Inequality involving Monotone likelihood ratio and CDF ratio
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This problem has really been bothering me and I have no idea whether the statement is true or not. So any help is appreciated.
Suppose $f(x)/g(x)$ satisfies monotone likelihood ratio property and have CDF $F$ and $G$, respectively. Assume $f$ and $g$ have support in $[0,1]$. Let $C leq 1$ be a constant, and some $1 geq x_1 geq x_2geq 0 $, such that the following condition holds:
$$ Ccdot frac{f(x_1)}{g(x_1)} = frac{f(x_2)}{g(x_2)}.$$
Prove or give a counterexample
$$ Ccdot frac{1-F(x_1)}{1-G(x_1)} leq frac{1-F(x_2)}{1-G(x_2)}$$
If false, is there a simple condition that could be added to $f$ and $g$ such that the statement is true?
Attempt to showing proof: A natural place to start was using the inequality
$$frac{f}{g}(x) leq frac{1-F(x)}{1-G(x)}$$
but that ended up going nowhere.
Attempt to construct counterexample
If there is a counterexample, it could be that $1-F(x_2)$ is extremely close to $0$. But that means $1-F(x_1)$ is also close to $0$ since $x_1 > x_2$. It's also not immediately clear that the construction is valid. I also attempted a simple case of $F = x^2$ and $G = x$ but it was not a counterexample.
statistics density-function ratio
asked Dec 30 '18 at 6:34
Jack FJack F
1112
$endgroup$
add a comment |
$begingroup$
This problem has really been bothering me and I have no idea whether the statement is true or not. So any help is appreciated.
Suppose $f(x)/g(x)$ satisfies monotone likelihood ratio property and have CDF $F$ and $G$, respectively. Assume $f$ and $g$ have support in $[0,1]$. Let $C leq 1$ be a constant, and some $1 geq x_1 geq x_2geq 0 $, such that the following condition holds:
$$ Ccdot frac{f(x_1)}{g(x_1)} = frac{f(x_2)}{g(x_2)}.$$
Prove or give a counterexample
$$ Ccdot frac{1-F(x_1)}{1-G(x_1)} leq frac{1-F(x_2)}{1-G(x_2)}$$
If false, is there a simple condition that could be added to $f$ and $g$ such that the statement is true?
Attempt to showing proof: A natural place to start was using the inequality
$$frac{f}{g}(x) leq frac{1-F(x)}{1-G(x)}$$
but that ended up going nowhere.
Attempt to construct counterexample
If there is a counterexample, it could be that $1-F(x_2)$ is extremely close to $0$. But that means $1-F(x_1)$ is also close to $0$ since $x_1 > x_2$. It's also not immediately clear that the construction is valid. I also attempted a simple case of $F = x^2$ and $G = x$ but it was not a counterexample.
statistics density-function ratio
asked Dec 30 '18 at 6:34
Jack FJack F
1112
$endgroup$
add a comment |
$begingroup$
This problem has really been bothering me and I have no idea whether the statement is true or not. So any help is appreciated.
Suppose $f(x)/g(x)$ satisfies monotone likelihood ratio property and have CDF $F$ and $G$, respectively. Assume $f$ and $g$ have support in $[0,1]$. Let $C leq 1$ be a constant, and some $1 geq x_1 geq x_2geq 0 $, such that the following condition holds:
$$ Ccdot frac{f(x_1)}{g(x_1)} = frac{f(x_2)}{g(x_2)}.$$
Prove or give a counterexample
$$ Ccdot frac{1-F(x_1)}{1-G(x_1)} leq frac{1-F(x_2)}{1-G(x_2)}$$
If false, is there a simple condition that could be added to $f$ and $g$ such that the statement is true?
Attempt to showing proof: A natural place to start was using the inequality
$$frac{f}{g}(x) leq frac{1-F(x)}{1-G(x)}$$
but that ended up going nowhere.
Attempt to construct counterexample
If there is a counterexample, it could be that $1-F(x_2)$ is extremely close to $0$. But that means $1-F(x_1)$ is also close to $0$ since $x_1 > x_2$. It's also not immediately clear that the construction is valid. I also attempted a simple case of $F = x^2$ and $G = x$ but it was not a counterexample.
statistics density-function ratio
asked Dec 30 '18 at 6:34
Jack FJack F
1112
$endgroup$
This problem has really been bothering me and I have no idea whether the statement is true or not. So any help is appreciated.
Suppose $f(x)/g(x)$ satisfies monotone likelihood ratio property and have CDF $F$ and $G$, respectively. Assume $f$ and $g$ have support in $[0,1]$. Let $C leq 1$ be a constant, and some $1 geq x_1 geq x_2geq 0 $, such that the following condition holds:
$$ Ccdot frac{f(x_1)}{g(x_1)} = frac{f(x_2)}{g(x_2)}.$$
Prove or give a counterexample
$$ Ccdot frac{1-F(x_1)}{1-G(x_1)} leq frac{1-F(x_2)}{1-G(x_2)}$$
If false, is there a simple condition that could be added to $f$ and $g$ such that the statement is true?
