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Suppose that X is a random variable for which \(E\left( X \right) = \mu \), and\(Var\left( X \right) = {\sigma ^2}\). Show that\(E\left( {X\left( {X - 1} \right)} \right) = \mu \left( {\mu - 1} \right) + {\sigma ^2}\).

Short Answer

Expert verified

\(E\left( {X\left( {X - 1} \right)} \right) = \mu \left( {\mu - 1} \right) + {\sigma ^2}\).

Step by step solution

01

Given information

X is a random variable for which \(E\left( X \right) = \mu \), and \(Var\left( X \right) = {\sigma ^2}\).

02

Show that \(E\left( {X\left( {X - 1} \right)} \right) = \mu \left( {\mu  - 1} \right) + {\sigma ^2}\)

\(\begin{aligned}{}E\left( {X\left( {X - 1} \right)} \right) = E\left( {{X^2} - X} \right)\\ = E\left( {{X^2}} \right) - E\left( X \right) \ldots \ldots \ldots \left( 1 \right)\end{aligned}\)

Now, we know that \(\begin{aligned}{}Var\left( X \right) &= E\left( {{X^2}} \right) - {\left( {E\left( X \right)} \right)^2}\\ \Rightarrow E\left( {{X^2}} \right) &= Var\left( X \right) + {\left( {E\left( X \right)} \right)^2}\end{aligned}\)

If we use the value of\(E\left( {{X^2}} \right)\)in equation (1), we get,

\(\begin{aligned}{}E\left( {X\left( {X - 1} \right)} \right) &= E\left( {{X^2}} \right) - E\left( X \right)\\ &= Var\left( X \right) + {\left( {E\left( X \right)} \right)^2} - E\left( X \right)\\ = {\sigma ^2} + {\mu ^2} - \mu \\ = \mu \left( {\mu - 1} \right) + {\sigma ^2}\end{aligned}\)

Hence, \(E\left( {X\left( {X - 1} \right)} \right) = \mu \left( {\mu - 1} \right) + {\sigma ^2}\).

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