কোনও "স্ট্যান্ডার্ড" বিতরণ দ্বারা নির্ধারিত কোনও অবস্থান-স্কেল পরিবার বিবেচনা করুন ,F
ΩF={F(μ,σ):x→F(x−μσ)∣σ>0}.
Assuming F differentiable we readily find that the PDFs are 1σf((x−μ)/σ)dx.
Truncating these distributions to restrict their support between a and b, a<b, means that the PDFs are replaced by
f(μ,σ;a,b)(x)=f(x−μσ)dxσC(μ,σ,a,b),a≤x≤b
(and are zero for all other values of x) where C(μ,σ,a,b)=F(μ,σ)(b)−F(μ,σ)(a) is the normalizing factor needed to ensure that f(μ,σ;a,b) integrates to unity. (Note that C is identically 1 in the absence of truncation.) The log likelihood for iid data xi therefore is
Λ(μ,σ)=∑i[logf(xi−μσ)−logσ−logC(μ,σ,a,b)].
Critical points (including any global minima) are found where either σ=0 (a special case I will ignore here) or the gradient vanishes. Using subscripts to denote derivatives, we may formally compute the gradient and write the likelihood equations as
00=∂Λ∂μ=∂Λ∂σ=∑i⎡⎣⎢−fμ(xi−μσ)f(xi−μσ)−Cμ(μ,σ,a,b)C(μ,σ,a,b)⎤⎦⎥=∑i⎡⎣⎢−fσ(xi−μσ)σ2f(xi−μσ)−1σ−Cσ(μ,σ,a,b)C(μ,σ,a,b)⎤⎦⎥
Because a and b are fixed, drop them from the notation and write nCμ(μ,σ,a,b)/C(μ,σ,a,b) as A(μ,σ) and nCσ(μ,σ,a,b)/C(μ,σ,a,b) as B(μ,σ). (With no truncation, both functions would be identically zero.) Separating the terms involving the data from the rest gives
−A(μ,σ)−σ2B(μ,σ)−nσ=∑ifμ(xi−μσ)f(xi−μσ)=∑ifσ(xi−μσ)f(xi−μσ)
By comparing these to the no-truncation situation it is evident that
Any sufficient statistics for the original problem are sufficient for the truncated problem (because the right hand sides have not changed).
Our ability to find closed-form solutions relies on the tractability of A and B. If these do not involve μ and σ in simple ways, we cannot hope to obtain closed-form solutions in general.
For the case of a normal family, C(μ,σ,a,b) of course is given by the cumulative normal PDF, which is a difference of error functions: there is no chance that a closed-form solution can be obtained in general. However, there are only two sufficient statistics (the sample mean and variance will do) and the CDF is as smooth as can be, so numerical solutions will be relatively easy to obtain.