Attempt to showing proof: A natural place to start was using the inequality
$$frac{f}{g}(x) leq frac{1-F(x)}{1-G(x)}$$
but that ended up going nowhere.
Attempt to construct counterexample
If there is a counterexample, it could be that $1-F(x_2)$ is extremely close to $0$. But that means $1-F(x_1)$ is also close to $0$ since $x_1 > x_2$. It's also not immediately clear that the construction is valid. I also attempted a simple case of $F = x^2$ and $G = x$ but it was not a counterexample.
statistics density-function ratio
statistics density-function ratio
asked Dec 30 '18 at 6:34
Jack FJack F
1112
asked Dec 30 '18 at 6:34
Jack FJack F
1112
asked Dec 30 '18 at 6:34
Jack FJack F
1112
asked Dec 30 '18 at 6:34
Jack FJack F
1112
asked Dec 30 '18 at 6:34
Jack FJack F
1112
1112
add a comment |
add a comment |
1 Answer
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The answer should be false. A counter example can be constructed attacking $G(x)$ instead. Suppose $G$ rises extremely fast initially with a no change in slope until $G(x)$ is close to $1$, while $F$ is relatively flat initially with no change in slope until $G(x)$ changes slope as well.
Then $x_1$ can be picked just after the change in slope such that $f(x_1)/g(x_1)$ is very close to $f(x_2)/g(x_2)$. Pick $x_2$ near $0$, so that $(1-F(x_2))/(1-G(x_2))$ is close to $1$, but by the construction of $F$ and $G$, $(1-F(x_1))/(1-G(x_1)) >> 1$. This then implies the desired inequality is false.
answered Dec 30 '18 at 23:02
Jack FJack F
1112
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add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
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votes
active
oldest
votes
$begingroup$
The answer should be false. A counter example can be constructed attacking $G(x)$ instead. Suppose $G$ rises extremely fast initially with a no change in slope until $G(x)$ is close to $1$, while $F$ is relatively flat initially with no change in slope until $G(x)$ changes slope as well.
Then $x_1$ can be picked just after the change in slope such that $f(x_1)/g(x_1)$ is very close to $f(x_2)/g(x_2)$. Pick $x_2$ near $0$, so that $(1-F(x_2))/(1-G(x_2))$ is close to $1$, but by the construction of $F$ and $G$, $(1-F(x_1))/(1-G(x_1)) >> 1$. This then implies the desired inequality is false.
answered Dec 30 '18 at 23:02
Jack FJack F
1112
$endgroup$
add a comment |
$begingroup$
The answer should be false. A counter example can be constructed attacking $G(x)$ instead. Suppose $G$ rises extremely fast initially with a no change in slope until $G(x)$ is close to $1$, while $F$ is relatively flat initially with no change in slope until $G(x)$ changes slope as well.
Then $x_1$ can be picked just after the change in slope such that $f(x_1)/g(x_1)$ is very close to $f(x_2)/g(x_2)$. Pick $x_2$ near $0$, so that $(1-F(x_2))/(1-G(x_2))$ is close to $1$, but by the construction of $F$ and $G$, $(1-F(x_1))/(1-G(x_1)) >> 1$. This then implies the desired inequality is false.
answered Dec 30 '18 at 23:02
Jack FJack F
1112
$endgroup$
add a comment |
$begingroup$
The answer should be false. A counter example can be constructed attacking $G(x)$ instead. Suppose $G$ rises extremely fast initially with a no change in slope until $G(x)$ is close to $1$, while $F$ is relatively flat initially with no change in slope until $G(x)$ changes slope as well.
Then $x_1$ can be picked just after the change in slope such that $f(x_1)/g(x_1)$ is very close to $f(x_2)/g(x_2)$. Pick $x_2$ near $0$, so that $(1-F(x_2))/(1-G(x_2))$ is close to $1$, but by the construction of $F$ and $G$, $(1-F(x_1))/(1-G(x_1)) >> 1$. This then implies the desired inequality is false.
answered Dec 30 '18 at 23:02
Jack FJack F
1112
$endgroup$
The answer should be false. A counter example can be constructed attacking $G(x)$ instead. Suppose $G$ rises extremely fast initially with a no change in slope until $G(x)$ is close to $1$, while $F$ is relatively flat initially with no change in slope until $G(x)$ changes slope as well.
Then $x_1$ can be picked just after the change in slope such that $f(x_1)/g(x_1)$ is very close to $f(x_2)/g(x_2)$. Pick $x_2$ near $0$, so that $(1-F(x_2))/(1-G(x_2))$ is close to $1$, but by the construction of $F$ and $G$, $(1-F(x_1))/(1-G(x_1)) >> 1$. This then implies the desired inequality is false.
answered Dec 30 '18 at 23:02
Jack FJack F
1112
answered Dec 30 '18 at 23:02
Jack FJack F
1112
answered Dec 30 '18 at 23:02
Jack FJack F
1112
answered Dec 30 '18 at 23:02
Jack FJack F
1112
1112
add a comment |
add a comment |
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