I am reading inference statistics by casella and berger. They are deriving the general formula for the probability distribution like that:

$$EX=\sum_{x=1}^{N}xP(X=x|N)=\sum_{x=1}^{N}x\frac{1}{N}=\frac{N+1}{2} $$

I do not get the step before the result. I would appreciate your explanation on this step!

  • $\begingroup$ The last equality? $\endgroup$ Jul 16, 2014 at 6:49
  • $\begingroup$ @StefanHansen I mean, how to get the $\frac {N+1} {2}$? $\endgroup$
    – Carol.Kar
    Jul 16, 2014 at 7:35
  • 2
    $\begingroup$ This follows from the fact that the sum of the first $N$ integers is $N(N+1)/2$. This can easily be shown using induction in $N$. $\endgroup$ Jul 16, 2014 at 7:50
  • $\begingroup$ Your question should specify which particular distribution is under discussion. The result is not general, but specific. You should also give the correct title of the book, capitalize it, and capitalize the names of the authors. $\endgroup$
    – Glen_b
    Jul 16, 2014 at 23:59

1 Answer 1


I have not read the article/book but the last part of the formula can be derived as follows:

\begin{align*} &\displaystyle \sum_{x = 1}^N x \frac{1}{N} = \frac{1}{N}\displaystyle \sum_{x = 1}^N x = \frac{1}{N}\left( 1 + 2 + \cdots+N\right) = \frac{1}{N} \left[ \frac{N}{2} \left(1+N\right)\right]\\ \Rightarrow &\displaystyle \sum_{x = 1}^N x \frac{1}{N} =\frac{N+1}{2} \end{align*}

The derivation for the arithmetic series formula here can help.


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