A question about convex analysis of expectation value
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In general case, the expectation value is defined like this
$$E[X]=int_Omega X(w)dP(w)$$
and in absolutely continuous or countably case the expectation value is defined:
$$E[X]=int_Bbb R xf(x)dx$$ or $$E[X]=sum xf(x)$$
if $w_1,...w_ngeq0$ and $f_1,...f_n$ are convex, then function $w_1f_1+...w_nf_n$ is also convex
and this can extends to the integrals.
So,My problem is what if today the density function $f(x)$ is doesn't exist
or in general case $E[X]=int_Omega X(w)dP(w)$
Expectation value $$E[X]=int_Omega X(w)dP(w)$$
and Expectation value of convex function $$E[g(X)]=int_Omega g(X(w))dP(w)$$
are convex?
probability-theory convex-analysis
asked Nov 23 at 2:24
Vergil Chan
334
add a comment |
up vote
-1
down vote
favorite
In general case, the expectation value is defined like this
$$E[X]=int_Omega X(w)dP(w)$$
and in absolutely continuous or countably case the expectation value is defined:
$$E[X]=int_Bbb R xf(x)dx$$ or $$E[X]=sum xf(x)$$
if $w_1,...w_ngeq0$ and $f_1,...f_n$ are convex, then function $w_1f_1+...w_nf_n$ is also convex
and this can extends to the integrals.
So,My problem is what if today the density function $f(x)$ is doesn't exist
or in general case $E[X]=int_Omega X(w)dP(w)$
Expectation value $$E[X]=int_Omega X(w)dP(w)$$
and Expectation value of convex function $$E[g(X)]=int_Omega g(X(w))dP(w)$$
are convex?
probability-theory convex-analysis
asked Nov 23 at 2:24
Vergil Chan
334
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
In general case, the expectation value is defined like this
$$E[X]=int_Omega X(w)dP(w)$$
and in absolutely continuous or countably case the expectation value is defined:
$$E[X]=int_Bbb R xf(x)dx$$ or $$E[X]=sum xf(x)$$
if $w_1,...w_ngeq0$ and $f_1,...f_n$ are convex, then function $w_1f_1+...w_nf_n$ is also convex
and this can extends to the integrals.
So,My problem is what if today the density function $f(x)$ is doesn't exist
or in general case $E[X]=int_Omega X(w)dP(w)$
Expectation value $$E[X]=int_Omega X(w)dP(w)$$
and Expectation value of convex function $$E[g(X)]=int_Omega g(X(w))dP(w)$$
are convex?
probability-theory convex-analysis
asked Nov 23 at 2:24
Vergil Chan
334
In general case, the expectation value is defined like this
$$E[X]=int_Omega X(w)dP(w)$$
and in absolutely continuous or countably case the expectation value is defined:
$$E[X]=int_Bbb R xf(x)dx$$ or $$E[X]=sum xf(x)$$
if $w_1,...w_ngeq0$ and $f_1,...f_n$ are convex, then function $w_1f_1+...w_nf_n$ is also convex
and this can extends to the integrals.
So,My problem is what if today the density function $f(x)$ is doesn't exist
or in general case $E[X]=int_Omega X(w)dP(w)$
Expectation value $$E[X]=int_Omega X(w)dP(w)$$
and Expectation value of convex function $$E[g(X)]=int_Omega g(X(w))dP(w)$$
are convex?
probability-theory convex-analysis
probability-theory convex-analysis
asked Nov 23 at 2:24
Vergil Chan
334
asked Nov 23 at 2:24
Vergil Chan
334
asked Nov 23 at 2:24
Vergil Chan
334
asked Nov 23 at 2:24
Vergil Chan
334
asked Nov 23 at 2:24
Vergil Chan
334
334
add a comment |
add a comment |
1 Answer
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3
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I think you're confusing many things at once.
If $w_1, ldots, w_n ge 0$ and $f_1, ldots, f_n$ are convex functions then yes $sum_i w_i f_i$ is a convex function.
I think you are looking at the sum $E[X] = sum_x x f(x)$ and drawing some sort of connection when there is none. There is only one function $f$ (the PMF of the distribution) and the sum is over values of $x$. The expectation $E[X]$ is a number, not a function.
It does not make sense to ask whether $E[X]$ or $E[g(X)]$ is "convex," since they are not functions.
answered Nov 23 at 2:32
angryavian
38k23180
add a comment |
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1 Answer
1
active
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
I think you're confusing many things at once.
If $w_1, ldots, w_n ge 0$ and $f_1, ldots, f_n$ are convex functions then yes $sum_i w_i f_i$ is a convex function.
I think you are looking at the sum $E[X] = sum_x x f(x)$ and drawing some sort of connection when there is none. There is only one function $f$ (the PMF of the distribution) and the sum is over values of $x$. The expectation $E[X]$ is a number, not a function.
It does not make sense to ask whether $E[X]$ or $E[g(X)]$ is "convex," since they are not functions.
answered Nov 23 at 2:32
angryavian
38k23180
add a comment |
up vote
3
down vote
accepted
I think you're confusing many things at once.
If $w_1, ldots, w_n ge 0$ and $f_1, ldots, f_n$ are convex functions then yes $sum_i w_i f_i$ is a convex function.
I think you are looking at the sum $E[X] = sum_x x f(x)$ and drawing some sort of connection when there is none. There is only one function $f$ (the PMF of the distribution) and the sum is over values of $x$. The expectation $E[X]$ is a number, not a function.
It does not make sense to ask whether $E[X]$ or $E[g(X)]$ is "convex," since they are not functions.
answered Nov 23 at 2:32
angryavian
38k23180
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
I think you're confusing many things at once.
If $w_1, ldots, w_n ge 0$ and $f_1, ldots, f_n$ are convex functions then yes $sum_i w_i f_i$ is a convex function.
I think you are looking at the sum $E[X] = sum_x x f(x)$ and drawing some sort of connection when there is none. There is only one function $f$ (the PMF of the distribution) and the sum is over values of $x$. The expectation $E[X]$ is a number, not a function.
It does not make sense to ask whether $E[X]$ or $E[g(X)]$ is "convex," since they are not functions.
answered Nov 23 at 2:32
angryavian
38k23180
I think you're confusing many things at once.
If $w_1, ldots, w_n ge 0$ and $f_1, ldots, f_n$ are convex functions then yes $sum_i w_i f_i$ is a convex function.
I think you are looking at the sum $E[X] = sum_x x f(x)$ and drawing some sort of connection when there is none. There is only one function $f$ (the PMF of the distribution) and the sum is over values of $x$. The expectation $E[X]$ is a number, not a function.
It does not make sense to ask whether $E[X]$ or $E[g(X)]$ is "convex," since they are not functions.
answered Nov 23 at 2:32
angryavian
38k23180
answered Nov 23 at 2:32
angryavian
38k23180
answered Nov 23 at 2:32
angryavian
38k23180
answered Nov 23 at 2:32
angryavian
38k23180
38k23180
add a comment |
add a comment |
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