Upper bound for probability of truncated normal random variable
$begingroup$
Suppose that $X$ and $Z$ are iid with distribution ${mathcal N}(mu,sigma^2)$. $X$ conditional on $d<X<+infty$ has a truncated normal distribution with support $(d,+infty)$. Letting $Y$ denote that random variable, its CDF is
$F(x,mu,sigma,d) = dfrac{PhiBigl(dfrac{x-mu}{sigma}Bigr)-PhiBigl(dfrac{d-mu}{sigma}Bigr)}{1-PhiBigl(dfrac{d-mu}{sigma}Bigr)}$.
I've computed $ {rm P} (Y > c+ {rm E}[Z,|, Z< d])$ and I'm trying to find an upper bound that does not depend on $d$. Any help?
begin{align*}
{rm P} (Y > c+ {rm E}[Z,|, Z< d]) = dfrac{Phileft(dfrac{c}{sigma}-dfrac{phiBigl(dfrac{d-mu}{sigma}Bigr)}{PhiBigl(dfrac{d-mu}{sigma}Bigr)}right)-PhiBigl(dfrac{d-mu}{sigma}Bigr)}{1-PhiBigl(dfrac{d-mu}{sigma}Bigr)}
end{align*}
where $Phi(cdot)$ and $phi(cdot)$ denote the standard normal CDF and PDF respectively.
probability normal-distribution conditional-probability upper-lower-bounds
asked Dec 11 '18 at 15:38
capadociacapadocia
325
$endgroup$
add a comment |
$begingroup$
Suppose that $X$ and $Z$ are iid with distribution ${mathcal N}(mu,sigma^2)$. $X$ conditional on $d<X<+infty$ has a truncated normal distribution with support $(d,+infty)$. Letting $Y$ denote that random variable, its CDF is
$F(x,mu,sigma,d) = dfrac{PhiBigl(dfrac{x-mu}{sigma}Bigr)-PhiBigl(dfrac{d-mu}{sigma}Bigr)}{1-PhiBigl(dfrac{d-mu}{sigma}Bigr)}$.
I've computed $ {rm P} (Y > c+ {rm E}[Z,|, Z< d])$ and I'm trying to find an upper bound that does not depend on $d$. Any help?
begin{align*}
{rm P} (Y > c+ {rm E}[Z,|, Z< d]) = dfrac{Phileft(dfrac{c}{sigma}-dfrac{phiBigl(dfrac{d-mu}{sigma}Bigr)}{PhiBigl(dfrac{d-mu}{sigma}Bigr)}right)-PhiBigl(dfrac{d-mu}{sigma}Bigr)}{1-PhiBigl(dfrac{d-mu}{sigma}Bigr)}
end{align*}
where $Phi(cdot)$ and $phi(cdot)$ denote the standard normal CDF and PDF respectively.
probability normal-distribution conditional-probability upper-lower-bounds
asked Dec 11 '18 at 15:38
capadociacapadocia
325
$endgroup$
2
$begingroup$
$X mid X > d$ is not a random variable. I think you mean $Y = X$ if $X > d$. But what is it if $X le d$?
$endgroup$
– Robert Israel
Dec 11 '18 at 15:50
$begingroup$
$Y$ has a truncated normal distribution with support $(d,+infty)$. I've edited the question to convey that idea.
$endgroup$
– capadocia
Dec 11 '18 at 17:05
add a comment |
$begingroup$
Suppose that $X$ and $Z$ are iid with distribution ${mathcal N}(mu,sigma^2)$. $X$ conditional on $d<X<+infty$ has a truncated normal distribution with support $(d,+infty)$. Letting $Y$ denote that random variable, its CDF is
$F(x,mu,sigma,d) = dfrac{PhiBigl(dfrac{x-mu}{sigma}Bigr)-PhiBigl(dfrac{d-mu}{sigma}Bigr)}{1-PhiBigl(dfrac{d-mu}{sigma}Bigr)}$.
I've computed $ {rm P} (Y > c+ {rm E}[Z,|, Z< d])$ and I'm trying to find an upper bound that does not depend on $d$. Any help?
begin{align*}
{rm P} (Y > c+ {rm E}[Z,|, Z< d]) = dfrac{Phileft(dfrac{c}{sigma}-dfrac{phiBigl(dfrac{d-mu}{sigma}Bigr)}{PhiBigl(dfrac{d-mu}{sigma}Bigr)}right)-PhiBigl(dfrac{d-mu}{sigma}Bigr)}{1-PhiBigl(dfrac{d-mu}{sigma}Bigr)}
end{align*}
where $Phi(cdot)$ and $phi(cdot)$ denote the standard normal CDF and PDF respectively.
probability normal-distribution conditional-probability upper-lower-bounds
asked Dec 11 '18 at 15:38
capadociacapadocia
325
$endgroup$
Suppose that $X$ and $Z$ are iid with distribution ${mathcal N}(mu,sigma^2)$. $X$ conditional on $d<X<+infty$ has a truncated normal distribution with support $(d,+infty)$. Letting $Y$ denote that random variable, its CDF is
$F(x,mu,sigma,d) = dfrac{PhiBigl(dfrac{x-mu}{sigma}Bigr)-PhiBigl(dfrac{d-mu}{sigma}Bigr)}{1-PhiBigl(dfrac{d-mu}{sigma}Bigr)}$.
