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Thread: Probability Integral

  1. #1
    Global Moderator

    Nov 2005
    New York City

    Probability Integral

    In probability theory the important Gaussian Distribution often leads to the integral $\displaystyle X=\int_{-\infty}^{\infty} e^{-x^2} dx $.

    In case you ever wondered where it comes from, here (informal but nice*) is a non-complex analysis demonstration.

    We begin by considering the integral,
    $\displaystyle Y=\iint_{\mathbb{R}^2} e^{-(x^2+y^2)} \ dA$

    This is the integral over the entire plane $\displaystyle \mathbb{R}^2$.

    Now there are two ways to get the entire plane.

    Method 1: For $\displaystyle a,b>0$ we can create a rectangle $\displaystyle -a\leq x\leq a \mbox{ and }-b\leq x\leq b$. Now we make $\displaystyle a\to \infty \mbox{ and }b\to \infty$. That infinite rectangle will take the entire plane.
    In other words, $\displaystyle Y=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2+y^2)} \ dy \ dx$.
    Now, the important realization is that this double integral has "seperated variables" meaning $\displaystyle f(x,y)=g(x)h(y)$ and we can write $\displaystyle \int_{-\infty}^{\infty} e^{-x^2} dx \cdot \int_{-\infty}^{\infty} e^{-y^2} dx$. Hence, $\displaystyle Y = X^2 \implies X = \sqrt{Y}$ (because it cannot be negative).

    Method 2: We notice the expression $\displaystyle x^2+y^2$ in the double integral and try a polar coordinates substitution. If we let $\displaystyle r^2 = x^2+y^2$, we can let $\displaystyle 0\leq \theta \leq 2\pi $ and $\displaystyle r\to \infty$. In other words, we draw a circle at the orgin with infinite radius, and this will take on values on the entire $\displaystyle \mathbb{R}^2$ plane.
    Thus, we end up with after a change to polar coordinates,
    $\displaystyle Y = \int_0^{\infty} \int_0^{2\pi} e^{-r^2} \ r d\theta \ dr = \lim_{r\to \infty} 2\pi \int_0^r re^{-r^2}dr = 2\pi \left( \frac{1}{2} \right) = \pi$

    Thus, $\displaystyle Y=\pi$ which implies $\displaystyle X=\sqrt{\pi}$.

    And hence,
    $\displaystyle \int_{-\infty}^{\infty} e^{-x^2} \ dx = \sqrt{\pi}$

    *)I am sure it can be made formal. But I do not know how do it because I am not familar with the theory of real multiple integration.
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  2. #2
    May 2007
    In my calculus textbook, a squeeze technique is used to prove the result via double integrals in polar coordinate.
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