Integral with respect to measures
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Suppose there are two non-negative random variables X and Y with their cumulative distribution functions $F_X$ and $F_Y$. Further, we know $F_X(t) leq F_Y(t)$ for all $t>0$. Then, for a non-negative function $phi(t)$, do we always have
$$int_0^inftyphi(t)dF_X(t) leq int_0^inftyphi(t)dF_Y(t)$$
For continuous case, I also tried to show density functions $f_X(t) leq f_Y(t)$, but do not know how. My feeling is this is not right. Since if PDFs cross once, one CDF is always greater or equal to the other one.
Finally, can we replace $F_X$ and $F_Y$ to any measures?
I saw the proof somewhere while I cannot remember which reference it is.
measure-theory lebesgue-integral
asked Jan 24 at 2:32
gouwangzhangdonggouwangzhangdong
888
$endgroup$
add a comment |
$begingroup$
Suppose there are two non-negative random variables X and Y with their cumulative distribution functions $F_X$ and $F_Y$. Further, we know $F_X(t) leq F_Y(t)$ for all $t>0$. Then, for a non-negative function $phi(t)$, do we always have
$$int_0^inftyphi(t)dF_X(t) leq int_0^inftyphi(t)dF_Y(t)$$
For continuous case, I also tried to show density functions $f_X(t) leq f_Y(t)$, but do not know how. My feeling is this is not right. Since if PDFs cross once, one CDF is always greater or equal to the other one.
Finally, can we replace $F_X$ and $F_Y$ to any measures?
I saw the proof somewhere while I cannot remember which reference it is.
measure-theory lebesgue-integral
asked Jan 24 at 2:32
gouwangzhangdonggouwangzhangdong
888
$endgroup$
$begingroup$
What if one increases faster in some interval and you sift that interval through the $phi$?
$endgroup$
– lightxbulb
Jan 24 at 2:41
$begingroup$
Could you please give me an example?
$endgroup$
– gouwangzhangdong
Jan 24 at 2:51
2
$begingroup$
The pdf is the derivative of the cdf, you've only provided a constraint such that the cdf of the first is smaller than that of the second. Now imagine that at point $a$ we have $F_Y(a) - F_X(a) = d$, and that at point $a+delta$ we have $F_Y(a+delta) = F_Y(a) + h$. Then in between $a$ and $a+delta$, $F_X$ is allowed to grow by $d+h$ which is more than $h$ as long as $d>0$, that means that the derivative is larger there, so all you need to do is set $phi$ to $0$ everywhere else, and your inequality won't hold anymore I believe. That is $F_X(a+delta)-F_X(a)>F_Y(a+delta)-F_Y(a)$.
$endgroup$
– lightxbulb
Jan 24 at 2:57
add a comment |
$begingroup$
Suppose there are two non-negative random variables X and Y with their cumulative distribution functions $F_X$ and $F_Y$. Further, we know $F_X(t) leq F_Y(t)$ for all $t>0$. Then, for a non-negative function $phi(t)$, do we always have
$$int_0^inftyphi(t)dF_X(t) leq int_0^inftyphi(t)dF_Y(t)$$
For continuous case, I also tried to show density functions $f_X(t) leq f_Y(t)$, but do not know how. My feeling is this is not right. Since if PDFs cross once, one CDF is always greater or equal to the other one.
Finally, can we replace $F_X$ and $F_Y$ to any measures?
I saw the proof somewhere while I cannot remember which reference it is.
measure-theory lebesgue-integral
asked Jan 24 at 2:32
gouwangzhangdonggouwangzhangdong
888
$endgroup$
Suppose there are two non-negative random variables X and Y with their cumulative distribution functions $F_X$ and $F_Y$. Further, we know $F_X(t) leq F_Y(t)$ for all $t>0$. Then, for a non-negative function $phi(t)$, do we always have
$$int_0^inftyphi(t)dF_X(t) leq int_0^inftyphi(t)dF_Y(t)$$
For continuous case, I also tried to show density functions $f_X(t) leq f_Y(t)$, but do not know how. My feeling is this is not right. Since if PDFs cross once, one CDF is always greater or equal to the other one.
