why does the area under the bell curve involve pi?

Answer 1

Just so we're on the same page, the Gaussian distribution is:

p(x) = [1/(√(2π)*σ)]*e^[-(x - µ)²/(2σ²)]

Where µ = the mean x-value
and σ = the standard deviation

The bell curve, or the Gaussian distribution is a probability distribution function. A fundamental axiom (law) of probability theory is that the sum of all probabilities must be 1. For the bell curve, the sum of all probabilities is the total area under the curve.

The √(2π) in the denominator is there to "normalize" the distribution: that is, to make the integral from -∞ to +∞ be exactly 1. To see it yourself, take out the [1/(√(2π)*σ] and integrate what's left. It will be exactly √(2π)*σ.

Here is a really awesome trick to integrating p(x):

p(x) = e^[-(x - µ)²/(2σ²)]

For simplicity, let µ = 0 (peak is at x=0) and let σ = 1:
p(x) = e^(-x²/2)

Here's the trick:

P(x) = int(e^(-x²/2),dx) = √(int(e^(-x²/2),dx)²)

(That's just like saying 4 = √(4²))

= √[int(e^(-x²/2),dx) * int(e^(-x²/2),dx)]

Now let one of the integrals in x be an integral in y (you can do this because x could just be any variable, the integral will be the same):

= √[int(e^(-x²/2),dx) * int(e^(-y²/2),dy)]

Now combine the x and y integrals into a double integral:

= √[int(int(e^(-x²/2)e^(-y²/2),dx,dy)]
= √[int(int(e^(-(x² +y²)/2),dx,dy)]

Now switch to polar coordinates:
r² = x² +y²
dxdy = da = rdrdφ

= √[int(int(r*e^(-r²/2),dr,dφ)]

Now instead of integrating x and y from -∞ ──> +∞, you integrate r from 0 ──> ∞ and φ from 0 ──> 2π.

P(x) comes out to be exactly √(2π). If σ had been anything other than 1, P(x) would be √(2π)*σ.

I'm not sure this is a complete answer...but you DO see how π comes into play, right (at least from a computational standpoint)?