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Is it possible to write $\sum_{i=1}^k y_i\log p_i +(n_i-y_i)\log(1-p_i)+\log \binom{n_i}{y_i}$ as one mathematician said it is correct but another said that one should write $\sum_{i=1}^k\left ( y_i\log p_i +(n_i-y_i)\log(1-p_i)+\log \binom{n_i}{y_i} \right )$

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To expand on @vonbrand’s answer, the fact that the index $i$ appears in all three terms almost guarantees that they all belong to the summation, but you shouldn’t make the reader work that hard: it’s much better to enclose the entire summand in parentheses, so that it’s immediately obvious what is being summed. –  Brian M. Scott Apr 19 '13 at 9:46

3 Answers 3

up vote 5 down vote accepted

If in doubt, add parentesis. The expression under the sum is too large in this case, so I'd add them to make clear what is being summed.

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Expressions like $$\tag?\sum_{i=1}^n a_i + 1$$ have two possible interpretations: $$\tag1\sum_{i=1}^n (a_i + 1)$$ or $$\tag2\left(\sum_{i=1}^n a_i\right) + 1.$$ Among others because of the uglyness of $(2)$ it is customary to interprete $(?)$ as $(2)$ and use explicit parentheses if one wants to have $(1)$. Note however, that no parentheses are necessary/customary for $$\sum_{i=1}^n a_i\cdot 2=\sum_{i=1}^n (a_i\cdot 2)= \left(\sum_{i=1}^n a_i\right)\cdot 2=2\sum_{i=1}^n a_i$$

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$1+\sum\limits_{i=1}^n a_i$ is clearer than $\sum\limits_{i=1}^n a_i+1$ with the same number of parentheses. –  Jonas Meyer Aug 8 at 2:18

The only thing in $$ \sum_{i=1}^k y_i\log p_i +(n_i-y_i)\log(1-p_i)+\log \binom{n_i}{y_i} $$ that tells us that the summation should involve all terms is that all terms have an index $_i$ on them. For example, the following example is much unclearer: $$ \sum_{i=1}^n a_i+b $$ which could either mean $$ \left(\sum_{i=1}^n a_i\right)+b\quad\text{or}\quad\sum_{i=1}^n \left(a_i+b\right)=\left(\sum_{i=1}^n a_i\right)+nb, $$ which is two completely different things. So, it's a good habit to include parentheses as @vonbrand also answers.

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