Recently, I am wondering how can I solve the following problem.
X is a variable with no randomness，while $\xi$ is a stochastic variable which follows a normal distribution $\mathcal{N}(\mu,\sigma^2)$.
In a stochastic problem, one of the constraints is $\mathbb{P}(|X+\xi|\leq C)\geq\eta$, where $C$ is a positive constant value.
In some papers, this constraint is shown to be equal to $|X+\mu|+\sigma Z_{1-\eta}\leq C$, where $Z_x$ is the 100 $\times (1-x)$th percentile of the 2 tailed standard normal distribution which can be determined by taking the inverse CDF of the standard normal distribution evaluated at $x$.
But from my viewpoint, I don't think it is correct. I think the probability problem should be solved as follows:
$\mathbb{P}(|X+\xi|\leq C) \Rightarrow \mathbb{P}(-C-X\leq \xi \leq C-X )\Rightarrow \mathbb{P}(\frac{-C-X-\mu}{\sigma}\leq \frac{\xi-\mu}{\sigma}\leq \frac{C-X-\mu}{\sigma})$.
The constraint should be equal to $\Phi \left( \frac{C-X-\mu}{\sigma}\right)-\Phi \left( \frac{-C-X-\mu}{\sigma}\right)\geq\eta$.
Therefore, I don't think it is equal to the form shown in many papers. Is there any problem with my thought?
Looking forward to receiving your help! Thanks a lot!
Xin
You missed a $1\over 2$ factor in the calculation. Specifically, making $\Phi ({C-X-\mu\over{ \sigma}})$ equal to $1-\Phi ({-C-X-\mu\over{ \sigma}})$ gives you $$\Phi ({C-X-\mu\over{ \sigma}})=\Phi ({-C-X-\mu\over{ \sigma}})=1-{\eta \over 2}.$$ This means ${C-X-\mu\over{ \sigma}}=Z_{1-\eta / 2}$ and ${-C-X-\mu\over{ \sigma}}=-Z_{1-\eta / 2}$. So we get $$C-\sigma Z_{1-\eta / 2}\le X+\mu\le\sigma Z_{1-\eta / 2}-C.$$