Subtracting very small probabilities - How to compute?











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This question is an extension of a related question about adding small probabilities. Suppose you have log-probabilities $ell_1 geqslant ell_2$, where the corresponding probabilities $exp(ell_1)$ and $exp(ell_2)$ are too small to be distinguished from zero in the initial computational facility being used (e.g., base R). We want to find the log-diffference of these probabilities, which we denote by:



$$ell_- equiv ln big( exp(ell_1) - exp(ell_2) big)$$



Questions: How can you effectively compute this log-difference? Can this be done in the base R? If not, what is the simplest way to do it with package extensions?










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    Possible duplicate of Computation of likelihood when $n$ is very large, so likelihood gets very small?
    – Xi'an
    1 hour ago










  • "where the corresponding probabilities exp(ℓ1) and exp(ℓ2) are too small to be distinguished from zero" - I think you mean one instead of zero?
    – Don Hatch
    3 mins ago















up vote
3
down vote

favorite
2












This question is an extension of a related question about adding small probabilities. Suppose you have log-probabilities $ell_1 geqslant ell_2$, where the corresponding probabilities $exp(ell_1)$ and $exp(ell_2)$ are too small to be distinguished from zero in the initial computational facility being used (e.g., base R). We want to find the log-diffference of these probabilities, which we denote by:



$$ell_- equiv ln big( exp(ell_1) - exp(ell_2) big)$$



Questions: How can you effectively compute this log-difference? Can this be done in the base R? If not, what is the simplest way to do it with package extensions?










share|cite|improve this question


















  • 1




    Possible duplicate of Computation of likelihood when $n$ is very large, so likelihood gets very small?
    – Xi'an
    1 hour ago










  • "where the corresponding probabilities exp(ℓ1) and exp(ℓ2) are too small to be distinguished from zero" - I think you mean one instead of zero?
    – Don Hatch
    3 mins ago













up vote
3
down vote

favorite
2









up vote
3
down vote

favorite
2






2





This question is an extension of a related question about adding small probabilities. Suppose you have log-probabilities $ell_1 geqslant ell_2$, where the corresponding probabilities $exp(ell_1)$ and $exp(ell_2)$ are too small to be distinguished from zero in the initial computational facility being used (e.g., base R). We want to find the log-diffference of these probabilities, which we denote by:



$$ell_- equiv ln big( exp(ell_1) - exp(ell_2) big)$$



Questions: How can you effectively compute this log-difference? Can this be done in the base R? If not, what is the simplest way to do it with package extensions?










share|cite|improve this question













This question is an extension of a related question about adding small probabilities. Suppose you have log-probabilities $ell_1 geqslant ell_2$, where the corresponding probabilities $exp(ell_1)$ and $exp(ell_2)$ are too small to be distinguished from zero in the initial computational facility being used (e.g., base R). We want to find the log-diffference of these probabilities, which we denote by:



$$ell_- equiv ln big( exp(ell_1) - exp(ell_2) big)$$



Questions: How can you effectively compute this log-difference? Can this be done in the base R? If not, what is the simplest way to do it with package extensions?







r computational-statistics underflow






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share|cite|improve this question











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asked 3 hours ago









Ben

21k22499




21k22499








  • 1




    Possible duplicate of Computation of likelihood when $n$ is very large, so likelihood gets very small?
    – Xi'an
    1 hour ago










  • "where the corresponding probabilities exp(ℓ1) and exp(ℓ2) are too small to be distinguished from zero" - I think you mean one instead of zero?
    – Don Hatch
    3 mins ago














  • 1




    Possible duplicate of Computation of likelihood when $n$ is very large, so likelihood gets very small?
    – Xi'an
    1 hour ago










  • "where the corresponding probabilities exp(ℓ1) and exp(ℓ2) are too small to be distinguished from zero" - I think you mean one instead of zero?
    – Don Hatch
    3 mins ago








1




1




Possible duplicate of Computation of likelihood when $n$ is very large, so likelihood gets very small?
– Xi'an
1 hour ago




Possible duplicate of Computation of likelihood when $n$ is very large, so likelihood gets very small?
– Xi'an
1 hour ago












