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enter image description here

I'm having some trouble understanding this slide. It's mentioned in the context of gaussian distributions.

I sort of understand the Dirac delta "function". The main difficulty I'm having is with the summation over the delta functions.

1) How does it replace the probability distribution p(x)? Is it not just a scalar?

2) How do we go from the left equation to the one on the right?

Tim
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Anonymous
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  • I really loved the way you pasted the slide... no latex, no code, no ascii... pure and simple whiteboard, the old way always win.... There is a mistake on the slide. I will write it and reply soon.... – Brethlosze May 27 '15 at 17:44
  • I recall answering almost the same question at http://stats.stackexchange.com/a/73626/919. Would that perhaps answer yours? Since you exhibit six equations, exactly which one are you asking about in (2)? – whuber May 27 '15 at 17:55
  • Well. that question is a good approach.... – Brethlosze May 27 '15 at 18:31
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    @hypfco this is **not** the recommended way of posting equations: (a) it makes them impossible to search, (b) they are totally unreadable for visually impaired people. In case of $\TeX$ it is the opposite for both cases. – Tim Nov 21 '16 at 08:46
  • @Tim. The visually impaired will have not problem to see a computer image. The minded impaired, will never see it even in the most beautiful math system available – Brethlosze Nov 23 '16 at 12:45
  • @hypfco so you are saying that blind people are able to read equations provided as images of handwriting..? – Tim Nov 23 '16 at 12:47
  • Hi, sir, can you provide the source link of this picture? – GoingMyWay Aug 17 '20 at 04:24

1 Answers1

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Proposition: $$ \mathbb{E}(x)=\int x p(x) dx=\mu \approx\frac{1}{N}\sum_{i=1}^Nx^{(i)}\\ $$ Proof: Using the fact that $p(x)$ is an Histogram Estimator, that is, for each $x^{(i)}$ it adds a frequency point, with $N$ the total number of points: $$ p(x) \approx \frac{1}{N}\sum_{i=1}^{N}\delta(x-x^{(i)}) $$ and from the properties of the $\delta$ function, with $f(x)=x$ and $a=x{(i)}$: $$ \int f(x)\delta(x-a)dx=f(a)\\ \int x\delta(x-x^{(i)})dx=x^{(i)}\\ $$ we have: $$ \mathbb{E}(x) = \int x p(x) dx \approx \int x \frac{1}{N}\sum_{i=1}^{N}\delta(x-x^{(i)}) dx\\ =\frac{1}{N}\sum_{i=1}^{N}x^{(i)} $$

Brethlosze
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