Express a Probability through $\Phi(t)$

Question

Using Poisson i.i.d. random variables and having $Z$ as standard Gaussian r.v. $~N(0,1)$, I am struggling to express $P(|Z|≤t)$ in terms of $Φ(r)=P(Z≤r)$ for $t>0$.

I understand that the absolute value of Z means non-negativity and therefore the range is bound by 0 and t, but this doesn't seem to help.

My guess was $P(|Z|≤t) = 2*\Phi(t)$, but this is incorrect.

I assume the solution is simple, but I cannot get there. Any input appreciated.

Answer 1

Note, first of all, that this question doesn't have anything to do with Poisson variables.

Anyway, for $t>0$:

$$ P(|Z|<t)=P(Z<t,Z>-t)=P(Z<t)-P(Z<-t)=\Phi(t)-\Phi(-t) $$ If you want to get even more clever about it, you might note that $$P(Z<-t)=P(-Z>t)=1-P(-Z<t)=1-\Phi(t),$$ since $-Z$ is also standard Gaussian.

Hence, it is also true that $$ P(|Z|<t)=2\Phi(t)-1 $$