"where the corresponding probabilities exp(ℓ1) and exp(ℓ2) are too small to be distinguished from zero" - I think you mean one instead of zero?
– Don Hatch
3 mins ago




"where the corresponding probabilities exp(ℓ1) and exp(ℓ2) are too small to be distinguished from zero" - I think you mean one instead of zero?
– Don Hatch
3 mins ago










1 Answer
1






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up vote
2
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To see how to deal with differences of this kind, we first note a useful mathematical result concerning differences of exponentials:



$$begin{equation} begin{aligned}
exp(ell_1) - exp(ell_2)
&= exp(ell_1) (1 - exp(-(ell_1 - ell_2))). \[6pt]
end{aligned} end{equation}$$



This result converts the difference to a product, which allows us to present the log-difference as:



$$begin{equation} begin{aligned}
ell_-
&= ln big( exp(ell_1) - exp(ell_2) big) \[6pt]
&= ln big( exp(ell_1) (1 - exp(-(ell_1 - ell_2))) big) \[6pt]
&= ell_1 + ln (1 - exp(-(ell_1 - ell_2))). \[6pt]
end{aligned} end{equation}$$



In the case where $ell_1 = ell_2$ we obtain the expression $ell_+ = ell_1 + ln 0 = -infty$. Using the Maclaurin series expansion for $ln(1-x)$ we obtain the formula:



$$begin{equation} begin{aligned}
ell_-
&= ell_1 - sum_{k=1}^infty frac{exp(-k(ell_1 - ell_2))}{k} quad quad quad text{for } ell_1 neq ell_2. \[6pt]
end{aligned} end{equation}$$



Since $exp(-(ell_1 - ell_2)) < 1$ the terms in this expansion diminish rapidly (faster than exponential decay). If $ell_1 - ell_2$ is large then the terms diminish particularly rapid. In any case, this expression allows us to compute the log-sum to any desired level of accuracy by truncating the infinite sum to a desired number of terms.





Implementation in base R: It is possible to compute this log-difference accurately in base R using the log1p function. This is a primitive function in the base package that computes the value of $ln(1+x)$ for an argument $x$ (with accurate computation even for $x ll 1$). This primitive function can be used to give a simple function for the log-difference:



logdiff <- function(l1, l2) { l1 + log1p(-exp(-(l1-l2))); }


Implementation with VGAM package: Machler (2012) analyses accuracy issues in evaluating the function $ln(1-exp(-|x|))$, and suggests that use of the base R functions may involve a loss of accuracy. It is possible to compute this log-difference more accurately in using the log1mexp function in the VGAM package. This gives you the an alternative function for the log-difference:



logdiff <- function(l1, l2) { l1 + VGAM::log1mexp(l1-l2); }





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    up vote
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    down vote













    To see how to deal with differences of this kind, we first note a useful mathematical result concerning differences of exponentials:



    $$begin{equation} begin{aligned}
    exp(ell_1) - exp(ell_2)
    &= exp(ell_1) (1 - exp(-(ell_1 - ell_2))). \[6pt]
    end{aligned} end{equation}$$



    This result converts the difference to a product, which allows us to present the log-difference as:



    $$begin{equation} begin{aligned}
    ell_-
    &= ln big( exp(ell_1) - exp(ell_2) big) \[6pt]
    &= ln big( exp(ell_1) (1 - exp(-(ell_1 - ell_2))) big) \[6pt]
    &= ell_1 + ln (1 - exp(-(ell_1 - ell_2))). \[6pt]
    end{aligned} end{equation}$$



    In the case where $ell_1 = ell_2$ we obtain the expression $ell_+ = ell_1 + ln 0 = -infty$. Using the Maclaurin series expansion for $ln(1-x)$ we obtain the formula:



    $$begin{equation} begin{aligned}
    ell_-
    &= ell_1 - sum_{k=1}^infty frac{exp(-k(ell_1 - ell_2))}{k} quad quad quad text{for } ell_1 neq ell_2. \[6pt]
    end{aligned} end{equation}$$



    Since $exp(-(ell_1 - ell_2)) < 1$ the terms in this expansion diminish rapidly (faster than exponential decay). If $ell_1 - ell_2$ is large then the terms diminish particularly rapid. In any case, this expression allows us to compute the log-sum to any desired level of accuracy by truncating the infinite sum to a desired number of terms.





