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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

$F(x,\mu,\sigma,d) = \dfrac{\Phi\Bigl(\dfrac{x-\mu}{\sigma}\Bigr)-\Phi\Bigl(\dfrac{d-\mu}{\sigma}\Bigr)}{1-\Phi\Bigl(\dfrac{d-\mu}{\sigma}\Bigr)}$.

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{\Phi\left(\dfrac{c}{\sigma}-\dfrac{\phi\Bigl(\dfrac{d-\mu}{\sigma}\Bigr)}{\Phi\Bigl(\dfrac{d-\mu}{\sigma}\Bigr)}\right)-\Phi\Bigl(\dfrac{d-\mu}{\sigma}\Bigr)}{1-\Phi\Bigl(\dfrac{d-\mu}{\sigma}\Bigr)} \end{align*} where $\Phi(\cdot)$ and $\phi(\cdot)$ denote the standard normal CDF and PDF respectively.

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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 $X \le d$? $\endgroup$ Dec 11, 2018 at 15:50
  • $\begingroup$ $Y$ has a truncated normal distribution with support $(d,+\infty)$. I've edited the question to convey that idea. $\endgroup$
    – Ararat
    Dec 11, 2018 at 17:05

1 Answer 1

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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 $1$.

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