How can I show this property about the quantile function?
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Let $F:mathbb{R}rightarrow [0,1]$ be a non-decreasing right continuous function, and it's generalized inverse function $F^{-1}:[0,1]rightarrow mathbb{R}$ defined as $F^{-1}(y)=inf {xin mathbb{R}:F(x)geq y}$.
How can I show that if
$F^{-1}$ is continuous at $y$ and if $F(x-)leq yleq F(x)$ then $F^{-1}(y)=x$?
probability probability-theory probability-distributions
asked Dec 6 '18 at 0:56
Enzo GiannottaEnzo Giannotta
13011
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add a comment |
$begingroup$
Let $F:mathbb{R}rightarrow [0,1]$ be a non-decreasing right continuous function, and it's generalized inverse function $F^{-1}:[0,1]rightarrow mathbb{R}$ defined as $F^{-1}(y)=inf {xin mathbb{R}:F(x)geq y}$.
How can I show that if
$F^{-1}$ is continuous at $y$ and if $F(x-)leq yleq F(x)$ then $F^{-1}(y)=x$?
probability probability-theory probability-distributions
asked Dec 6 '18 at 0:56
Enzo GiannottaEnzo Giannotta
13011
$endgroup$
1
$begingroup$
If you take $F$ left continuous, why $F(x-)$ and $F(x)$ differ?
$endgroup$
– NCh
Dec 6 '18 at 1:19
$begingroup$
I edited the post, I meant right continuous, my bad, thank you.
$endgroup$
– Enzo Giannotta
Dec 6 '18 at 16:22
add a comment |
$begingroup$
Let $F:mathbb{R}rightarrow [0,1]$ be a non-decreasing right continuous function, and it's generalized inverse function $F^{-1}:[0,1]rightarrow mathbb{R}$ defined as $F^{-1}(y)=inf {xin mathbb{R}:F(x)geq y}$.
How can I show that if
$F^{-1}$ is continuous at $y$ and if $F(x-)leq yleq F(x)$ then $F^{-1}(y)=x$?
probability probability-theory probability-distributions
asked Dec 6 '18 at 0:56
Enzo GiannottaEnzo Giannotta
13011
$endgroup$
Let $F:mathbb{R}rightarrow [0,1]$ be a non-decreasing right continuous function, and it's generalized inverse function $F^{-1}:[0,1]rightarrow mathbb{R}$ defined as $F^{-1}(y)=inf {xin mathbb{R}:F(x)geq y}$.
How can I show that if
$F^{-1}$ is continuous at $y$ and if $F(x-)leq yleq F(x)$ then $F^{-1}(y)=x$?
probability probability-theory probability-distributions
probability probability-theory probability-distributions
asked Dec 6 '18 at 0:56
Enzo GiannottaEnzo Giannotta
13011
asked Dec 6 '18 at 0:56
Enzo GiannottaEnzo Giannotta
13011
edited Dec 6 '18 at 16:21
Enzo Giannotta
asked Dec 6 '18 at 0:56
Enzo GiannottaEnzo Giannotta
13011
asked Dec 6 '18 at 0:56
Enzo GiannottaEnzo Giannotta
13011
asked Dec 6 '18 at 0:56
Enzo GiannottaEnzo Giannotta
13011
13011
1
$begingroup$
If you take $F$ left continuous, why $F(x-)$ and $F(x)$ differ?
$endgroup$
– NCh
Dec 6 '18 at 1:19
$begingroup$
I edited the post, I meant right continuous, my bad, thank you.
$endgroup$
– Enzo Giannotta
Dec 6 '18 at 16:22
add a comment |
1
$begingroup$
If you take $F$ left continuous, why $F(x-)$ and $F(x)$ differ?
$endgroup$
– NCh
Dec 6 '18 at 1:19
$begingroup$
I edited the post, I meant right continuous, my bad, thank you.
$endgroup$
– Enzo Giannotta
Dec 6 '18 at 16:22
1
1
$begingroup$
If you take $F$ left continuous, why $F(x-)$ and $F(x)$ differ?
$endgroup$
– NCh
Dec 6 '18 at 1:19
$begingroup$
If you take $F$ left continuous, why $F(x-)$ and $F(x)$ differ?
$endgroup$
– NCh
Dec 6 '18 at 1:19
$begingroup$
I edited the post, I meant right continuous, my bad, thank you.
$endgroup$
– Enzo Giannotta
Dec 6 '18 at 16:22
$begingroup$
I edited the post, I meant right continuous, my bad, thank you.
$endgroup$
– Enzo Giannotta
Dec 6 '18 at 16:22
add a comment |
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1
$begingroup$
If you take $F$ left continuous, why $F(x-)$ and $F(x)$ differ?
$endgroup$
– NCh
Dec 6 '18 at 1:19
$begingroup$
I edited the post, I meant right continuous, my bad, thank you.
$endgroup$
– Enzo Giannotta
Dec 6 '18 at 16:22