Important results
Covariance properties​
Cov(X,a)Cov(X,X)Cov(X,Y)Cov(aX,bY)Cov(X+a,Y+b)Cov(aX+bY,cW+dV)​=0=Var(X)=Cov(Y,X)=abCov(X,Y)=Cov(X,Y)=acCov(X,W)+adCov(X,V)+bcCov(Y,W)+bdCov(Y,V)​
Mean independence and Covariance​
Prove: If X and U are mean independent and E[U]=0 then Cov(X,U)=0.
Note: Mean independence is defined as E[U∣X]=E[U].
Proof:
Cov(X,U)​=E[(X−E[X])(U−E[U])]=E[XU−XE[U]−E[X]U+E[X]E[U]]=E[XU]−E[X]E[U]−E[X]E[U]+E[X]E[U]=E[XU]−E[X]E[U]=E[XU]−E[X]E[E[U∣X]]=E[XU]−E[X]E[0]=E[XU]=E[E[XU∣X]]=E[XE[U∣X]]=E[X⋅0]=0​
Unbiased estimator of Variance​
Let the population variance and mean are σ2 and μ. The sample variance is given by:
S2=n−11​i=1∑n​(Xi​−Xˉ)2
To prove the unbiasedness of S2 we have to prove that E[S2]=σ2.
S2=n−11​i=1∑n​(Xi2​+Xˉ2−2Xi​Xˉ)
Taking expectation both sides:
E[S2]​=E[n−11​i=1∑n​(Xi2​+Xˉ2−2Xi​Xˉ)]=n−11​E[i=1∑n​(Xi2​)+i=1∑n​(Xˉ2)−2Xˉi=1∑n​(Xi​)]=n−11​E[i=1∑n​(Xi2​)+n(Xˉ2)−2Xˉn(Xˉ)]=n−11​E[i=1∑n​(Xi2​)−n(Xˉ2)]=n−11​[i=1∑n​E[Xi2​]−nE[Xˉ2]] as E operator is linear.​
We know that Var[Xi​]=E[Xi2​]−(E[Xi​])2, that is