Evaluating the hazard function when the CDF is close to 1?

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I need to evaluate a hazard function $h(t;theta) = dfracf(t;theta)1-F(t;theta)$, where $f$ and $F$ are a pdf and a cdf, respectively, at many values of $t$ (and for several values of the parameter $theta$). In some cases, when I evaluate $F(t;theta)$, it returns the value $1$ for some values of $t$, making $h$ infinite.



For example, in R pweibull(100,1,1) returns 1.



Is there any trick to avoid this problem?



  • I wasn't sure if I should ask this question on stackoverflow instead, but since the question is related to a function that is widely used in statistics, I thought crossvalidated was a better place as some people may know of a "classical" solution.





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    I need to evaluate a hazard function $h(t;theta) = dfracf(t;theta)1-F(t;theta)$, where $f$ and $F$ are a pdf and a cdf, respectively, at many values of $t$ (and for several values of the parameter $theta$). In some cases, when I evaluate $F(t;theta)$, it returns the value $1$ for some values of $t$, making $h$ infinite.



    For example, in R pweibull(100,1,1) returns 1.



    Is there any trick to avoid this problem?



    • I wasn't sure if I should ask this question on stackoverflow instead, but since the question is related to a function that is widely used in statistics, I thought crossvalidated was a better place as some people may know of a "classical" solution.





    r cdf hazard







    share|cite|improve this question









    New contributor




    Hazardous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.





















      up vote
      4
      down vote

      favorite









      up vote
      4
      down vote

      favorite











      I need to evaluate a hazard function $h(t;theta) = dfracf(t;theta)1-F(t;theta)$, where $f$ and $F$ are a pdf and a cdf, respectively, at many values of $t$ (and for several values of the parameter $theta$). In some cases, when I evaluate $F(t;theta)$, it returns the value $1$ for some values of $t$, making $h$ infinite.



      For example, in R pweibull(100,1,1) returns 1.



      Is there any trick to avoid this problem?



      • I wasn't sure if I should ask this question on stackoverflow instead, but since the question is related to a function that is widely used in statistics, I thought crossvalidated was a better place as some people may know of a "classical" solution.





      r cdf hazard







      share|cite|improve this question









      New contributor




      Hazardous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      I need to evaluate a hazard function $h(t;theta) = dfracf(t;theta)1-F(t;theta)$, where $f$ and $F$ are a pdf and a cdf, respectively, at many values of $t$ (and for several values of the parameter $theta$). In some cases, when I evaluate $F(t;theta)$, it returns the value $1$ for some values of $t$, making $h$ infinite.



      For example, in R pweibull(100,1,1) returns 1.



      Is there any trick to avoid this problem?



      • I wasn't sure if I should ask this question on stackoverflow instead, but since the question is related to a function that is widely used in statistics, I thought crossvalidated was a better place as some people may know of a "classical" solution.





      r cdf hazard




      r cdf hazard






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      Hazardous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











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      Ferdi

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          If the matter is numerical stability, you could look at the log of the hazard function:



          $$log(h(t; theta)) = log(f(t;theta)) - log(1-F(t;theta))$$



          You could use the log / log.p = TRUE flag in R for log values and the lower.tail flag for obtaining $log(1 - F(t;theta))$ values:



          dweibull(100,1,1, log = T) # -100
          pweibull(100, 1, 1, log.p = TRUE, lower.tail = FALSE) # -100


          Which gives you an estimate: $h(t;theta) = exp(-100 + 100) = 1$



          Edit: By the way, when you have a $Weibull(1, 1)$ distribution, I believe that this is an $Exponential(1)$, so it has a constant hazard function.






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            active

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            up vote
            2
            down vote













            If the matter is numerical stability, you could look at the log of the hazard function:



            $$log(h(t; theta)) = log(f(t;theta)) - log(1-F(t;theta))$$



            You could use the log / log.p = TRUE flag in R for log values and the lower.tail flag for obtaining $log(1 - F(t;theta))$ values:



            dweibull(100,1,1, log = T) # -100
            pweibull(100, 1, 1, log.p = TRUE, lower.tail = FALSE) # -100


            Which gives you an estimate: $h(t;theta) = exp(-100 + 100) = 1$



            Edit: By the way, when you have a $Weibull(1, 1)$ distribution, I believe that this is an $Exponential(1)$, so it has a constant hazard function.






            share|cite|improve this answer


























              up vote
              2
              down vote













              If the matter is numerical stability, you could look at the log of the hazard function:



              $$log(h(t; theta)) = log(f(t;theta)) - log(1-F(t;theta))$$



              You could use the log / log.p = TRUE flag in R for log values and the lower.tail flag for obtaining $log(1 - F(t;theta))$ values:



              dweibull(100,1,1, log = T) # -100
              pweibull(100, 1, 1, log.p = TRUE, lower.tail = FALSE) # -100


              Which gives you an estimate: $h(t;theta) = exp(-100 + 100) = 1$



              Edit: By the way, when you have a $Weibull(1, 1)$ distribution, I believe that this is an $Exponential(1)$, so it has a constant hazard function.






              share|cite|improve this answer
























                up vote
                2
                down vote










                up vote
                2
                down vote









                If the matter is numerical stability, you could look at the log of the hazard function:



                $$log(h(t; theta)) = log(f(t;theta)) - log(1-F(t;theta))$$



                You could use the log / log.p = TRUE flag in R for log values and the lower.tail flag for obtaining $log(1 - F(t;theta))$ values:



                dweibull(100,1,1, log = T) # -100
                pweibull(100, 1, 1, log.p = TRUE, lower.tail = FALSE) # -100


                Which gives you an estimate: $h(t;theta) = exp(-100 + 100) = 1$



                Edit: By the way, when you have a $Weibull(1, 1)$ distribution, I believe that this is an $Exponential(1)$, so it has a constant hazard function.






                share|cite|improve this answer














                If the matter is numerical stability, you could look at the log of the hazard function:



                $$log(h(t; theta)) = log(f(t;theta)) - log(1-F(t;theta))$$



                You could use the log / log.p = TRUE flag in R for log values and the lower.tail flag for obtaining $log(1 - F(t;theta))$ values:



                dweibull(100,1,1, log = T) # -100
                pweibull(100, 1, 1, log.p = TRUE, lower.tail = FALSE) # -100


                Which gives you an estimate: $h(t;theta) = exp(-100 + 100) = 1$



                Edit: By the way, when you have a $Weibull(1, 1)$ distribution, I believe that this is an $Exponential(1)$, so it has a constant hazard function.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








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                InfProbSciX

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