# Upper bound for probability of truncated normal random variable

Suppose that $$X$$ and $$Z$$ are iid with distribution $${\mathcal N}(\mu,\sigma^2)$$. $$X$$ conditional on $$d has a truncated normal distribution with support $$(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.

• $X \mid X > d$ is not a random variable. I think you mean $Y = X$ if $X > d$. But what is it if $X \le d$? Dec 11, 2018 at 15:50
• $Y$ has a truncated normal distribution with support $(d,+\infty)$. I've edited the question to convey that idea. Dec 11, 2018 at 17:05

$$E [Z|Z, so $$P(Y > c + E[Z|Z 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$$.