    Implementation in base R: It is possible to compute this log-difference accurately in base R using the log1p function. This is a primitive function in the base package that computes the value of $ln(1+x)$ for an argument $x$ (with accurate computation even for $x ll 1$). This primitive function can be used to give a simple function for the log-difference:



    logdiff <- function(l1, l2) { l1 + log1p(-exp(-(l1-l2))); }


    Implementation with VGAM package: Machler (2012) analyses accuracy issues in evaluating the function $ln(1-exp(-|x|))$, and suggests that use of the base R functions may involve a loss of accuracy. It is possible to compute this log-difference more accurately in using the log1mexp function in the VGAM package. This gives you the an alternative function for the log-difference:



    logdiff <- function(l1, l2) { l1 + VGAM::log1mexp(l1-l2); }





    share|cite|improve this answer

























      up vote
      2
      down vote













      To see how to deal with differences of this kind, we first note a useful mathematical result concerning differences of exponentials:



      $$begin{equation} begin{aligned}
      exp(ell_1) - exp(ell_2)
      &= exp(ell_1) (1 - exp(-(ell_1 - ell_2))). \[6pt]
      end{aligned} end{equation}$$



      This result converts the difference to a product, which allows us to present the log-difference as:



      $$begin{equation} begin{aligned}
      ell_-
      &= ln big( exp(ell_1) - exp(ell_2) big) \[6pt]
      &= ln big( exp(ell_1) (1 - exp(-(ell_1 - ell_2))) big) \[6pt]
      &= ell_1 + ln (1 - exp(-(ell_1 - ell_2))). \[6pt]
      end{aligned} end{equation}$$



      In the case where $ell_1 = ell_2$ we obtain the expression $ell_+ = ell_1 + ln 0 = -infty$. Using the Maclaurin series expansion for $ln(1-x)$ we obtain the formula:



      $$begin{equation} begin{aligned}
      ell_-
      &= ell_1 - sum_{k=1}^infty frac{exp(-k(ell_1 - ell_2))}{k} quad quad quad text{for } ell_1 neq ell_2. \[6pt]
      end{aligned} end{equation}$$



      Since $exp(-(ell_1 - ell_2)) < 1$ the terms in this expansion diminish rapidly (faster than exponential decay). If $ell_1 - ell_2$ is large then the terms diminish particularly rapid. In any case, this expression allows us to compute the log-sum to any desired level of accuracy by truncating the infinite sum to a desired number of terms.





      Implementation in base R: It is possible to compute this log-difference accurately in base R using the log1p function. This is a primitive function in the base package that computes the value of $ln(1+x)$ for an argument $x$ (with accurate computation even for $x ll 1$). This primitive function can be used to give a simple function for the log-difference:



      logdiff <- function(l1, l2) { l1 + log1p(-exp(-(l1-l2))); }


      Implementation with VGAM package: Machler (2012) analyses accuracy issues in evaluating the function $ln(1-exp(-|x|))$, and suggests that use of the base R functions may involve a loss of accuracy. It is possible to compute this log-difference more accurately in using the log1mexp function in the VGAM package. This gives you the an alternative function for the log-difference:



      logdiff <- function(l1, l2) { l1 + VGAM::log1mexp(l1-l2); }





      share|cite|improve this answer























        up vote
        2
        down vote










        up vote
        2
        down vote









        To see how to deal with differences of this kind, we first note a useful mathematical result concerning differences of exponentials:



        $$begin{equation} begin{aligned}
        exp(ell_1) - exp(ell_2)
        &= exp(ell_1) (1 - exp(-(ell_1 - ell_2))). \[6pt]
        end{aligned} end{equation}$$



        This result converts the difference to a product, which allows us to present the log-difference as:



        $$begin{equation} begin{aligned}
        ell_-
        &= ln big( exp(ell_1) - exp(ell_2) big) \[6pt]
        &= ln big( exp(ell_1) (1 - exp(-(ell_1 - ell_2))) big) \[6pt]
        &= ell_1 + ln (1 - exp(-(ell_1 - ell_2))). \[6pt]
        end{aligned} end{equation}$$



