Mixing
Functions that cause Jensen's Inequality to be strict
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I have recently been working on a problem concerning the relation between the expected values of random variables and the expected value of a function of these variables. This relation can be described by Jensen's Inequality which states that for a convex function $f$ and integrable random variables $X_1, ..., X_n$, the following holds:
$$ f(E(X_1),...,E(X_n)) leq E(f(X_1,...,X_n))$$
In my specific case, $X_1,...,X_n$ are independent, normally distributed random variables.
I am interested in functions for which this inequality becomes strict. The function that caused me to consider this problem in the first place is the maximum function: For two (or more) independent, normally distributed random variables $X_1$ and $X_2$, $max(E(X_1), E(X_2)) < E(max(X_1, X_2))$ (this has been explained here for identically distributed variables).
In short, my question is: Given the type of random variables described above, what properties does a function $f$ need to have in order for Jensen's Inequality to be strict.
Edit: After reading and thinking about the first comment, I realized that this question is not the correct one to address my actual problem, but I'm lacking the expertise to correctly formulate the right question (I'm comming from computer science, so my knowledge about probablility theory is rather limited). I hope I can change this and come back with the right question.
probability normal-distribution expected-value
asked Jan 30 at 7:16
FneugeFneuge
12
$endgroup$
add a comment |
$begingroup$
I have recently been working on a problem concerning the relation between the expected values of random variables and the expected value of a function of these variables. This relation can be described by Jensen's Inequality which states that for a convex function $f$ and integrable random variables $X_1, ..., X_n$, the following holds:
$$ f(E(X_1),...,E(X_n)) leq E(f(X_1,...,X_n))$$
In my specific case, $X_1,...,X_n$ are independent, normally distributed random variables.
I am interested in functions for which this inequality becomes strict. The function that caused me to consider this problem in the first place is the maximum function: For two (or more) independent, normally distributed random variables $X_1$ and $X_2$, $max(E(X_1), E(X_2)) < E(max(X_1, X_2))$ (this has been explained here for identically distributed variables).
In short, my question is: Given the type of random variables described above, what properties does a function $f$ need to have in order for Jensen's Inequality to be strict.
Edit: After reading and thinking about the first comment, I realized that this question is not the correct one to address my actual problem, but I'm lacking the expertise to correctly formulate the right question (I'm comming from computer science, so my knowledge about probablility theory is rather limited). I hope I can change this and come back with the right question.
probability normal-distribution expected-value
asked Jan 30 at 7:16
FneugeFneuge
12
$endgroup$
3
$begingroup$
You have to impose conditions both on $f$ and on the random variables. Whatever $f$ may be you get equality if $X_n=0$ for all $n$. In the case of independent normal variables a sufficient condition is strict convexity of $f$.
$endgroup$
– Kavi Rama Murthy
Jan 30 at 7:20
$begingroup$
Related, maybe duplicate: math.stackexchange.com/questions/628386/…
$endgroup$
– Nate Eldredge
Jan 31 at 4:51
add a comment |
$begingroup$
I have recently been working on a problem concerning the relation between the expected values of random variables and the expected value of a function of these variables. This relation can be described by Jensen's Inequality which states that for a convex function $f$ and integrable random variables $X_1, ..., X_n$, the following holds:
$$ f(E(X_1),...,E(X_n)) leq E(f(X_1,...,X_n))$$
In my specific case, $X_1,...,X_n$ are independent, normally distributed random variables.
I am interested in functions for which this inequality becomes strict. The function that caused me to consider this problem in the first place is the maximum function: For two (or more) independent, normally distributed random variables $X_1$ and $X_2$, $max(E(X_1), E(X_2)) < E(max(X_1, X_2))$ (this has been explained here for identically distributed variables).
In short, my question is: Given the type of random variables described above, what properties does a function $f$ need to have in order for Jensen's Inequality to be strict.
