1.12 Properties of \(E(X)\)

1.

If \(X\geq 0\), then \(E(X)> 0\) and \(E(X)=0\implies P(X=0)=1\).

2.

\(E(X+a)=E(X)+a\), where \(a\) is a constant.

Proof. \begin {align*} E(X+a) &=\frac {1}{n}\sum (X+a)=\frac {1}{n}\Bigg [\sum X+\sum a\Bigg ]\\ &=\frac {1}{n}\sum X +\frac {1}{n}(na)\\ &=E(X)+a. \qedhere \end {align*} □

Adding a constant to every observation shifts the mean by that constant and nothing else – which is why an assumed mean may be subtracted before computing and added back afterwards. If \(C\) is a constant then \(E(CX)=CE(X)\) since \(E(C)=C\).

3.

For the random variable \(X\) and \(Y\) then \(E(X+Y)=E(X)+E(Y)\) property 3 and 4 shows the important property that the operation \(E(.)\) is a linear operator.

4.

\(E(g(X))=\sum \limits _1^ng(x)P(X=x)\)

5.

\(E(XY)=E(X)E(Y)\) if and only if \(X\) and \(Y\) are independent random variables.

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