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Chapter 7: Q.7.43 (page 362)

Show that for random variables X and Z.

E(X-Y)2=EX2-EY2

whereY=E[X∣Z]

Short Answer

Expert verified

X-E(X∣Z) is orthogonal to E(X∣Z).

Step by step solution

01

Given Information

Y=E[X∣Z]

02

Explanation

Write out the left side. We have that

E(X-Y)2=EX2+Y2-2XY

=EX2+EY2-2E(XY)

Now, what is E(XY)? Let's prove that it is equal to EY2.

EXY-Y2=E(Y(X-Y))

=E(E(X∣Z)·(X-E(X∣Z)))=0

03

Explanation

The expression above is equal to zero, since we know that the projection E(X∣Z)and the connection between the point Xand the projection E(X∣Z)are orthogonal. Therefore E(XY)=EY2which implies

E(X-Y)2=EX2-EY2.

04

Final Answer

X-E(X∣Z) is orthogonal to E(X∣Z).

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