Let X Be Standard Normal and Y = X²: Expectations and Covariance

Let $X\sim N(0,1)$ and define $Y=X^2$. This setup has a useful contrast: the conditional mean of $X$ given $Y$ is zero, and $X$ and $Y$ have covariance zero, even though $Y$ is determined by $X$ and so the variables are dependent.

Distribution and moments of $Y$

Squaring a standard normal random variable gives a chi-square random variable with one degree of freedom. Its density is

$$f_Y(y)=\frac{1}{\sqrt{2\pi y}}e^{-y/2},\qquad y>0.$$

The density is zero for $y\leq 0$. Since a standard normal variable has $\mathbb{E}[X^2]=1$ and $\mathbb{E}[X^4]=3$,

$$\mathbb{E}[Y]=1,\qquad \operatorname{Var}(Y)=\mathbb{E}[X^4]-\mathbb{E}[X^2]^2=3-1=2.$$

Conditional expectations

Because $Y=X^2$, knowing $X$ tells you $Y$ exactly:

$$\mathbb{E}[Y\mid X]=X^2.$$

In the other direction, knowing $Y=y$ tells you that $X$ is either $\sqrt y$ or $-\sqrt y$. The standard normal density is symmetric about zero, so these two possibilities have equal conditional probability. Thus, for $y>0$, the conditional distribution of $X$ given $Y=y$ puts probability $1/2$ on each of $\sqrt y$ and $-\sqrt y$. It follows that

$$\mathbb{E}[X\mid Y]=\frac{\sqrt Y+(-\sqrt Y)}{2}=0.$$

This is a statement about a conditional expectation, which is defined up to events of probability zero. Since $Y$ is continuous, the event $Y=y$ itself has probability zero for any single value $y$; the conditional distribution above is the usual regular conditional distribution.

The same reasoning also gives

$$\mathbb{E}[X^2\mid Y]=Y, \qquad \operatorname{Var}(X\mid Y)=Y.$$

More generally, for an integrable function $g$,

$$\mathbb{E}[g(X)\mid Y] =\frac{g(\sqrt Y)+g(-\sqrt Y)}{2}.$$

The conditional expectation keeps the even part of $g$ and cancels its odd part.

Covariance: zero, but not independence

By definition,

$$\begin{aligned} \operatorname{Cov}(X,Y) &=\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]\\ &=\mathbb{E}[X^3]-(0)(1)\\ &=0. \end{aligned}$$

The third moment is zero because the standard normal distribution is symmetric and $x^3$ is odd. Both variables have finite, nonzero variance, so their correlation is also zero.

Zero covariance does not imply independence in general. Here dependence is direct: once you know $X$, you know $Y=X^2$. For example, the event $\{|X|\leq 1\}$ is exactly the event $\{Y\leq 1\}$. Therefore

$$\mathbb{P}(Y\leq 1\mid |X|\leq 1)=1,$$

whereas $\mathbb{P}(Y\leq 1)<1$. These events are not independent, confirming that $X$ and $Y$ are dependent.

Another way to see that the dependence has not disappeared is to compare $Y$ with $X^2$: since they are the same random variable,

$$\operatorname{Cov}(X^2,Y)=\operatorname{Var}(Y)=2.$$

A concrete conditional example

Suppose you are told that $Y=4$. Then $X$ can only be $2$ or $-2$, and symmetry assigns probability $1/2$ to each. Hence

$$\mathbb{E}[X\mid Y=4]=\tfrac12(2)+\tfrac12(-2)=0, \qquad \mathbb{E}[X^2\mid Y=4]=4.$$

This illustrates why conditioning on the square does not identify the sign of $X$, even though it fixes its magnitude.

Common mistakes

• Claiming $X$ and $Y$ are independent because their covariance is zero. Zero covariance captures only linear association; it does not rule out a nonlinear relationship.
• Writing $\mathbb{E}[X\mid Y]=X$. Once you know $Y$, you know $|X|$, not its sign. Symmetry makes the conditional mean zero.
• Confusing $\mathbb{E}[Y\mid X]$ with $\mathbb{E}[Y]$. The conditional expectation is $X^2$, while the unconditional expectation is the number $1$.