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Why inverse Mills ratio for normal distribution is 1-Lipschitz continuous?

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0

$begingroup$

The inverse Mill ratio for a standard normal distribution is:
$$
IMR(x) = fracphi(x)Phi(x),
$$
where $phi(x)$ is the pdf of standard normal distribution and $Phi(x)$ is the cdf of standard normal distribution.
The paper "Nonasymptotic analysis of semiparametric regression models with high-dimensional parametric coefficients" by Ying Zhu suggests that this function is 1-Liphshits continuous (see note at page 2290):
$$
|IMR(x) - IMR(y)| leq |x - y|.
$$

How one should rigorously prove this statement?

probability normal-distribution lipschitz-functions gaussian-integral

share|cite|improve this question

asked Mar 29 at 11:21

Alexey ZaytsevAlexey Zaytsev

20019

$endgroup$

add a comment | 

0

$begingroup$

The inverse Mill ratio for a standard normal distribution is:
$$
IMR(x) = fracphi(x)Phi(x),
$$
where $phi(x)$ is the pdf of standard normal distribution and $Phi(x)$ is the cdf of standard normal distribution.
The paper "Nonasymptotic analysis of semiparametric regression models with high-dimensional parametric coefficients" by Ying Zhu suggests that this function is 1-Liphshits continuous (see note at page 2290):
$$
|IMR(x) - IMR(y)| leq |x - y|.
$$

How one should rigorously prove this statement?

probability normal-distribution lipschitz-functions gaussian-integral

share|cite|improve this question

asked Mar 29 at 11:21

Alexey ZaytsevAlexey Zaytsev

20019

$endgroup$

add a comment | 

$begingroup$

The inverse Mill ratio for a standard normal distribution is:
$$
IMR(x) = fracphi(x)Phi(x),
$$
where $phi(x)$ is the pdf of standard normal distribution and $Phi(x)$ is the cdf of standard normal distribution.
The paper "Nonasymptotic analysis of semiparametric regression models with high-dimensional parametric coefficients" by Ying Zhu suggests that this function is 1-Liphshits continuous (see note at page 2290):
$$
|IMR(x) - IMR(y)| leq |x - y|.
$$

How one should rigorously prove this statement?

probability normal-distribution lipschitz-functions gaussian-integral

share|cite|improve this question

asked Mar 29 at 11:21

Alexey ZaytsevAlexey Zaytsev

20019

$endgroup$

share|cite|improve this question

asked Mar 29 at 11:21

Alexey ZaytsevAlexey Zaytsev

20019

share|cite|improve this question

asked Mar 29 at 11:21

Alexey ZaytsevAlexey Zaytsev

20019

asked Mar 29 at 11:21

Alexey ZaytsevAlexey Zaytsev

20019

asked Mar 29 at 11:21

Alexey ZaytsevAlexey Zaytsev

20019