For a random variable $X$ we know that $Var(X) = E(X^2) - E^2(X)$. If we think of a measurable function $g(X)$, since we can calculate $E(g(X^2))$ and $E^2(g(X))$ would it also be true that $Var(g(X)) = E(g(X^2)) - E^2(g(X))$ for any $g$?

  • $\begingroup$ I assume X is a scalar rather than a vector. $\endgroup$ Dec 8, 2016 at 15:31

1 Answer 1


You can look at $Y=g(X)$ as another random variable and use the definition of the variance to obtain the following formula: $$Var(g(X))=E[g(X)^2]- E[g(X)]^2$$

Be careful your square was misplaced !

  • $\begingroup$ Was the question edited? It looks to me that the square is in a proper place for your equations to agree. $\endgroup$ Dec 8, 2016 at 15:32
  • 1
    $\begingroup$ @MichaelChernick I was refering to the square in $E[g(X^2)]$ $\endgroup$
    – RUser4512
    Dec 8, 2016 at 15:36

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