Finally, can we replace $F_X$ and $F_Y$ to any measures?
I saw the proof somewhere while I cannot remember which reference it is.
measure-theory lebesgue-integral
measure-theory lebesgue-integral
asked Jan 24 at 2:32
gouwangzhangdonggouwangzhangdong
888
asked Jan 24 at 2:32
gouwangzhangdonggouwangzhangdong
888
edited Jan 24 at 3:38
gouwangzhangdong
asked Jan 24 at 2:32
gouwangzhangdonggouwangzhangdong
888
asked Jan 24 at 2:32
gouwangzhangdonggouwangzhangdong
888
asked Jan 24 at 2:32
gouwangzhangdonggouwangzhangdong
888
888
$begingroup$
What if one increases faster in some interval and you sift that interval through the $phi$?
$endgroup$
– lightxbulb
Jan 24 at 2:41
$begingroup$
Could you please give me an example?
$endgroup$
– gouwangzhangdong
Jan 24 at 2:51
2
$begingroup$
The pdf is the derivative of the cdf, you've only provided a constraint such that the cdf of the first is smaller than that of the second. Now imagine that at point $a$ we have $F_Y(a) - F_X(a) = d$, and that at point $a+delta$ we have $F_Y(a+delta) = F_Y(a) + h$. Then in between $a$ and $a+delta$, $F_X$ is allowed to grow by $d+h$ which is more than $h$ as long as $d>0$, that means that the derivative is larger there, so all you need to do is set $phi$ to $0$ everywhere else, and your inequality won't hold anymore I believe. That is $F_X(a+delta)-F_X(a)>F_Y(a+delta)-F_Y(a)$.
$endgroup$
– lightxbulb
Jan 24 at 2:57
add a comment |
$begingroup$
What if one increases faster in some interval and you sift that interval through the $phi$?
$endgroup$
– lightxbulb
Jan 24 at 2:41
$begingroup$
Could you please give me an example?
$endgroup$
– gouwangzhangdong
Jan 24 at 2:51
2
$begingroup$
The pdf is the derivative of the cdf, you've only provided a constraint such that the cdf of the first is smaller than that of the second. Now imagine that at point $a$ we have $F_Y(a) - F_X(a) = d$, and that at point $a+delta$ we have $F_Y(a+delta) = F_Y(a) + h$. Then in between $a$ and $a+delta$, $F_X$ is allowed to grow by $d+h$ which is more than $h$ as long as $d>0$, that means that the derivative is larger there, so all you need to do is set $phi$ to $0$ everywhere else, and your inequality won't hold anymore I believe. That is $F_X(a+delta)-F_X(a)>F_Y(a+delta)-F_Y(a)$.
$endgroup$
– lightxbulb
Jan 24 at 2:57
$begingroup$
What if one increases faster in some interval and you sift that interval through the $phi$?
$endgroup$
– lightxbulb
Jan 24 at 2:41
$begingroup$
What if one increases faster in some interval and you sift that interval through the $phi$?
$endgroup$
– lightxbulb
Jan 24 at 2:41
$begingroup$
Could you please give me an example?
$endgroup$
– gouwangzhangdong
Jan 24 at 2:51
$begingroup$
Could you please give me an example?
$endgroup$
– gouwangzhangdong
Jan 24 at 2:51
2
2
$begingroup$
The pdf is the derivative of the cdf, you've only provided a constraint such that the cdf of the first is smaller than that of the second. Now imagine that at point $a$ we have $F_Y(a) - F_X(a) = d$, and that at point $a+delta$ we have $F_Y(a+delta) = F_Y(a) + h$. Then in between $a$ and $a+delta$, $F_X$ is allowed to grow by $d+h$ which is more than $h$ as long as $d>0$, that means that the derivative is larger there, so all you need to do is set $phi$ to $0$ everywhere else, and your inequality won't hold anymore I believe. That is $F_X(a+delta)-F_X(a)>F_Y(a+delta)-F_Y(a)$.
