Given $Θ$ is an unknown, and $X$~$U(Θ-0.5,Θ+0.5)$
Is $[X-2,X+2]$ an 80% interval?
Apparently, yes it is:
$$P(X-2≤Θ≤X+2)=P(Θ-2≤X≤Θ+2)=1$$
The transition from $P(X-2≤Θ≤X+2)$ to $P(Θ-2≤X≤Θ+2)$ is straightforward.
After rearranging the inequality in terms of $Θ$, how does $P(Θ-2≤X≤Θ+2)=1$?
Is it okay to assume continuous uniform distribution?
I mean, $Θ$ can be any value between the intervals with infinite precision of decimal places.
Anyways, after applying the formula provided by wikipedia, this is what I get
$$P(Θ-2≤X≤Θ+2)=\frac{(Θ+2)-(Θ-2)}{(Θ+0.5)-(Θ-0.5)}=4≠1$$
Can anyone help me figure this out? Thanks!