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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.

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this is not a paradox since special cases you've described does not contradict with a general case: all probabilities are above $0.81$; "in general" means that something always hold, but of course in each particular case you can have better result. – Ilya May 25 '12 at 10:59
I'm just wondering what was wrong with the first paragraphs logic that asserts there exists a 90% confidence interval $(i,j)$ that properly contains $(x_1/y_2,x_2/y_1)$. This obviously isn't true if the the interval itself has confidence greater than 90% already. – Steven-Owen May 25 '12 at 11:03
If your question is what's wrong with the first paragraph's logic, you'll have to first state that logic. Currently all it says is "However, I would imagine", and all we can say about that is, "Well, you're imagining incorrectly". – joriki May 25 '12 at 11:23
up vote 1 down vote accepted

It is true (enough) that you have 81% confidence that $X \in (x_1,x_2)$ and $Y \in (y_1,y_2)$. It is true (assuming all positive numbers) that if $X \in (x_1,x_2)$ and $Y \in (y_1,y_2)$, then $X/Y \in (x_1/y_2, x_2/y_1)$.

It is not generally true that if $X/Y \in (x_1/y_2, x_2/y_1)$, then $X \in (x_1,x_2)$ and $Y \in (y_1,y_2)$.

Basically, you only counted some of the ways in which you could have $X/Y \in (x_1/y_2, x_2/y_1)$, and consequently, all you have shown is that 81% is a lower bound on the probability of it happening.

share|cite|improve this answer
Is that the best lower bound? – Steven-Owen May 25 '12 at 20:06
It shouldn't be hard to construct sample spaces where there are no other ways for $X/Y$ to be in that interval. e.g. the only outcomes of $X$ are in $(x_1, x_2)$ and some other interval $I$ of very large numbers where as $Y$ is limited to $(y_1,y_2)$ and some other interval $J$ of very tiny numbers. If you don't like the disjointness, you should be able to still get arbitrarily close to 81%. – Hurkyl May 25 '12 at 23:03

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