# Summing over constants [closed]

I'm following some notes and I came across an example which said that for this x_n with a proposed estimator of the mean, that it is an unbiased estimator for the DC level.

I'm trying to do it out myself

But as you can see, I'm ending up with an empty sum. I know the answer should be theta.

## closed as off-topic by Marcus Müller, MBaz, lennon310, Peter K.♦Dec 4 '18 at 15:32

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• stick a one in the “empty” spot because it there’s a one there – Stanley Pawlukiewicz Dec 1 '18 at 19:03
• um, this is really basic math: your sum symbol in $\frac1N\sum\theta$ is just a shorthand for $\frac1N \left(\theta + \theta + \ldots\right)$. So, you can "drag out" the common factor of $\theta$ from all the elements in that sum. So, what's $\frac\theta\theta$? Right, it's 1. Really, if this is news to you, practice a bit of working with sums; you'll need that more often if you're already doing estimator theory (did you skip some classes?). – Marcus Müller Dec 1 '18 at 19:30
• I'm voting to close this question as off-topic because it's not really a signal processing question, but a school-level math question. – Marcus Müller Dec 1 '18 at 20:26

you factored $$\theta$$ as $$\theta \times 1$$ to pull it out of the summation leaving $$\frac{1}{N} \sum_{i=0}^{N-1}\;=\frac{1}{N} \sum_{i=0}^{N-1} \underbrace{1}_{\text{implicit}}\,=1$$
It should have been written as $$\frac{1}{N}\sum_{i=0}^{N-1}\theta=\frac{1}{N}\times N\,\theta=\theta$$
• Don't you think this fits the OP's approach better? $$\frac{1}{N}\sum_{i=0}^{N-1}\theta= \theta \cdot \left( \frac{1}{N} \sum_{i=0}^{N-1} 1 \right) = \theta \cdot \left( \frac{1}{N} \cdot N \right) = \theta \cdot 1 = \theta$$ – Cedron Dawg Dec 2 '18 at 15:19