Estimating Weibull Distribution parameters by Least Square I'm working on something about Least square estimation for Weibull distribution. After reviewing this paper, I'm pretty confused by the derivation of least square method. 
In this paper, the author rank the data before calculating the least square of the "linear equation $(Y_i = aX_i+b)$" , why is that?
My second question is: why does the author use $i/(n+1)$ as the estimated value of cdf $F(x_{(i)})$? Is there any mathematical explanation on this problem? 
Looking forward to any feedback and suggestions. 
 A: I'll start with your second question first. Often, it is simpler to use $i/(n+1)$ (or $(i-1/2)/n$) instead of $i/n$ to avoid boundary issues. Let's say you've sorted your data so $Z_1,\ldots, Z_n$ refers to the smallest up to the biggest values in the data. Then, $Z_i$ is a pretty good estimate of the $(i/n)$th quantile of the distribution. But $i/n$ and $i/(n+1)$ are really close values and typically the true quantiles of $i/n$ and $i/(n+1)$ are also really close. Thus, we can argue that $Z_i$ is a pretty good estimate of the $(i/(n+1))$th quantile of the distribution. So, why does all this matter? Consider the biggest value in the data set, $Z_n$. Would this ever be a good estimate of quantile $n/n=1$? Of course not. For a Weibull distribution, quantile $1$ is always equal to $\infty$ yet $Z_n$ is always finite. That said, $Z_n$ is still a good approximation of quantile $n/(n+1)$.
As for your primary question, "why estimate by least squares?", you simply need to understand a simple fact of Weibull distributions. The Weibull CDF is given by 
$$F(x) = 1-\exp(-(x/\delta)^c)$$
If we rearrange this expression, we get
$$\ln(x) = \dfrac 1 c \ln⁡(-\ln⁡(1-F(x))) + \ln⁡(\delta)$$
Now, if we define $Y_i = \ln(Z_i)$ and $X_i=\ln⁡(-\ln⁡(1-[i/(n+1)]))$, then we would expect to get the relationship, $Y_i = a X_i + b$, where $a=1/c$ and $b=\ln(\delta)$. All you have to do then is run a least squares regression of $X$ vs $Y$ and convert the parameter estimates into $c$ and $\delta$.
