What's the probability distribution of the minimum of dependent variables?

Question

I have $n$ variables defined by $$ Y_j = |X-c_j| \ \ \ \ (j=1,...,n) $$ where $c_j$ are constants and $X$ is a gaussian variable (X ~ $N(\mu,\sigma^2$)). What would be $$ P(\min_j \ Y_j > 1) \ ? $$ The minimum is on dependent variables, how could I get rid of it? Can I just work on the joint variable to isolate $X$ ?

Answer 1

$\min Y_j >1$ iff $|X-c_j| >1$ for each $j \in \{1,2,...,n\}$. This is true iff $X> \max_j c_j +1$ or $X < \min_j c_j -1$. So the answer is $\Phi (\mu + \sigma (\min_j c_j -1))+1-\Phi (\mu + \sigma (\max_j c_j +1))$.

Answer 2

$$ P(\{\min_j Y_j > 1\}) = P(\{\text{all} \, Y_j > 1\}) $$

Now this is easy to calculate (just a product) in case $Y_j$ is independent. Else if dependent but Normal you need the covariance matrix and generally for this case the evaluation will be of a $N(O,I)$ random variable over a non simple region. To evaluate this for any covariance matrix $\Sigma$ you will need numerical analysis see https://stackoverflow.com/questions/28946719/finding-probability-in-multivariate-normal-distribution