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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. – Michael Hardy Sep 14 '12 at 17:07
up vote 2 down vote accepted

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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So if I calculate: $y(x) = \frac{1}{\int_0^500 b + \frac{a}{c\sqrt{\frac{\pi}{2}}} \cdot e^{\frac{-2(u-d)^{2}}{c^{2}}} du} \cdot (b + \frac{a}{c\sqrt{\frac{\pi}{2}}} \cdot e^{\frac{-2(x-d)^{2}}{c^{2}}})$ I can normalize the function yes? Also since the integral is between 0 and 500 that means that my x values will be bounded? Final question does this mean that I will be able to change the standard deviation and still maintain a curve with an approximately consistent area below the curve? Thank you very much for your clear and concise answer – Filipe Teixeira Sep 14 '12 at 9:58
$\int_{-\infty}^\infty bdx$ is not defined. Look here for a precise definition of the normal distribution. – vanna Sep 14 '12 at 10:04
Yes you right. My mistake. I just copied the formula directly without thinking. – Filipe Teixeira Sep 14 '12 at 10:14
This is helping me greatly. Thank you. Could you recommend any reading material on normalizing functions. I am particularly interested in how you managed to calculate $$ y(x) = \frac{1}{\phi\left(\frac{500-d}{c}\right)-\phi\left(\frac{-d}{c}\right)}\frac{1}‌​{c\sqrt{2\pi}}e^{-\frac{1}{2}\frac{(x-d)^2}{c^2}} 1_{\{0\le x \le 500\}}$$ – Filipe Teixeira Sep 14 '12 at 13:52
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. – vanna Sep 14 '12 at 14:32

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