Reweighting average of increasing function
$begingroup$
Consider the entity $A=frac{int_a^b w(x)xf(x)dmu(x)}{int_a^b w(x)xdmu(x)}$, where $f$ is a smooth, monotonously increasing function with positive values: $f,f^prime>0$, $mu$ is the measure used in the integration, and $w$ is a smooth function with positive values: $w>0$. The derivative of $w$ is assumed to have the same sign everywhere: $w^prime=0$ everywhere, $w^prime>0$ everywhere or $w^prime<0$ everywhere.
Now consider the entity $B=frac{int_a^b frac{x}{w(x)}f(x)dmu(x)}{int_a^b frac{x}{w(x)}dmu(x)}$.
I have a strong intuition that the sign of $w^prime$ determines which is greater, $A$ or $B$. If $w^prime=0$ everywhere, then clearly $A=B$.
Now my intuition tells me the following:
$Alessgtr BLeftrightarrow w^primelessgtr 0$.
The intuition is clear: given that $f$ increases, then if $w$ increases, $A$ gives a higher weight to larger values of $f$ than $B$ does. If $w$ decreases, $A$ gives a lower weight to larger values of $f$ than $B$ does.
Only, I haved tried for a long time, I am unable to prove this intuition more formally. Is my intuition correct, and why? Or is my intuition wrong, and is there an easy counterexample?
real-analysis
asked Jan 8 at 13:14
ScindapsusScindapsus
335
$endgroup$
add a comment |
$begingroup$
Consider the entity $A=frac{int_a^b w(x)xf(x)dmu(x)}{int_a^b w(x)xdmu(x)}$, where $f$ is a smooth, monotonously increasing function with positive values: $f,f^prime>0$, $mu$ is the measure used in the integration, and $w$ is a smooth function with positive values: $w>0$. The derivative of $w$ is assumed to have the same sign everywhere: $w^prime=0$ everywhere, $w^prime>0$ everywhere or $w^prime<0$ everywhere.
Now consider the entity $B=frac{int_a^b frac{x}{w(x)}f(x)dmu(x)}{int_a^b frac{x}{w(x)}dmu(x)}$.
I have a strong intuition that the sign of $w^prime$ determines which is greater, $A$ or $B$. If $w^prime=0$ everywhere, then clearly $A=B$.
Now my intuition tells me the following:
$Alessgtr BLeftrightarrow w^primelessgtr 0$.
The intuition is clear: given that $f$ increases, then if $w$ increases, $A$ gives a higher weight to larger values of $f$ than $B$ does. If $w$ decreases, $A$ gives a lower weight to larger values of $f$ than $B$ does.
Only, I haved tried for a long time, I am unable to prove this intuition more formally. Is my intuition correct, and why? Or is my intuition wrong, and is there an easy counterexample?
real-analysis
asked Jan 8 at 13:14
ScindapsusScindapsus
335
$endgroup$
add a comment |
$begingroup$
Consider the entity $A=frac{int_a^b w(x)xf(x)dmu(x)}{int_a^b w(x)xdmu(x)}$, where $f$ is a smooth, monotonously increasing function with positive values: $f,f^prime>0$, $mu$ is the measure used in the integration, and $w$ is a smooth function with positive values: $w>0$. The derivative of $w$ is assumed to have the same sign everywhere: $w^prime=0$ everywhere, $w^prime>0$ everywhere or $w^prime<0$ everywhere.
Now consider the entity $B=frac{int_a^b frac{x}{w(x)}f(x)dmu(x)}{int_a^b frac{x}{w(x)}dmu(x)}$.
I have a strong intuition that the sign of $w^prime$ determines which is greater, $A$ or $B$. If $w^prime=0$ everywhere, then clearly $A=B$.
Now my intuition tells me the following:
$Alessgtr BLeftrightarrow w^primelessgtr 0$.
The intuition is clear: given that $f$ increases, then if $w$ increases, $A$ gives a higher weight to larger values of $f$ than $B$ does. If $w$ decreases, $A$ gives a lower weight to larger values of $f$ than $B$ does.
Only, I haved tried for a long time, I am unable to prove this intuition more formally. Is my intuition correct, and why? Or is my intuition wrong, and is there an easy counterexample?
real-analysis
asked Jan 8 at 13:14
ScindapsusScindapsus
335
$endgroup$
Consider the entity $A=frac{int_a^b w(x)xf(x)dmu(x)}{int_a^b w(x)xdmu(x)}$, where $f$ is a smooth, monotonously increasing function with positive values: $f,f^prime>0$, $mu$ is the measure used in the integration, and $w$ is a smooth function with positive values: $w>0$. The derivative of $w$ is assumed to have the same sign everywhere: $w^prime=0$ everywhere, $w^prime>0$ everywhere or $w^prime<0$ everywhere.
Now consider the entity $B=frac{int_a^b frac{x}{w(x)}f(x)dmu(x)}{int_a^b frac{x}{w(x)}dmu(x)}$.
I have a strong intuition that the sign of $w^prime$ determines which is greater, $A$ or $B$. If $w^prime=0$ everywhere, then clearly $A=B$.
Now my intuition tells me the following:
$Alessgtr BLeftrightarrow w^primelessgtr 0$.
The intuition is clear: given that $f$ increases, then if $w$ increases, $A$ gives a higher weight to larger values of $f$ than $B$ does. If $w$ decreases, $A$ gives a lower weight to larger values of $f$ than $B$ does.
Only, I haved tried for a long time, I am unable to prove this intuition more formally. Is my intuition correct, and why? Or is my intuition wrong, and is there an easy counterexample?
real-analysis
real-analysis
asked Jan 8 at 13:14
ScindapsusScindapsus
335
asked Jan 8 at 13:14
ScindapsusScindapsus
335
edited Jan 8 at 13:20
Scindapsus
asked Jan 8 at 13:14
ScindapsusScindapsus
335
asked Jan 8 at 13:14
ScindapsusScindapsus
335
asked Jan 8 at 13:14
ScindapsusScindapsus
335
335
add a comment |
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