Problem 12.15 Easy Free

Gaussian Share of a Sum

Let X1∼N(0,9)X_1\sim\mathcal{N}(0,9) and X2∼N(0,16)X_2\sim\mathcal{N}(0,16) be independent. Given that X1+X2=5X_1+X_2=5, find E[X1∣X1+X2=5]\mathbb{E}[X_1\mid X_1+X_2=5].

Work it out, then compare with the answer.

SolutionShow the worked solution

Key idea. Subtract from X1X_1 the multiple of the sum that is uncorrelated with the sum. For jointly normal variables, uncorrelated means independent.

Let S=X1+X2S=X_1+X_2. The pair (X1,S)(X_1,S) is a linear image of the independent normals (X1,X2)(X_1,X_2), so it is jointly normal. We have

Var⁡(S)=9+16=25,Cov⁡(X1,S)=Var⁡(X1)=9.\operatorname{Var}(S)=9+16=25,\qquad \operatorname{Cov}(X_1,S)=\operatorname{Var}(X_1)=9 .

Put R=X1−925SR=X_1-\tfrac{9}{25}S. Then

Cov⁡(R,S)=9−925⋅25=0.\operatorname{Cov}(R,S)=9-\tfrac{9}{25}\cdot 25=0 .

Since (R,S)(R,S) is jointly normal and uncorrelated, RR is independent of SS. Also E[R]=0\mathbb{E}[R]=0. Therefore

E[X1∣S]=925S+E[R]=925S.\mathbb{E}[X_1\mid S]=\tfrac{9}{25}S+\mathbb{E}[R]=\tfrac{9}{25}S .

So each summand takes a share of the sum proportional to its variance.

At S=5S=5 this gives 925⋅5=95\tfrac{9}{25}\cdot 5=\tfrac95.

The conditional variance is Var⁡(R)=9−8125=14425\operatorname{Var}(R)=9-\tfrac{81}{25}=\tfrac{144}{25}. Hence

X1∣S=5 ∼ N ⁣(95,14425).X_1\mid S=5\ \sim\ \mathcal{N}\!\left(\tfrac95,\tfrac{144}{25}\right).
E[X1∣X1+X2=5]=95\boxed{\mathbb{E}[X_1\mid X_1+X_2=5]=\tfrac{9}{5}}