My math is quite limited so please bear with me. I will get to the point: Is there a way to transform a continuous function into a bounded one?
In essence I have a normalized Gaussian distribution defined by:
$$y = b + \frac{a}{c\sqrt{\frac{\pi}{2}}} \cdot e^{\frac{-2(x-d)^2}{c^2}}$$
I am only interested in values of $y$ for $0 \leq x \leq 500$. a,b,c and d are parameters.
I would like to maintain the ability of the curve to be a normalised probability distribution but only for $0 \leq x \leq 500$.
Is there a way of doing this or am I trying the impossible?
Thanks to anyone that can render assistance.
Firstly $y(x)$ you defined is not a Gaussian distribution. $$y(x)=\frac{1}{c\sqrt{2\pi}}e^{-\frac{1}{2}\frac{(x-d)^2}{c^2}}$$ is one with $c > 0$ and $d\in \mathbb{R}$ (respectively mean and standard deviation).
That being said, if you change the expression you won't get a Gaussian distribution. However you can change $y$ any way you want. Just ensure that $\int_{-\infty}^\infty y(x)dx = 1$. For example if you define
$$ y(x) = \frac{1}{c\sqrt{2\pi}}e^{-\frac{1}{2}\frac{(x-d)^2}{c^2}} 1_{\{0\le x \le 500\}}$$ Then it is not a probability density function. But if you normalize it then you actually get a pdf : $$ y(x) = \frac{1}{\phi\left(\frac{500-d}{c}\right)-\phi\left(\frac{-d}{c}\right)}\frac{1}{c\sqrt{2\pi}}e^{-\frac{1}{2}\frac{(x-d)^2}{c^2}} 1_{\{0\le x \le 500\}}$$ where $\phi$ is the cummulative distribution function of the standard normal law.