How could you show that the normal density integrates to 1?
$$ \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi \sigma^2}} e^{-(x+\mu)^2 / \sigma^2} dx = 1 $$
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$\exp(-x^2)$ is a so called Schwartz-function for which you can use Fubini in 2-dimension. It is $$\int_{-\infty}^{+\infty} \exp(-x^2)dx\cdot\int_{-\infty}^{+\infty}\exp(-y^2)dx = \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\exp(-x^2-y^2)dxdy$$ This can be solved in polar-coordinates easilly and its value is $\pi$. Because of $\exp(-x^2)\gt0\,\forall\,x$ you have $\int_{-\infty}^{+\infty}\exp(-x^2)\gt0$ and finally $\int_{-\infty}^{+\infty}\exp(-x^2) = \sqrt{\pi}$ |
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