{i} tv_i X_i)^{2k}right]^{frac{1}{2k}}=t Eleft[(sum_{i} v_i X_i)^{2k}right]^{frac{1}{2k}} = tg(v),quadforall tgeq 0.
$$ This proves $$
g(tv + (1-t)w) leq g(tv) + g((1-t)w) =tg(v) + (1-t)g(w) ,$$ for all $tin [0,1]$ and $v,winmathbb{R}^n$. Since $g$ is a non-negative convex function, and the map $tin [0,infty)mapsto t^{2k}$ is non-decreasing convex, we have $f(v) = g(v)^{2k}$ is also convex.
answered Dec 21 '18 at 19:28
SongSong
11.3k628
$endgroup$
$begingroup$
Convexity came without any requirement on $x$! I guess @Michael was right that convexity is not really the property I need, since my variables live on the sphere. I really appreciate the answer, but do you have any ideas how to show that the sum is maximized when $v=[1, 0,dots]$ as in the question?
$endgroup$
– Thomas Ahle
Dec 22 '18 at 12:22
add a comment |
$begingroup$
In fact, we can show that $$
g(v) = Eleft[(sum_{i} v_i X_i)^{2k}right]^{frac{1}{2k}}$$ is subadditive, homogeneous of degree $1$ over $vinmathbb{R}^n$ (this implies convexity.) Subadditivity comes from Minkowski's inequality which is implying that
$$begin{eqnarray}
g(v+w) &=& Eleft[(sum_{i} v_i X_i +sum_{i} w_i X_i)^{2k}right]^{frac{1}{2k}}\&leq& Eleft[(sum_{i} v_i X_i)^{2k}right]^{frac{1}{2k}} + Eleft[(sum_{i} w_i X_i)^{2k}right]^{frac{1}{2k}}\&=&g(v) + g(w).
end{eqnarray}$$ Homogeneity can be shown easily:
$$
g(tv) = Eleft[(sum_{i} tv_i X_i)^{2k}right]^{frac{1}{2k}}=t Eleft[(sum_{i} v_i X_i)^{2k}right]^{frac{1}{2k}} = tg(v),quadforall tgeq 0.
$$ This proves $$
g(tv + (1-t)w) leq g(tv) + g((1-t)w) =tg(v) + (1-t)g(w) ,$$ for all $tin [0,1]$ and $v,winmathbb{R}^n$. Since $g$ is a non-negative convex function, and the map $tin [0,infty)mapsto t^{2k}$ is non-decreasing convex, we have $f(v) = g(v)^{2k}$ is also convex.
answered Dec 21 '18 at 19:28
SongSong
11.3k628
$endgroup$
In fact, we can show that $$
g(v) = Eleft[(sum_{i} v_i X_i)^{2k}right]^{frac{1}{2k}}$$ is subadditive, homogeneous of degree $1$ over $vinmathbb{R}^n$ (this implies convexity.) Subadditivity comes from Minkowski's inequality which is implying that
$$begin{eqnarray}
g(v+w) &=& Eleft[(sum_{i} v_i X_i +sum_{i} w_i X_i)^{2k}right]^{frac{1}{2k}}\&leq& Eleft[(sum_{i} v_i X_i)^{2k}right]^{frac{1}{2k}} + Eleft[(sum_{i} w_i X_i)^{2k}right]^{frac{1}{2k}}\&=&g(v) + g(w).
end{eqnarray}$$ Homogeneity can be shown easily:
$$
g(tv) = Eleft[(sum_{i} tv_i X_i)^{2k}right]^{frac{1}{2k}}=t Eleft[(sum_{i} v_i X_i)^{2k}right]^{frac{1}{2k}} = tg(v),quadforall tgeq 0.
$$ This proves $$
g(tv + (1-t)w) leq g(tv) + g((1-t)w) =tg(v) + (1-t)g(w) ,$$ for all $tin [0,1]$ and $v,winmathbb{R}^n$. Since $g$ is a non-negative convex function, and the map $tin [0,infty)mapsto t^{2k}$ is non-decreasing convex, we have $f(v) = g(v)^{2k}$ is also convex.
answered Dec 21 '18 at 19:28
SongSong
11.3k628
answered Dec 21 '18 at 19:28
SongSong
11.3k628
answered Dec 21 '18 at 19:28
SongSong
11.3k628
answered Dec 21 '18 at 19:28
SongSong
11.3k628
11.3k628
$begingroup$
Convexity came without any requirement on $x$! I guess @Michael was right that convexity is not really the property I need, since my variables live on the sphere. I really appreciate the answer, but do you have any ideas how to show that the sum is maximized when $v=[1, 0,dots]$ as in the question?
$endgroup$
– Thomas Ahle
Dec 22 '18 at 12:22
add a comment |
$begingroup$
Convexity came without any requirement on $x$! I guess @Michael was right that convexity is not really the property I need, since my variables live on the sphere. I really appreciate the answer, but do you have any ideas how to show that the sum is maximized when $v=[1, 0,dots]$ as in the question?
$endgroup$
– Thomas Ahle
Dec 22 '18 at 12:22
$begingroup$
Convexity came without any requirement on $x$! I guess @Michael was right that convexity is not really the property I need, since my variables live on the sphere. I really appreciate the answer, but do you have any ideas how to show that the sum is maximized when $v=[1, 0,dots]$ as in the question?
$endgroup$
– Thomas Ahle
Dec 22 '18 at 12:22
$begingroup$
Convexity came without any requirement on $x$! I guess @Michael was right that convexity is not really the property I need, since my variables live on the sphere. I really appreciate the answer, but do you have any ideas how to show that the sum is maximized when $v=[1, 0,dots]$ as in the question?
$endgroup$
– Thomas Ahle
Dec 22 '18 at 12:22
add a comment |
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2
$begingroup$
$f(v_1,...,v_n)$ is convex (for any random variables for which the expectations are finite) over $mathbb{R}^n$, but not over the domain ${v:||v||=1}$. For example take the convex function $g(x_1,x_2)=(x_1+x_2)^{2k}$. Then $g(1,0)=1 < g(1/sqrt{2},1/sqrt{2})=sqrt{2}^{2k}$.
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– Michael
Dec 21 '18 at 18:37
1
$begingroup$
In other words, convexity seems to have nothing to do with the problem. It would also help to clarify how many terms your sum is over (i.e., the dimension of $v$ is what?), and if the square is inside or outside the expectation (I assume inside). That is, does $E(sum X_i)^2$ mean $E[(sum X_i)^2]$ or $E[sum X_i]^2$?
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– Michael
Dec 21 '18 at 18:48
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@Michael Thank you, I've added more brackets to my expectations. Regarding convexity, I do indeed mean convex over the domain. Is there a better way to state this?
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– Thomas Ahle
Dec 21 '18 at 19:01
1
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A convex function must have a convex domain. There is no function that is convex over the domain ${v: ||v||=1}$.
$endgroup$
– Michael
Dec 21 '18 at 19:02
1
$begingroup$
You likely want the $X_i$ to have mean zero since if $E[X_i]=m$ we get $f(1,0,0,..,0)=E[X_1^{2k}]$ and $f(1/sqrt{n}, ..., 1/sqrt{n})geq (msqrt{n})^{2k}$, which goes to infinity with $n$ whenever $mneq 0$. It would also help to know whether or not $n=k$.
$endgroup$
– Michael
Dec 21 '18 at 19:06