Can a Gaussian distribution be represented as a sum of Dirac Deltas?

[tworitdash]

We know that Dirac Delta is not a function. However, I just talk about the numerical version of it that we use every day. We can simply represent the Dirac delta function as a limiting case of Gaussian distribution when the width of the distribution $\sigma->0$.

$$
\delta(x - \mu) = lim_{\sigma -> 0} \frac{1}{\sqrt{2\pi \sigma^2}} e^{\frac{-(x - \mu)^2}{2\sigma^2}}
$$

Is it possible to also say the reverse with a weighted sum of Dirac Deltas to construct a Gaussian spectrum?

$$
\frac{1}{\sqrt{2\pi \sigma^2}} e^{\frac{-(x - mu)^2}{2\sigma^2}} = \sum_{i} w_i \delta(x - i)
$$

Where, somehow the weights $w_i$ constitute how it is distributed ($\sigma$). If yes, how do we decide these weights?

 

[stevendaryl]

Any function can be represented as a sum of Dirac delta functions:

Let $f(x)$ be an arbitrary function of $x$. Then you can represent it as:

$\int f(y) \delta(x-y) dy$

So that's a weighted sum (well, integral) of delta functions.

 

[stevendaryl]

Any function can be represented as a sum of Dirac delta functions:

Let $f(x)$ be an arbitrary function of $x$. Then you can represent it as:

$\int f(y) \delta(x-y) dy$

So that's a weighted sum (well, integral) of delta functions.

If you really want a discrete sum, instead of an integral, then it can't be done for most functions. But I guess for some purposes, you can approximate a function by delta functions: Pick a small positive x increment $\Delta x$ and define $\tilde{f}(x, \Delta x)$ by:

$\tilde{f}(x, \Delta x) = \sum_j f(j \Delta x) \delta(x- j\Delta x) \Delta x$

where $\Delta x$ is some small real number. This approximation works in an integration sense: For any other smooth function $g(x)$, we have:

$lim_{\Delta x \Rightarrow 0} \int \tilde{f}(x, \Delta x) g(x) dx = \int f(x) g(x) dx$