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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

enter image description here

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

Marcus Müller
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AlfroJang80
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    stick a one in the “empty” spot because it there’s a one there –  Dec 01 '18 at 19:03
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    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 01 '18 at 19:30
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    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 01 '18 at 20:26

2 Answers2

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you factored $\theta$ as $\theta \times 1$ to pull it out of the summation leaving $$ \frac{1}{N} \sum_{i=0}^{N-1}\;=\frac{1}{N} \sum_{i=0}^{N-1} \underbrace{1}_{\text{implicit}}\,=1 $$

  • Wow. Didn't know that. I was thinking of what is it actually even summing. – AlfroJang80 Dec 01 '18 at 19:16
  • An implicit 1 in an empty summation is a convention I have never seen. So, I went searching for confirmation and have been unable find any. This leaves me a bit puzzled. Can any one cite anywhere else this convention is used or defined? – Cedron Dawg Dec 03 '18 at 15:36
  • pulling the theta out left a one. the empty summation is unorthodox but the sum is pretty self evident. are you really puzzled? its just manipulation of symbols that “fixed” the OP’s problem. people create notation all the time –  Dec 03 '18 at 20:19
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It should have been written as $$\frac{1}{N}\sum_{i=0}^{N-1}\theta=\frac{1}{N}\times N\,\theta=\theta$$

BlackMath
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    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 02 '18 at 15:19