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I'm looking over Binomial random variables and I understand that

$ \sum\limits_{k=0}^n k\binom{n}{k} p^k (1-p)^{n-k} = np $ from $\mathrm{Bin}(n,p)$

However, I don't understand how, if $S_n = u^{2T-n}$, why we get

$E[S_n] = \sum\limits_{k=0}^n u^{2k-n}\binom{n}{k} p^k (1-p)^{n-k}$

Thank You

user131983
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    Read about the law of the unconscious statistician. You can find this on Wikipedia if you like. – Dilip Sarwate Sep 04 '14 at 02:38
  • @DilipSarwate Thank You! I understand how we obtained the expression for $E[S_n]$. However, I am less clear as to why the first term of the expression is $u^{2k-n}$ rather than $u^{2T-n}$. Thanks again. – user131983 Sep 04 '14 at 12:24
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    Re-examine your understanding of the law of the unconscious statistician (LOTUS). For a discrete random variable $X$ taking on values $a_1, a_2, \ldots$ with probabilities $p_1, p_2\ldots$ respectively, LOTUS says $$E[g(X)] = \sum_i g(a_i)\cdot p_i.$$ Where is there an $X$ on the right side? What did it get replaced by? – Dilip Sarwate Sep 04 '14 at 12:56
  • @DilipSarwate Thank You! That was very silly of me. – user131983 Sep 04 '14 at 13:11

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