# Thread: Convolution theorem and problem about variables

1. ## Convolution theorem and problem about variables

I have not added the full proof of proposition 3.2 because it is not relevant for my problem. In (3.1) they have the cdf $F_X (a-y)$. In (3.2) they have differntiated it with respect to a and they call it $f_X(a-y)$. In proposition 3.2 they use (3.2). It seems they use the normal distribution for $X$ with $\mu=0$ and $var=\sigma ^2$ and with $x=a-y$. Now my problem is if they integrate with respect to y and have integration limits a-y in (3.1) on the normal disitribution. And afterwards differntiate with respect to a in (3.2) and a which clearly is not the same y as a variable. How can they then end up with the normal distribution for $f_X(a-y)$? If you integrate the normal distribution on an online integral calculator you end up with:

$\frac{1}{\sqrt{2 \pi}\sigma} \int_{-\infty}^a e^{-\frac{(x-\mu)^2}{2\sigma^2}} dy=\frac{erf (\frac{x-\mu}{\sqrt{2}\sigma})}{2}$

Where erf is the error function. I dont know how to differentiate the error function because it is defined as an integral.

2. ## Re: Convolution theorem and problem about variables

$\dfrac{d}{dx}~ erf(x) = \dfrac {2}{\sqrt{\pi}} e^{-x^2}$

3. ## Re: Convolution theorem and problem about variables

I dont know how to differentiate the error function because it is defined as an integral.
The fact that it is "defined as an integral" makes it easy!

Fundamental Theorem of Calculus: $\frac{d}{dx}\left(\int_a^x f(t)dt\right)= f(x)$.