Edit: After reading and thinking about the first comment, I realized that this question is not the correct one to address my actual problem, but I'm lacking the expertise to correctly formulate the right question (I'm comming from computer science, so my knowledge about probablility theory is rather limited). I hope I can change this and come back with the right question.
probability normal-distribution expected-value
asked Jan 30 at 7:16
FneugeFneuge
12
$endgroup$
I have recently been working on a problem concerning the relation between the expected values of random variables and the expected value of a function of these variables. This relation can be described by Jensen's Inequality which states that for a convex function $f$ and integrable random variables $X_1, ..., X_n$, the following holds:
$$ f(E(X_1),...,E(X_n)) leq E(f(X_1,...,X_n))$$
In my specific case, $X_1,...,X_n$ are independent, normally distributed random variables.
I am interested in functions for which this inequality becomes strict. The function that caused me to consider this problem in the first place is the maximum function: For two (or more) independent, normally distributed random variables $X_1$ and $X_2$, $max(E(X_1), E(X_2)) < E(max(X_1, X_2))$ (this has been explained here for identically distributed variables).
In short, my question is: Given the type of random variables described above, what properties does a function $f$ need to have in order for Jensen's Inequality to be strict.
Edit: After reading and thinking about the first comment, I realized that this question is not the correct one to address my actual problem, but I'm lacking the expertise to correctly formulate the right question (I'm comming from computer science, so my knowledge about probablility theory is rather limited). I hope I can change this and come back with the right question.
probability normal-distribution expected-value
probability normal-distribution expected-value
asked Jan 30 at 7:16
FneugeFneuge
12
asked Jan 30 at 7:16
FneugeFneuge
12
edited Jan 31 at 2:35
Fneuge
asked Jan 30 at 7:16
FneugeFneuge
12
asked Jan 30 at 7:16
FneugeFneuge
12
asked Jan 30 at 7:16
FneugeFneuge
12
12
3
$begingroup$
You have to impose conditions both on $f$ and on the random variables. Whatever $f$ may be you get equality if $X_n=0$ for all $n$. In the case of independent normal variables a sufficient condition is strict convexity of $f$.
$endgroup$
– Kavi Rama Murthy
Jan 30 at 7:20
$begingroup$
Related, maybe duplicate: math.stackexchange.com/questions/628386/…
$endgroup$
– Nate Eldredge
Jan 31 at 4:51
add a comment |
3
$begingroup$
You have to impose conditions both on $f$ and on the random variables. Whatever $f$ may be you get equality if $X_n=0$ for all $n$. In the case of independent normal variables a sufficient condition is strict convexity of $f$.
$endgroup$
– Kavi Rama Murthy
Jan 30 at 7:20
$begingroup$
Related, maybe duplicate: math.stackexchange.com/questions/628386/…
$endgroup$
– Nate Eldredge
Jan 31 at 4:51
3
3
$begingroup$
You have to impose conditions both on $f$ and on the random variables. Whatever $f$ may be you get equality if $X_n=0$ for all $n$. In the case of independent normal variables a sufficient condition is strict convexity of $f$.
$endgroup$
– Kavi Rama Murthy
Jan 30 at 7:20
$begingroup$
You have to impose conditions both on $f$ and on the random variables. Whatever $f$ may be you get equality if $X_n=0$ for all $n$. In the case of independent normal variables a sufficient condition is strict convexity of $f$.
$endgroup$
– Kavi Rama Murthy
Jan 30 at 7:20
$begingroup$
Related, maybe duplicate: math.stackexchange.com/questions/628386/…
$endgroup$
– Nate Eldredge
Jan 31 at 4:51
$begingroup$
Related, maybe duplicate: math.stackexchange.com/questions/628386/…
$endgroup$
– Nate Eldredge
Jan 31 at 4:51
add a comment |
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$begingroup$
You have to impose conditions both on $f$ and on the random variables. Whatever $f$ may be you get equality if $X_n=0$ for all $n$. In the case of independent normal variables a sufficient condition is strict convexity of $f$.
$endgroup$
– Kavi Rama Murthy
Jan 30 at 7:20
$begingroup$
Related, maybe duplicate: math.stackexchange.com/questions/628386/…
$endgroup$
– Nate Eldredge
Jan 31 at 4:51