Double sum subtraction with one missing term

Question

If I have some sum and subtract another why is the following relation true?

$$\sum_{h}^{n} \sum_{i}^{n} f(h,i) - \sum_{h \neq \theta}^{n} \sum_{i \neq \theta}^{n} f(h,i) = \sum_{i}^{n} f(\theta,i) $$

Answer 1

Let $[n] = \{1, 2, \ldots, n\}, \theta \in [n]$ and $N_\theta = [n] \setminus \{\theta\}$.

I am assuming

• $\sum_h^n$ is really $\sum_{h=1}^n$ and
• $\sum_{h \ne \theta}^n$ is really $\sum_{h \in N_\theta}$

Then,

• the first sum on the left includes all indices $(h,i)$ where $h,i \in [n]$
• the second sum on the left includes all indices $(h,i)$ where $h,i \in N_\theta$

The difference includes all terms where $h=\theta$ or $i = \theta$, i.e. $$ \sum_{h\ne \theta} f(h, \theta) + \sum_{i\ne \theta} f(\theta, i) + f(\theta,\theta) = \sum_{h\ne \theta} f(h, \theta) + \sum_i^n f(\theta, i) $$

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A slightly modified version of your expression would be true: $$\sum_{h}^{n} \sum_{i}^{n} f(h,i) - \sum_{h \neq \theta}^{n} \sum_{i}^n f(h,i) = \sum_{i}^{n} f(\theta,i) $$