Understanding how $\rho=0$ implies independence in bivariate normal distribution (intuition) I understand that if $\rho=0$ we can factor the joint pdf of a bivariate normal distribution into the product of the marginal pdfs of $X$ and $Y$ (where $X$ and $Y$ are standard normal random variables). And that this implies that $X$ and $Y$ are independent. 
However, I don't understand how (intuitively) knowing a given value of $X$ tells us nothing about the value of $Y$.
For instance, if one looks at the level curves they are circular:

So wouldn't knowing that $X$ is an extreme value limit the possibilities of $Y$? For instance, I can clearly see how two random variables $X'$ and $Y'$ would be indepdent if $1 \geq x \geq 0$ and $1 \geq y \geq 0$. Since knowing that $X'=.5$ tells us nothing and $Y'\in [0,1]$ with equal likelihood still. Is there an easy way to see this independence geometrically in the case of the bivariate normal distribution?
 A: I'm not sure if this will help, but very roughly, the fact that the curves are circular says that there are three ways to get a particular value of the joint pdf:


*

*Likely value of $X$, unlikely value of $Y$;

*Moderately likely values of both $X$ and $Y$;

*Unlikely value of $X$, likely value of $Y$.


An extreme value of $X$, therefore, would restrict the possible values of $Y$ if you knew a priori that your result was inside a particular level curve. But you don't actually know this a priori; it's less likely to get extreme values of $X$ and $Y$ at once than it is to get only one, but one doesn't affect the other.
Here's another example which should prime your intuition: Imagine flipping a coin 10 times; $X$ is the number of heads in the first 5 flips, and $Y$ the number of heads in the second 5. It's a lot less likely that $X = Y = 5$ than that, say, $X = 5$ and $Y = 2$, but this is purely because $Y = 5$ is unlikely in the first place. It doesn't mean that a lot of initial heads make later heads less likely.
A: You shouldn't be looking at the level curves. Look instead at slices: when you learn that $X = a$ you've now restricted your attention to the slice $X = a$ of your plot. All of these slices look the same, namely they are all Gaussians. It's true that as $|a|$ gets bigger these slices get smaller, but that just means that the prior probability of $X = a$ gets smaller; once you condition on $X = a$ this doesn't matter anymore, and the shapes get rescaled to be exactly the same (namely, Gaussians with the same mean and variance).
(Ignoring for now the subtleties involved in conditioning on an event of zero probability; you can instead condition on the event $X \in [a - \epsilon, a + \epsilon]$ for small $\epsilon$.) 
