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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.

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    The question, "Is there a way of doing this?", has nothing to do with transforming a continuous function to a bounded function.2012-09-14

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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.

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    Normalizing a continuous function $p$ to get a density (pdf) is not a specific topic. Just divide by $\int_{-\infty}^\infty p(x)dx$. Here the denominator is just $\int_0^{500} \frac{1}{c\sqrt{2\pi}} e^{-\frac{1}{2} \frac{(x-d)^2}{c^2}}dx$ which is $P(0\le X\le 500)$ where $X \sim \mathcal{N}(d,c)$. I just wrote $P(0\le X\le 500) = P(X\le 500) - P(X \le 0)$ and used the fact that $Z := \frac{X-d}{c} \sim \mathcal{N}(0,1)$. Read the wikipedia page about the normal distribution to get additional material about all this.2012-09-14