Conditional Expectation of $X_2^2$ given $X_1$

Question

Consider two Normally distributed random variables, $X_1 \sim N(0,T_1)$ and $X_2 \sim N(0,T_2)$, such that $X_2-X_1 \sim N(0,T_2-T_1)$. How to calculate $E[X_2^2\mid X_1]$?

Answer 1

General forecasting rule

Suppose $X_1$ has mean $\mu_1$, $X_2$ has mean $\mu_2$, and that they are Gaussian with covariance matrix

$$ \left[\begin{matrix} \Omega_{11},\Omega_{12} \\ \Omega_{21},\Omega_{22} \end{matrix} \right]$$

then

$$ X_2∣X_1 \sim N(\mu_2+(\Omega_{21}\Omega_{11}^{-1}(X_1-\mu_1),\Omega_{22}-\Omega_{21}\Omega_{11}^{-1}\Omega_{12}) $$

Applying it to your problem

All we need to do is find $\mu_1$, $\mu_2$, $\Omega_{11}$, $\Omega_{12}$, $\Omega_{21}$, $\Omega_{22}$. First, notice that

\begin{align} X_1 & \sim N(0,T_1) \implies \mu_1=0 \;\; \text{ and } \;\; \Omega_{11}=T_1 \\ X_2 & \sim N(0,T_2) \implies \mu_2=0 \;\; \text{ and } \;\; \Omega_{22}=T_2 \end{align}

Next, assuming $X_1$ and $X_2$ are scalars, it follows that $\Omega_{12}=\Omega_{21}$. We know that $\Omega_{12}=E[(X_1-E[X_1])(X_2-E[X_2])]=E[X_1 X_2]$. Notice that

$$ E[(X_2-X_1)^2] = E[X_1^2+X_2^2-2X_1 X_2] $$

and, since $X_2-X_1 \sim N(0,T_2-T1)$ the left hand side of this equation is equal to $T_2-T_1$. On the other hand, notice that $E[X_1^2]=T_1$ and $E[X_2^2]=T_2$ imply that the right hand side equals $T_1+T_2-2E[X_1 X_2]$. Hence,

$$ T_2-T_1 = T_1+T_2-2E[X_1 X_2] $$

which imples

$$ \Omega_{12}=E[X_1 X_2] = T_1 $$

Plugging this into the forecast equation leads to

$$ X_2∣X_1∼N(X_1,T_2-T_1) $$

which, in particular, implies that

$$ E[(X_2-X_1)^2\mid X_1]=T_2-T_1 $$

$$ E[X_2^2+X_1^2-2X_1X_2\mid X_1]=T_2-T_1 $$

$$ E[X_2^2\mid X_1]+X_1^2-2X_1E[X_2\mid X_1]=T_2-T_1 $$

and, using the fact that $E[X_2\mid X_1]= X_1$, it follows that $$ E[X_2^2\mid X_1]= X_1^2+T_2-T_1 $$