I was looking at the example from "Statistical Decision Theory and Bayesian Analysis" by Jim Berger (page 10) -- below.

And I had trouble understanding how the expression underlined by the upper redline gets expanded to the lower underlined expression.

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    $\begingroup$ There are only two tricks in mathematics: adding zero or multiplying by one. This is an application of the first. $\endgroup$ Jun 22 '17 at 0:09
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    $\begingroup$ I think Matthew is suggesting adding and subtracting c$\theta$ inside the brackets in the first underlined quantity and expanding out the square. $\endgroup$ Jun 22 '17 at 0:22
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    $\begingroup$ Note that the terms are reordered before squaring. $\endgroup$ Jun 22 '17 at 0:37
  • $\begingroup$ You don't have to obtain the result this way (although interpreting the terms as you go through Berger's derivation can provide some insight). From the definition of the variance of any random variable $Y$ you know that $E(Y^2)=\operatorname{Var}(Y)+E(Y)^2$. Obviously $Y=\theta-cX$ has a distribution with mean $\theta-c\theta=(1-c)\theta$ and variance $(-c)^2(1)=c^2$. This lets you write down the result without any further thought as $$E((\theta-cX)^2)=c^2+((1-c)\theta)^2,$$which obviously is equivalent to Berger's expression. $\endgroup$
    – whuber
    Jun 22 '17 at 20:47

This is the old mathematician's trick of adding zero then rearranging.

$$ \theta - c X = \theta - \underbrace{c \theta + c \theta}_{= 0}- c X = c(\theta - X) + (1 - c) \theta $$


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