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When combining two line shapes (for example a Gaussian and a Lorentzian), the effect of Both of them combined is the convolution of both (with a Gaussian and a Lorentizan,this is the Voigt function). I am wondering what is the reason that a convolution arises.

I saw the explanation here: https://phys.libretexts.org/Bookshelves/Astronomy__Cosmology/Stellar_Atmospheres_(Tatum)/10%3A_Line_Profiles/10.04%3A_Combination_of_Profiles

but it is not clear to me why

"At a distance $ξ $ from the line centre, the contribution to the line profile is the height of the function $𝑓(ξ)$ weighted by the function $𝑔(𝑥−ξ)$".

A clear explanation would be helpful.

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  • $\begingroup$ It might be from a point that the sum of two random variables give a distribution with probability density function which is the convolution of both. In case when two variables are independent. You can check this in a statistics book, for exemple, in [Papoulis "robability, Random Variables and Stochastic Processes", 2001] $\endgroup$ Commented Aug 26 at 9:23
  • $\begingroup$ This is mathematically a formal thing. Take a random variable $z$ as a sum of two independent random variables $z = x+y $ and get the distribution of $z$. There is a chapter on "two variables" in the mentioned book calculating the distribution of $z$ $\endgroup$ Commented Aug 26 at 9:28
  • $\begingroup$ I did not think about it this way. Thank you so much $\endgroup$
    – Goose
    Commented Aug 26 at 23:07

1 Answer 1

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Suppose we sample a function $f(x)$, but our resolution possesses a finite width $w=\Delta=const$: For each point $x_0$ we do not measure $f(x_0)$, but we obtain the moving average of the function $f$ with width $w$.

Next, we generalise this idea, and allow the finite resolution to be described by a function $w(x)$. We can think of it as a weighting function. At each point the same logic as above applies. However, instead of cutting out a finite part around $x=x_0$, and giving each point in this interval the same weight, the weight now depends of the distance to $x_0$. Thus, what we obtain is $$ g(x_0) = \int f(x-x_0) w(x) dx $$ which is identical to the convolution. Note, if we use a rectangular function to describe the finite resolution, we obtain the initially described moving average, where each point within a certain distance to $x_0$ gets the same weight. However, if we like to a continuous function to describe the contribution to each point, the Gaussian is a "natural" choice, because it possesses the largest entropy.

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