Template:Logistic probability paper

Logistic Distribution Probability Paper
The form of the Logistic probability paper is based on linearizing the $$cdf$$. From Eqn. (UnR fcn), $$z$$  can be calculated as a function of the  $$cdf$$   $$F$$  as follows:


 * $$z=\ln (F)-\ln (1-F)$$

or using Eqn. (z func of parameters)


 * $$\frac{T-\mu }{\sigma }=\ln (F)-\ln (1-F)$$


 * Then:


 * $$\ln (F)-\ln (1-F)=-\frac{\mu }{\sigma }+\frac{1}{\sigma }T$$


 * Now let:


 * $$y=\ln (F)-\ln (1-F)$$


 * $$x=T$$


 * and:


 * $$a=-\frac{\mu }{\sigma }$$


 * $$b=\frac{1}{\sigma }$$

which results in the following linear equation:


 * $$y=a+bx$$

The logistic probability paper resulting from this linearized $$cdf$$  function is shown next.



Since the logistic distribution is symmetrical, the area under the $$pdf$$  curve from  $$-\infty $$  to  $$\mu $$  is  $$0.5$$, as is the area from  $$\mu $$  to  $$+\infty $$. Consequently, the value of $$\mu $$  is said to be the point where  $$R(t)=Q(t)=50%$$. This means that the estimate of $$\mu $$  can be read from the point where the plotted line crosses the 50% unreliability line.

For  $$z=1$$,  $$\sigma =t-\mu $$  and  $$R(t)=\tfrac{1}{1+\exp (1)}\approx 0.2689.$$  Therefore,  $$\sigma $$  can be found by subtracting  $$\mu $$  from the time value where the plotted probability line crosses the 73.10% unreliability (26.89% reliability) horizontal line.