Analytical solution to mixed distribution fit to failure time data - Lambert W perhaps?I have a set of n device failure times { t i &gt; 0 } for i = 1... n and N − n devices which have not yet failed. Using maximum likelihood I am attempting to find a closed-form analytical solution to fit the data to the following cumulative distribution function: F ( t | λ , p ) = p ( 1 − e − λ t )where 0 &lt; p &lt; 1 is the asymptotic fraction of units to eventually fail and λ &gt; 0 the sub-population failure rate. The likelihood for this MLE attempt is given by: L = ( 1 − F ( t n ) ) N − n ∏ i = 1 n f ( t i )and ln ⁡ L = ( N − n ) ln ⁡ ( 1 − p + p e − λ t n ) + n λ p − λ ∑ i = 1 n t i with pdf of f ( t ) = d F / d t = λ p e − λ t . Here we take ∇ λ p L = 0 or ∇ λ p ln ⁡ L = 0 to solve for p and λ at max likelihood (or log likelihood). I've just recently learned a smidgen about the Lambert W function and was hoping that someone with a more nimble mind than mine might be able to derive a closed form solution using this and/or other cleverness.

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Analytical solution to mixed distribution fit to failure time data - Lambert W perhaps?I have a set of n device failure times   {      t    i    &amp;gt;  0  } for   i  =  1...  n and   N  −  n devices which have not yet failed. Using maximum likelihood I am attempting to find a closed-form analytical solution to fit the data to the following cumulative distribution function:  F  (  t      |    λ  ,  p  )  =  p  (  1  −      e          −      λ      t        )where   0  &amp;lt;  p  &amp;lt;  1 is the asymptotic fraction of units to eventually fail and   λ  &amp;gt;  0 the sub-population failure rate. The likelihood for this MLE attempt is given by:      L    =  (  1  −  F  (      t    n    )      )          N      −      n            ∏          i      =      1        n    f  (      t    i    )and  ln  ⁡            L        =  (  N  −  n  )  ln  ⁡            (      1      −      p      +      p              e                  −          λ                      t            n                              )        +  n  λ  p  −  λ      ∑          i      =      1        n        t    i  with pdf of   f  (  t  )  =  d  F      /    d  t  =  λ  p      e          −      λ      t      . Here we take       ∇          λ      p            L    =  0 or       ∇          λ      p        ln  ⁡      L    =  0 to solve for p and   λ at max likelihood (or log likelihood). I&#039;ve just recently learned a smidgen about the Lambert W function and was hoping that someone with a more nimble mind than mine might be able to derive a closed form solution using this and/or other cleverness.

Kayleigh Cross · Accepted Answer

Step 1Lambert W function can not help:      {                                                                                        d                                                  L                                                                              d                                p                                              =          −                                                    (                N                −                n                )                                  (                  1                  −                                      e                                          −                      λ                                              t                        n                                                                              )                                                            1                −                p                +                p                                  e                                      −                    λ                                          t                      n                                                                                                    +          n          λ          =          0          ,                                                                                                  d                                                  L                                                                              d                                λ                                              =          −                                                    (                N                −                n                )                p                                  t                  n                                                  e                                      −                    λ                                          t                      n                                                                                                  1                −                p                +                p                                  e                                      −                    λ                                          t                      n                                                                                                    +          n          p          −                      ∑                          t              =              1                        n                                t            i                    =          0                              {                            p          =                                    1                              1                −                                  e                                      −                    λ                                          t                      n                                                                                                    −                                                    n                λ                                            N                −                n                                                                                                                    (                N                −                n                )                p                                  t                  n                                                  (                  1                  −                                                            1                                              p                        +                                                                                                            n                              λ                                                                                      N                              −                              n                                                                                                                                                            )                                                            1                −                                                      p                                          p                      +                                                                                                    n                            λ                                                                                N                            −                            n                                                                                                                                                                            =          n          p          −                      ∑                          t              =              1                        n                                t            i                    ,                              {                            p          =                                    1                              1                −                                  e                                      −                    λ                                          t                      n                                                                                                    −                                                    n                λ                                            N                −                n                                                                                                                    (                N                −                n                )                p                                  t                  n                                                  (                  p                  −                  1                  +                                                                                    n                        λ                                                                    N                        −                        n                                                                              )                                                                                                  n                    λ                                                        N                    −                    n                                                                                =          n          p          −                      ∑                          t              =              1                        n                                t            i                    ,                                            (1)                              {                                                                                                                n                      λ                                                              N                      −                      n                                                                      =                                                                            (                      N                      −                      n                      )                                              t                        n                                            p                      (                      1                      −                      p                      )                                                                                      (                        (                        N                        −                        n                        )                                                  t                          n                                                −                        n                        )                                            p                      +                                              ∑                                                  t                          =                          1                                                n                                                                    t                        i                                                                                                                                                p                =                                                      1                                          1                      −                                              e                                                  −                          λ                                                      t                            n                                                                                                                                              −                                                                            n                      λ                                                              N                      −                      n                                                                                                                        Step 2I am not able to find a closed analytical solution of the system (1). At this time, it can be solved using iteration method or by eliminating of one unknown.

Analytical solution to mixed distribution fit to failure time data - Lambert W perhaps?

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