# Bivariate conditional probability confidence

I have two random variables following normal distributions, $X\sim N(\mu_{x},\sigma_{x})$ and $Y\sim N(\mu_{y},\sigma_{y})$

I know the covariance is $\operatorname{Cov}(X,Y) = c$.

And I believe (correct me if I'm wrong) that I can estimate the expected value of $Y$ given $X=\tau$ as follows:

$E(Y|X=\tau) = \rho \frac{\sigma_{y}}{\sigma_{x}}(\tau -\mu_{x})+\mu_{y}$

where $\rho = c/(\sigma_{x}\sigma_{y})$.

I believe this is a decent estimate. But now I want some way of estimating our "certainty" that our calculated value is correct. Obviously for $|\rho| \sim 1$ we can be very confident. For $\rho \sim 0.5$ a little bit confident. And for $\rho \sim 0$ I suppose our confidence isn't changed at all. i.e. since no correlation exists we have not changed any ideas about how confident we are about our expected $Y$ value.

How can I express this numerically?

• If the parameters you gave (means, sigmas, c and correlation) are known population parameters, then the estimate for the expected value of Y given X is exact - it is not an estimate, it's the true expected value. What are not exact are the occurrences of variable Y, that can be more or less spread out according to correlation, as stated in Dilip Sarwate's answer. – Pere Jan 20 '17 at 10:12

One measure of how "accurate" the estimate is is the variance (often called the residual variance) of the estimate which in this instance is $\sigma_y^2(1-\rho^2)$: the closer the residual variance to $\sigma_y^2$, the weaker our confidence that the estimate is significantly better than discarding the observed value $\tau$ of $X$ and just using $\mu_y$ as the estimate of $Y$. So our confidence in accuracy is a function of $\rho^2$, not $\rho$, which is why $\rho=0.5$ does not make us feel very confident, and even $\rho = \frac{1}{\sqrt{2}}\approx 0.707\ldots$ feels more like an even bet than a strong vote of confidence.
• I think that is a fair measure, but what about $\rho=0$? Then we have to say we have no confidence, which is a little unfair, because actually our confidence hasn't changed. – Solitude Jun 15 '13 at 14:39
• If $\rho = 0$, we are sure that our estimate $E(Y|X=\tau) = \rho \frac{\sigma_{y}}{\sigma_{x}}(\tau -\mu_{x})+\mu_{y}$ is the same as if we had never observed that $X$ has value $\tau$, and we have no confidence that the estimate is any better. – Dilip Sarwate Jun 15 '13 at 14:45