Best Prediction of \(X\) given \(X+Y=n\)
Suppose \(X\) and \(Y\) are independent normal random variables. What is the best guess for \(X\) given that \(X+Y=n\)?
Answer
Since \(X\) and \(Y\) are normal and they are independent, it follows that they are jointly normal.
Furthermore, \((X,\;X+Y)\) is a linear transformation of \((X,Y)\), hence it is also jointly normal.
For jointly normal random variables, we know that the conditional distribution
\[
Y \mid (X=x)
\sim
\mathcal{N}
\!\left(
\underbrace{\mu_Y - \frac{\operatorname{Cov}(X,Y)}{\operatorname{Var}(X)}\,\mu_X}_{\beta_0}
\;+\;
\underbrace{\frac{\operatorname{Cov}(X,Y)}{\operatorname{Var}(X)}}_{\beta_1} x,
\;
\sigma_Y^2
-
\frac{\operatorname{Cov}(X,Y)^2}{\operatorname{Var}(X)}
\right).
\]
Furthermore, the best prediction (w.r.t. MSE) of \(X\) given \(X+Y=n\) is
\[
\mathbb{E}[X \mid X+Y=n].
\]
Since \(X\) and \(Y\) are independent,
\[
\operatorname{Cov}(X,X+Y)=\operatorname{Var}(X)=\sigma_X^2,
\]
and
\[
\operatorname{Var}(X+Y)=\sigma_X^2+\sigma_Y^2.
\]
Therefore,
\[
\mathbb{E}[X \mid X+Y=n]
=
\underbrace{
\mu_X
-
\frac{\sigma_X^2}{\sigma_X^2+\sigma_Y^2}
(\mu_X+\mu_Y)
}_{\beta_0}
\;+\;
\underbrace{
\frac{\sigma_X^2}{\sigma_X^2+\sigma_Y^2}
}_{\beta_1}
\, n.
\]