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Apparently:

$$ E[X^2] = 1^2 \cdot \frac{1}{6} + 2^2 \cdot \frac{1}{6} + 3^2\cdot\frac{1}{6}+4^2\cdot\frac{1}{6}+5^2\cdot\frac{1}{6}+6^2\cdot\frac{1}{6} $$

where $X$ is the result of a die roll.

How come this expansion?

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  • $\begingroup$ This is called the second moment of $X$. Does the Wikipedia page help? Have a look at the Uses and application section. Particularly, it is useful for calculating the Variance. $\endgroup$ Jan 11, 2015 at 13:05
  • $\begingroup$ Kind of. So I would need to evaluate the moment generating function or so? Anyway, then for example evaluating $E[X^3]$ would require some work I guess. $\endgroup$
    – TMOTTM
    Jan 11, 2015 at 13:12
  • $\begingroup$ The expected value for discrete random variables is just the sum of the products of the outcome times its probability... $\endgroup$
    – mfnx
    May 11, 2022 at 22:36

2 Answers 2

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There are various ways to justify it.

For example, it follows from the definition of expectation and the law of the unconscious statistician.

Or consider the case $Y=X^2$ and computing $E(Y)$.

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Here's another way to compute $E[X^2]$.

If you know how to compute $E[X]$ and $Var(X)$ for a dice roll, then you can work out $E[X^2]$ using this equivalence of variance: $Var(X) = E[X^2] - (E[X])^2$.

While this is not a general answer (see @Glen_b), this equivalence comes in handy pretty often.

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