I've computed $ {rm P} (Y > c+ {rm E}[Z,|, Z< d])$ and I'm trying to find an upper bound that does not depend on $d$. Any help?
begin{align*}
{rm P} (Y > c+ {rm E}[Z,|, Z< d]) = dfrac{Phileft(dfrac{c}{sigma}-dfrac{phiBigl(dfrac{d-mu}{sigma}Bigr)}{PhiBigl(dfrac{d-mu}{sigma}Bigr)}right)-PhiBigl(dfrac{d-mu}{sigma}Bigr)}{1-PhiBigl(dfrac{d-mu}{sigma}Bigr)}
end{align*}
where $Phi(cdot)$ and $phi(cdot)$ denote the standard normal CDF and PDF respectively.
probability normal-distribution conditional-probability upper-lower-bounds
probability normal-distribution conditional-probability upper-lower-bounds
asked Dec 11 '18 at 15:38
capadociacapadocia
325
asked Dec 11 '18 at 15:38
capadociacapadocia
325
edited Dec 11 '18 at 17:01
capadocia
asked Dec 11 '18 at 15:38
capadociacapadocia
325
asked Dec 11 '18 at 15:38
capadociacapadocia
325
asked Dec 11 '18 at 15:38
capadociacapadocia
325
325
2
$begingroup$
$X mid X > d$ is not a random variable. I think you mean $Y = X$ if $X > d$. But what is it if $X le d$?
$endgroup$
– Robert Israel
Dec 11 '18 at 15:50
$begingroup$
$Y$ has a truncated normal distribution with support $(d,+infty)$. I've edited the question to convey that idea.
$endgroup$
– capadocia
Dec 11 '18 at 17:05
add a comment |
2
$begingroup$
$X mid X > d$ is not a random variable. I think you mean $Y = X$ if $X > d$. But what is it if $X le d$?
$endgroup$
– Robert Israel
Dec 11 '18 at 15:50
$begingroup$
$Y$ has a truncated normal distribution with support $(d,+infty)$. I've edited the question to convey that idea.
$endgroup$
– capadocia
Dec 11 '18 at 17:05
2
2
$begingroup$
$X mid X > d$ is not a random variable. I think you mean $Y = X$ if $X > d$. But what is it if $X le d$?
$endgroup$
– Robert Israel
Dec 11 '18 at 15:50
$begingroup$
$X mid X > d$ is not a random variable. I think you mean $Y = X$ if $X > d$. But what is it if $X le d$?
$endgroup$
– Robert Israel
Dec 11 '18 at 15:50
$begingroup$
$Y$ has a truncated normal distribution with support $(d,+infty)$. I've edited the question to convey that idea.
$endgroup$
– capadocia
Dec 11 '18 at 17:05
$begingroup$
$Y$ has a truncated normal distribution with support $(d,+infty)$. I've edited the question to convey that idea.
$endgroup$
– capadocia
Dec 11 '18 at 17:05
add a comment |
1 Answer
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$E [Z|Z<d] < mu$, so $P(Y > c + E[Z|Z<d]) ge P(Y > c+mu)$. But if $d > c+mu$, $P(Y > c+mu) = 1$. So the best upper bound you can have that doesn't depend on $d$ is $1$.
answered Dec 11 '18 at 17:20
Robert IsraelRobert Israel
322k23212465
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add a comment |
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$begingroup$
$E [Z|Z<d] < mu$, so $P(Y > c + E[Z|Z<d]) ge P(Y > c+mu)$. But if $d > c+mu$, $P(Y > c+mu) = 1$. So the best upper bound you can have that doesn't depend on $d$ is $1$.
answered Dec 11 '18 at 17:20
Robert IsraelRobert Israel
322k23212465
$endgroup$
add a comment |
$begingroup$
$E [Z|Z<d] < mu$, so $P(Y > c + E[Z|Z<d]) ge P(Y > c+mu)$. But if $d > c+mu$, $P(Y > c+mu) = 1$. So the best upper bound you can have that doesn't depend on $d$ is $1$.
answered Dec 11 '18 at 17:20
Robert IsraelRobert Israel
322k23212465
$endgroup$
add a comment |
$begingroup$
$E [Z|Z<d] < mu$, so $P(Y > c + E[Z|Z<d]) ge P(Y > c+mu)$. But if $d > c+mu$, $P(Y > c+mu) = 1$. So the best upper bound you can have that doesn't depend on $d$ is $1$.
answered Dec 11 '18 at 17:20
Robert IsraelRobert Israel
322k23212465
$endgroup$
$E [Z|Z<d] < mu$, so $P(Y > c + E[Z|Z<d]) ge P(Y > c+mu)$. But if $d > c+mu$, $P(Y > c+mu) = 1$. So the best upper bound you can have that doesn't depend on $d$ is $1$.
answered Dec 11 '18 at 17:20
Robert IsraelRobert Israel
322k23212465
answered Dec 11 '18 at 17:20
Robert IsraelRobert Israel
322k23212465
answered Dec 11 '18 at 17:20
Robert IsraelRobert Israel
322k23212465
answered Dec 11 '18 at 17:20
Robert IsraelRobert Israel
322k23212465
322k23212465
add a comment |
add a comment |
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$begingroup$
$X mid X > d$ is not a random variable. I think you mean $Y = X$ if $X > d$. But what is it if $X le d$?
$endgroup$
– Robert Israel
Dec 11 '18 at 15:50
$begingroup$
$Y$ has a truncated normal distribution with support $(d,+infty)$. I've edited the question to convey that idea.
$endgroup$
– capadocia
Dec 11 '18 at 17:05