I am trying to integrate Gaussian distribution from -m to m to find parametar A. I have done this so far:
$\int_{-m}^{m}\frac{A}{\sqrt(2\pi)\sigma}e^{-(x-m)^{2}/(2\sigma^{2})}dx=1$
after $u=\frac{x-m}{\sigma}$ I got:
$A \frac{1}{\sqrt(2\pi)} \int_{-2m/\sigma}^{0} e^{-0.5 u^{2}}du=1$
In order to solve this integral I want to make Q function but limits of integrals are not matching with limits of Q fnction.
Is anyone know how should solve this integral with Q function to get value of parametar A.
Thank you!
You cannot integrate the Normal Gaussian distribution, because you cannot express it in terms of elementary functions.
Instead, what you can try, is to express your Normal distribution as a Gaussian distibution with mean $0$ and variance $1$: from there you look up the integral table and find an esteem of your parameters.
In your case $$\frac{1}{\sqrt{2\pi}\sigma}e^{-(x-\mu)^{2}/(2\sigma^{2})} \sim N\left(\mu, \sigma^2\right) \sim X$$ but you can normalize it to get $$\frac{X - \mu}{\sigma} \sim Y \sim N(0,1)$$ so the integral you wanted to estimate now comes to be $$\begin{align} \int_{-\mu}^{\mu}\frac{A}{\sqrt{2\pi}\sigma}e^{-(x-\mu)^{2}/(2\sigma^{2})} dx &= A \cdot \int_{-\mu}^{\mu}\frac{1}{\sqrt{2\pi}\sigma}e^{-(x-\mu)^{2}/(2\sigma^{2})} dx\\ &= A \cdot P(-\mu \le X \le \mu)\\ &= A \cdot P\left(\frac{-\mu - \mu}{\sigma} \le \frac{X - \mu}{\sigma} \le \frac{\mu - \mu}{\sigma}\right)\\ &= A \cdot P\left(\frac{-2\mu}{\sigma} \le N(0,1) \le 0\right)\\ &= A \left( \Phi\left( 0 \right) - \Phi\left(\frac{-2\mu}{\sigma}\right) \right)\\ &= A \left( \frac{1}{2} - \Phi\left(\frac{-2\mu}{\sigma}\right) \right)= 1\\ &\Rightarrow A = \frac{1}{\frac{1}{2} - \Phi\left(\frac{-2\mu}{\sigma}\right)} \end{align}$$ where $$\Phi(x) = \text{integral of } N(0,1) = \int_{-\infty}^x \frac{1}{2\pi} e^{-t^2/2} dt = P(X \sim N(0,1) \le x)$$