In a bivariate normal distribution, what does conditioning on X = x do to the distribution of Y?

## The bivariate normal A pair (X, Y) is **jointly normal** (bivariate normal) when every linear combination `aX + bY` is univariate normal. It is parameterised by five numbers: the two means `mu_X, mu_Y`, the two standard deviations `sd_X, sd_Y`, and the correlation `rho` in (-1, 1). The level sets of its density are **ellipses** centred at `(mu_X, mu_Y)`, tilted upward when `rho > 0`, downward when `rho < 0`, and axis-aligned when `rho = 0`. The larger `|rho|`, the more the ellipses flatten into a cigar around a line; as `|rho|` approaches 1 the distribution degenerates onto that line. ## The conditional distribution The key structural fact: `Y | X = x ~ Normal( mu_Y + rho*(sd_Y/sd_X)*(x - mu_X), sd_Y^2 * (1 - rho^2) )` Three things to unpack. **It stays normal.** Slicing the joint density along a vertical line and renormalising gives another normal density. The family is closed under conditioning, which is a large part of why it is so tractable. **The conditional mean is linear in x.** Write it in standardised form: if `z = (x - mu_X)/sd_X` is how many standard deviations x sits above its mean, then the conditional mean of Y sits `rho * z` standard deviations above `mu_Y`. Because `|rho| < 1`, the prediction is always pulled *less far* from the centre than the conditioning value was — the shrinkage is proportional to `rho`. At `rho = 0.6`, a two-standard-deviation X corresponds to an expected Y only 1.2 standard deviations out. **The conditional variance does not depend on x.** `sd_Y^2 * (1 - rho^2)` is the same number for every conditioning value, so the scatter around the line is homoscedastic by construction. It is strictly less than the marginal variance `sd_Y^2` whenever `rho` is nonzero: learning X removes a fraction `rho^2` of the variance of Y and leaves the fraction `1 - rho^2`. At `rho = 0.8`, conditioning removes 64 percent of the variance; the residual standard deviation is `sd_Y * sqrt(1 - 0.64) = 0.6 * sd_Y`. ## Why zero correlation implies independence here Set `rho = 0` in the conditional above: the mean becomes `mu_Y` and the variance becomes `sd_Y^2`, i.e. the conditional distribution of Y is identical to its marginal for **every** x. A conditional that never changes with the conditioning value is precisely independence. Equivalently, the joint density factorises into the two marginal normal densities when `rho = 0`. This is a property of the **joint** family, not of the marginals. It is entirely possible for both marginals to be normal, for the covariance to be zero, and for the variables to be dependent — in which case the pair simply is not bivariate normal. Concretely: let `X ~ N(0,1)`, let S take the values +1 and -1 with probability 1/2 each, independently of X, and set `Y = S*X`. Then Y is also `N(0,1)`, and `Cov(X,Y) = E[S*X^2] = E[S]*E[X^2] = 0`. But `|X| = |Y|` always, so the two are strongly dependent. The pair fails joint normality, visible in the fact that `X + Y` equals either `0` or `2X` and so is not normal. ## The elliptical picture Everything above is readable off the contour plot. Draw an ellipse tilted upward. Slice it vertically at `x`: the chord you get is centred on the line `mu_Y + rho*(sd_Y/sd_X)*(x - mu_X)`, and — this is the part people find surprising — the *shape* of the density along that chord is the same everywhere, only its centre moves. That line of conditional means is not the major axis of the ellipse; it is shallower, which is the geometric statement of the shrinkage by `rho`. ## What this buys you in an interview 1. **A clean mental model of linear prediction.** The optimal predictor of Y from X under joint normality is exactly this linear conditional mean, and the leftover variance is `sd_Y^2*(1 - rho^2)`. This is where the interpretation of `rho^2` as the fraction of variance explained comes from at the population level. 2. **The regression-to-the-mean phenomenon.** Because `|rho| < 1`, extreme X values are paired with less extreme expected Y values. This is a mathematical necessity, not a causal effect, and it explains a large class of apparent "the best performers decline" observations. 3. **A precise statement of the exception.** When someone claims uncorrelated implies independent, the correct reply names the family: true under joint normality, false in general, and normal marginals alone are not sufficient. ## Cautions Joint normality is an assumption, and a strong one. Real bivariate data often has heavier joint tails than an ellipse allows: two variables can look mildly correlated in the bulk and move together violently in the extremes. Nothing in the bivariate normal can represent that — its tail dependence is zero for any `rho < 1`. Checking each variable's marginal for normality does not verify the joint assumption, precisely because normal marginals do not imply joint normality.