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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Ferdi

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up vote
4
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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.

edited just now

Ferdi

3,47842151

asked 27 mins ago

Hazardous

211

New contributor

add a commentÂ |Â

up vote
4
down vote

favorite

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.

edited just now

Ferdi

3,47842151

asked 27 mins ago

Hazardous

211

New contributor

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

edited just now

Ferdi

3,47842151

asked 27 mins ago

Hazardous

211

New contributor

edited just now

Ferdi

3,47842151

asked 27 mins ago

Hazardous

211

New contributor

edited just now

Ferdi

3,47842151

edited just now

Ferdi

3,47842151

edited just now

Ferdi

3,47842151

asked 27 mins ago

Hazardous

211

New contributor

asked 27 mins ago

Hazardous

211

asked 27 mins ago

Hazardous

211

New contributor

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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.

edited 7 mins ago

answered 13 mins ago

InfProbSciX

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1 Answer
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1 Answer
1

active

oldest

votes

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.

edited 7 mins ago

answered 13 mins ago

InfProbSciX

44910

add a commentÂ |Â

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.

edited 7 mins ago

answered 13 mins ago

InfProbSciX

44910

add a commentÂ |Â

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.

edited 7 mins ago

answered 13 mins ago

InfProbSciX

44910

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.

edited 7 mins ago

answered 13 mins ago

InfProbSciX

44910

edited 7 mins ago

answered 13 mins ago

InfProbSciX

44910

answered 13 mins ago

InfProbSciX

44910

answered 13 mins ago

InfProbSciX

44910

add a commentÂ |Â

Hazardous is a new contributor. Be nice, and check out our Code of Conduct.

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