Pro-Choice

 

I’m A Man

If I had my choice of matter
I'd would rather be with cats
All engrossed in mental chatter
Showing where your mind is at

If you believe in God, then you must also believe in choice and in tolerance.

The exponential distribution is defined for the range of x>0.  It is convenient to translate from an origin of 0 to a new location, μ, and thus it becomes a range of x>μ at that new location.  It is also convenient to define x<µ as the mirror, negative of this exponential distribution.  Otherwise, when x is between 0 and µ, the probability density function and the cumulative distribution function will be undefined.

Thus a translated mirrored exponential equation also has a maximum value of 1 at the discontinuity of µ, and the combination has a Probability Density Function, PDF, and a Cummulative Density Function, CDF, of

x>μ,  PDF= λ e^{- (x-µ)} and CDF =  1-e^-λ(x-µ) 

and

x<μ, -PDF=- λ e^-λ(x-µ) and CDF = -1+e^-λ(x-µ).

While the mean of an exponential distribution is 1/λ and the median is 1/λ, the mean of the combined exponential and negative exponential distribution is 0, the median is 0 and the mode is equal to 0 but a value of 1 occurs at µ.  But this is NOT a normal distribution, becuase it is very skewed. 

The CDF of the combined exponential and negative exponential distribution appears similar to a hyperbolic tangent that has been translated to a new origin of (μ,0).  That hyperbolic tangent would have an amplitude, a, and a period, p, and thus be

a*(tanh(p*(x-µ))+C) with a constant, C, of 0, and a of 1.

which can also be expressed exponentially as

a*((e^-p*(x-µ)- e^p*(x-µ))/(e^-p*(x-µ)+ e^p*(x-µ))+C), with a constant, C,  of 0, and a of  1.

However if this is the CDF for the entire range of x, then its integral, PDF, would be 
p*sech²(p*(x‑µ)).  This is not very different than the logistics distribution, 1/(4*s)*sech²((x‑u)/(2*s)).
Its CDF is ½*tanh(((x-u)/(2*s))+ ½  which can also be expressed as an exponential as 
½*((e ^(x-µ)/2-e^-(x-µ)/2s)/(e^-(x-µ)/2s+ e^(x-µ)/2s)+ 1).  This normal CDF is a hyperbolic trigonometric function with an amplitude of ½, a period of 1/(2s) and a constant of 1. The hyperbolic tangent repeats every p/(2πi) which means that it only repeats in imagainary planes.  Because the variance, σ², for this distribution can be expressed as s²π²/3, this means that the period, p, can be expressed as √3/(2πσ).  This means that with the discontinuity, the variance must be 0, which means that the period is undefined, the amplitude must be 1, and the constant must be 0.  In other words, having no difference, the variance; no choice, the amplitude; and no God, the constant; is not normal.  Having choice, variance AND God is normal.