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LetXbe a random variable for which \(E\left( X \right) = \mu \), and\(Var\left( X \right) = {\sigma ^2}\), and let c be an arbitrary constant. Show that \(E\left( {{{\left( {X - c} \right)}^2}} \right) = {\left( {\mu - c} \right)^2} + {\sigma ^2}\).

Short Answer

Expert verified

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

Step by step solution

01

Given information

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

02

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

\(\begin{aligned}{c}E\left\{ {{{\left( {X - c} \right)}^2}} \right\} &= E\left( {{X^2} - 2Xc + {c^2}} \right)\\ &= E\left( {{X^2}} \right) - 2cE\left( X \right) + {c^2} \ldots \ldots \ldots \left( 1 \right)\end{aligned}\)

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

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

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

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

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Most popular questions from this chapter

Consider a utility function U for which\({\bf{U}}\left( {\bf{0}} \right){\bf{ = 5}}\), \({\bf{U}}\left( {\bf{1}} \right){\bf{ = 8}}\), and \({\bf{U}}\left( {\bf{2}} \right){\bf{ = 10}}\). Suppose that a person who has this utility function is indifferent to either of two gambles X and Y, for which the probability distributions of the gains are as follows:

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