$endgroup$
– lightxbulb
Jan 24 at 2:57
$begingroup$
The pdf is the derivative of the cdf, you've only provided a constraint such that the cdf of the first is smaller than that of the second. Now imagine that at point $a$ we have $F_Y(a) - F_X(a) = d$, and that at point $a+delta$ we have $F_Y(a+delta) = F_Y(a) + h$. Then in between $a$ and $a+delta$, $F_X$ is allowed to grow by $d+h$ which is more than $h$ as long as $d>0$, that means that the derivative is larger there, so all you need to do is set $phi$ to $0$ everywhere else, and your inequality won't hold anymore I believe. That is $F_X(a+delta)-F_X(a)>F_Y(a+delta)-F_Y(a)$.
$endgroup$
– lightxbulb
Jan 24 at 2:57
add a comment |
1 Answer
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Certainly false. Let $X=2$ and $Y=1$. Then the hypothesis is true and you are asking if $phi (2) leq phi (1)%=$ for any non-negative $phi $ which is obviously false. However the conclusion is true if $phi$ is non-negative and decreasing.
answered Jan 24 at 6:37
Kavi Rama MurthyKavi Rama Murthy
65k42766
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add a comment |
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$begingroup$
Certainly false. Let $X=2$ and $Y=1$. Then the hypothesis is true and you are asking if $phi (2) leq phi (1)%=$ for any non-negative $phi $ which is obviously false. However the conclusion is true if $phi$ is non-negative and decreasing.
answered Jan 24 at 6:37
Kavi Rama MurthyKavi Rama Murthy
65k42766
$endgroup$
add a comment |
$begingroup$
Certainly false. Let $X=2$ and $Y=1$. Then the hypothesis is true and you are asking if $phi (2) leq phi (1)%=$ for any non-negative $phi $ which is obviously false. However the conclusion is true if $phi$ is non-negative and decreasing.
answered Jan 24 at 6:37
Kavi Rama MurthyKavi Rama Murthy
65k42766
$endgroup$
add a comment |
$begingroup$
Certainly false. Let $X=2$ and $Y=1$. Then the hypothesis is true and you are asking if $phi (2) leq phi (1)%=$ for any non-negative $phi $ which is obviously false. However the conclusion is true if $phi$ is non-negative and decreasing.
answered Jan 24 at 6:37
Kavi Rama MurthyKavi Rama Murthy
65k42766
$endgroup$
Certainly false. Let $X=2$ and $Y=1$. Then the hypothesis is true and you are asking if $phi (2) leq phi (1)%=$ for any non-negative $phi $ which is obviously false. However the conclusion is true if $phi$ is non-negative and decreasing.
answered Jan 24 at 6:37
Kavi Rama MurthyKavi Rama Murthy
65k42766
answered Jan 24 at 6:37
Kavi Rama MurthyKavi Rama Murthy
65k42766
answered Jan 24 at 6:37
Kavi Rama MurthyKavi Rama Murthy
65k42766
answered Jan 24 at 6:37
Kavi Rama MurthyKavi Rama Murthy
65k42766
65k42766
add a comment |
add a comment |
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$begingroup$
What if one increases faster in some interval and you sift that interval through the $phi$?
$endgroup$
– lightxbulb
Jan 24 at 2:41
$begingroup$
Could you please give me an example?
$endgroup$
– gouwangzhangdong
Jan 24 at 2:51
2
$begingroup$
The pdf is the derivative of the cdf, you've only provided a constraint such that the cdf of the first is smaller than that of the second. Now imagine that at point $a$ we have $F_Y(a) - F_X(a) = d$, and that at point $a+delta$ we have $F_Y(a+delta) = F_Y(a) + h$. Then in between $a$ and $a+delta$, $F_X$ is allowed to grow by $d+h$ which is more than $h$ as long as $d>0$, that means that the derivative is larger there, so all you need to do is set $phi$ to $0$ everywhere else, and your inequality won't hold anymore I believe. That is $F_X(a+delta)-F_X(a)>F_Y(a+delta)-F_Y(a)$.
$endgroup$
– lightxbulb
Jan 24 at 2:57