        In the case where $ell_1 = ell_2$ we obtain the expression $ell_+ = ell_1 + ln 0 = -infty$. Using the Maclaurin series expansion for $ln(1-x)$ we obtain the formula:



        $$begin{equation} begin{aligned}
        ell_-
        &= ell_1 - sum_{k=1}^infty frac{exp(-k(ell_1 - ell_2))}{k} quad quad quad text{for } ell_1 neq ell_2. \[6pt]
        end{aligned} end{equation}$$



        Since $exp(-(ell_1 - ell_2)) < 1$ the terms in this expansion diminish rapidly (faster than exponential decay). If $ell_1 - ell_2$ is large then the terms diminish particularly rapid. In any case, this expression allows us to compute the log-sum to any desired level of accuracy by truncating the infinite sum to a desired number of terms.





        Implementation in base R: It is possible to compute this log-difference accurately in base R using the log1p function. This is a primitive function in the base package that computes the value of $ln(1+x)$ for an argument $x$ (with accurate computation even for $x ll 1$). This primitive function can be used to give a simple function for the log-difference:



        logdiff <- function(l1, l2) { l1 + log1p(-exp(-(l1-l2))); }


        Implementation with VGAM package: Machler (2012) analyses accuracy issues in evaluating the function $ln(1-exp(-|x|))$, and suggests that use of the base R functions may involve a loss of accuracy. It is possible to compute this log-difference more accurately in using the log1mexp function in the VGAM package. This gives you the an alternative function for the log-difference:



        logdiff <- function(l1, l2) { l1 + VGAM::log1mexp(l1-l2); }





        share|cite|improve this answer












        To see how to deal with differences of this kind, we first note a useful mathematical result concerning differences of exponentials:



        $$begin{equation} begin{aligned}
        exp(ell_1) - exp(ell_2)
        &= exp(ell_1) (1 - exp(-(ell_1 - ell_2))). \[6pt]
        end{aligned} end{equation}$$



        This result converts the difference to a product, which allows us to present the log-difference as:



        $$begin{equation} begin{aligned}
        ell_-
        &= ln big( exp(ell_1) - exp(ell_2) big) \[6pt]
        &= ln big( exp(ell_1) (1 - exp(-(ell_1 - ell_2))) big) \[6pt]
        &= ell_1 + ln (1 - exp(-(ell_1 - ell_2))). \[6pt]
        end{aligned} end{equation}$$



        In the case where $ell_1 = ell_2$ we obtain the expression $ell_+ = ell_1 + ln 0 = -infty$. Using the Maclaurin series expansion for $ln(1-x)$ we obtain the formula:



        $$begin{equation} begin{aligned}
        ell_-
        &= ell_1 - sum_{k=1}^infty frac{exp(-k(ell_1 - ell_2))}{k} quad quad quad text{for } ell_1 neq ell_2. \[6pt]
        end{aligned} end{equation}$$



        Since $exp(-(ell_1 - ell_2)) < 1$ the terms in this expansion diminish rapidly (faster than exponential decay). If $ell_1 - ell_2$ is large then the terms diminish particularly rapid. In any case, this expression allows us to compute the log-sum to any desired level of accuracy by truncating the infinite sum to a desired number of terms.





        Implementation in base R: It is possible to compute this log-difference accurately in base R using the log1p function. This is a primitive function in the base package that computes the value of $ln(1+x)$ for an argument $x$ (with accurate computation even for $x ll 1$). This primitive function can be used to give a simple function for the log-difference:



        logdiff <- function(l1, l2) { l1 + log1p(-exp(-(l1-l2))); }


        Implementation with VGAM package: Machler (2012) analyses accuracy issues in evaluating the function $ln(1-exp(-|x|))$, and suggests that use of the base R functions may involve a loss of accuracy. It is possible to compute this log-difference more accurately in using the log1mexp function in the VGAM package. This gives you the an alternative function for the log-difference:



        logdiff <- function(l1, l2) { l1 + VGAM::log1mexp(l1-l2); }






        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 3 hours ago









        Ben

        21k22499




        21k22499






























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