Say I have two independent random variables, $X$ and $Y$. If I give you a 90% confidence interval on each $(x_1,x_2)$ and $(y_1,y_2)$, i.e. $P(X\in(x_1,x_2))=.9$ and $P(Y\in(y_1,y_2))=.9$. If I asked you for a 90% confidence interval on their ratio $Z=X/Y$, what would you do? My first step was to just start with the interval $(x_1/y_2,x_2/y_1)$. However, I would imagine that this is only an 81% confidence interval and thus my full answer is just that I know there exist 90% confidence which properly contains the interval $(x_1/y_2,x_2/y_1)$.
However there are many things wrong with this.
Case 1: $(x_1/y_2,x_2/y_1)$ may actually be a 90% confidence interval. If we let $X$ and $Y$ be i.i.d. uniform on $(0,1)$ and I pick $x_1,y_1=.1$ and $x_2,y_2=1$ I get the interval $(.1,10)$ which is a 90% confidence interval for $Z$. That is, somehow I got a exact interval even though there should only be an $.81$ probability that $X$ and $Y$ are in $(.1,1)$. However, this is where I realized that it's not an 81% confidence interval, because in the .01 case that both $X$ and $Y$ are not in the $(.1,1)$, there are non-negligible ways to produce some of the ratios in $(.1,10)$, however, this shouldn't account for nearly as much as it does to make the interval $(.1,10)$ a 90% confidence interval.
Case 2: Same thing, but I pick $x_1,y_1=0$ and $x_2,y_2=.9$, and thus using my flawed method get the confidence interval to be $(0,\infty)$ which is obviously not a 90% confidence interval for $Z$ but a 100% one. How could it possible be the case now that I actually have to shrink my computed interval to get a 90% confidence interval.
Case 3: Same thing but I pick the more innocuous $x_1,y_1=0.05$ and $x_2,y_2=.95$. In this case I get a 94.71% confidence interval.
I don't get it, what is making this so counterintuitive for me.