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The distribution: $P(x)=(\lambda / \pi)^{1/2} e^{- \lambda (x-a)^2}$

Using a definition for the expectation value of something, I set up the problem like this:

$$<x>=\sum_{x=0}^\infty xP(x)$$ $$ <x>=\sum_{x=0}^\infty x(\lambda / \pi)^{1/2} e^{- \lambda (x-a)^2} $$

After simplifying the summation, I arrived at:

$$ <x>= (\lambda / \pi)^{1/2} e^{-a^2 \lambda} \sum_{n=1}^\infty e^{-(n)^2} e^{2an} $$

But thinking over this, it doesn't make intuitive sense.

  1. I solved for the constant $ (\lambda / \pi)^{1/2} $ using an integral that went from $-\infty$ to $+\infty$ (not shown, this was in an earlier part of the problem). Therefore, shouldn't my summation do the same?

  2. I don't understand what the expectation value of a gaussian distribution even is. Whatever I get at the end should tell me something - what is that something?

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For the first question, yes, the expectation for a continuous distribution is indeed an integral and not a sum, i.e. in your notation it would be:

$$ \langle X \rangle = \int_{-\infty}^{+\infty} xP(x)dx $$

For the second question, the expectation of a Gaussian distribution is very intuitive since it also equal to its median and mode. In particular, if you plot the density of a given Gaussian, then its expectation is the position at which the density gets maximized! (In your case it gets maximized at $x=a$, which is the expectation).

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