Hi (apologies for my english), i have been interested in the CLT for the past weeks, and already have proved that the convolution of 2 normal distributions is another normal distribution using the Fourier transform and the convolution theorem (I know that it is not necessary to use the convolution theorem to prove that, you can do it from the definition of the convolution itself: https://www.youtube.com/watch?v=d_qvLDhkg00&t=3s). My problem comes when i try to use the Laplace transform.
So first i will get the Laplace transform of a normal distribution:
- Lets define:
$$ f(x) = e^{-x^2} $$
So we have that the laplace transform of that function is (https://www.youtube.com/watch?v=b8ph1p85-dI):
$$ F(s) = \frac{\sqrt{\pi}}{2}e^{\frac{s^2}{4}}\text{erfc}(\frac{s}{2}) \tag1 $$
Using the scaling property of the Laplace transform, we get:
$$ L\Big[f(ax)\Big] = \frac{1}{a}F(\frac{s}{a}) $$
So lets define a new function:
$$ g(x,a)=\frac{a}{\sqrt{\pi}}f(ax) $$
so using equation (1), we have that the Laplace transform of $g $ with respect to $ x $ is:
$$ L\Big[g(x,a)\Big] = L\Big[{\frac{a}{\sqrt{\pi}}f(ax)}\Big] $$ $$ L\Big[g(x,a)\Big] = \frac{a}{\sqrt{\pi}}L\Big[{f(ax)}\Big] $$ $$ L\Big[g(x,a)\Big] = \frac{a}{\sqrt{\pi}}\frac{1}{a}F(\frac{s}{a}) $$ $$ L\Big[g(x,a)\Big] = \frac{1}{\sqrt{\pi}}F(\frac{s}{a}) $$ $$ L\Big[g(x,a)\Big] = \frac{1}{\sqrt{\pi}}\frac{\sqrt{\pi}}{2}e^{\frac{(s/a)^2}{4}}\text{erfc}(\frac{(s/a)}{2}) $$
$$ L\Big[g(x,a)\Big] = \frac{1}{2}e^{\frac{(s/a)^2}{4}}\text{erfc}(\frac{(s/a)}{2}) $$
But if we replace: $$ a=\frac{1}{\sigma \sqrt{2}} $$
We have:
$$ L\Big[g(x,\frac{1}{\sigma \sqrt{2}})\Big] = \frac{1}{2}e^{\frac{(s\sigma \sqrt{2})^2}{4}}\text{erfc}(\frac{s\sigma \sqrt{2}}{2}) $$ $$ L\Big[g(x,\frac{1}{\sigma \sqrt{2}})\Big] = \frac{1}{2}e^{\frac{(s\sigma)^2}{2}}\text{erfc}(\frac{s\sigma}{\sqrt{2}}) \tag2$$
But the function $ g(x,\frac{1}{\sigma \sqrt{2}}) $ is also the density of a normal distribution with $ 0 $ mean and $\sigma$ standard deviations:
$$ g(x,a)=\frac{a}{\sqrt{\pi}}f(ax) $$
$$ g(x,\frac{1}{\sigma \sqrt{2}})=\frac{1/{\sigma \sqrt{2}}}{\sqrt{\pi}}f(\frac{x}{\sigma \sqrt{2}}) $$
$$ g(x,\frac{1}{\sigma \sqrt{2}})=\frac{1}{\sigma\sqrt{2\pi}} e^{-(\frac{x}{\sigma \sqrt{2}})^2}$$
$$ g(x,\frac{1}{\sigma \sqrt{2}})=\frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{x^2}{2\sigma^2}} := \varphi(x,\sigma) \tag3$$
But from equation (2) and (3), we know that: $$ L\Big[g(x,\frac{1}{\sigma \sqrt{2}})\Big]=\frac{1}{2}e^{\frac{(s\sigma)^2}{2}}\text{erfc}(\frac{s\sigma}{\sqrt{2}}) $$ $$ L\Big[\varphi(x,\sigma)\Big] = \frac{1}{2}e^{\frac{(s\sigma)^2}{2}}\text{erfc}(\frac{s\sigma}{\sqrt{2}}) \tag4$$
I checked this equality against the response of Laplace transform of a normal distribution and it seems to give the same result as mine.
- So using the known properties of the convolution of 2 normal distributions we have:
$$ \varphi(x,\sigma_0) * \varphi(x,\sigma_1) = \varphi(x,\sqrt{\sigma_0^2+\sigma_1^2}) \tag5$$
And the convolution theorem tells us that:
$$ L\Big[\varphi(x,\sigma_0)\Big]L\Big[\varphi(x,\sigma_1)\Big]=L\Big[{\varphi(x,\sigma_0) * \varphi(x,\sigma_1)}\Big]$$
Lets replace using equation (5): $$L\Big[\varphi(x,\sigma_0)\Big]L\Big[\varphi(x,\sigma_1)\Big]=L\Big[{\varphi(x,\sqrt{\sigma_0^2+\sigma_1^2})}\Big]$$
Lets just see what happens when $\sigma_0=\sigma_1=1$:
$$L\Big[\varphi(x,1)\Big]L\Big[\varphi(x,1)\Big]=L\Big[{\varphi(x,\sqrt{1^2+1^2})}\Big]$$ $$L\Big[\varphi(x,1)\Big]^2=L\Big[{\varphi(x,\sqrt{2})}\Big]$$
So replacing with equation (4) on both sides we get:
$$\Big({\frac{1}{2}e^{\frac{s^2}{2}}\text{erfc}(\frac{s}{\sqrt{2}})}\Big)^2=\frac{1}{2}e^{\frac{(s\sqrt{2})^2}{2}}\text{erfc}(\frac{s\sqrt{2}}{\sqrt{2}})$$
But using Geogebra to visualize both sides, the equality does not hold, for example lets set $ s=0$:
$$\Big({\frac{1}{2}e^{\frac{0^2}{2}}\text{erfc}(\frac{0}{\sqrt{2}})}\Big)^2=\frac{1}{2}e^{\frac{(0\sqrt{2})^2}{2}}\text{erfc}(\frac{0\sqrt{2}}{\sqrt{2}})$$
$$\Big({\frac{1}{2}}\Big)^2=\frac{1}{2}$$
$$\frac{1}{4}=\frac{1}{2}$$
I dont know where i went wrong, i am used to not being very good at algebra so i am searching for that sign that i forgot in some calculation, but cannot find it; or maybe it is a conceptual problem, since i am very new to the world of transforms, thanks in advance!!