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Science Forum Index » Statistics - Math Forum » Number of events
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| Alejandro113 |
 Posted: Thu Mar 08, 2007 12:13 pm |
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Hi, I`m a bit confused:
Giver the probability density funcion that a component will faill "f" (the cumulative distribution is F), an interval of time T, and a sub-interval R. Failures occur in an instant of time (have no duration), they are simply events in the interval. So, what`s wrong with this:
" There are T/R sub intervals of lenght R in the interval T. So T/R*(1-F(R)) is the mean number of intervals of length R in wich the component doesn`t fail. So
R*(T/R)*(1-F(R)) is the mean time when no failures are observed. But that doesn`t make sense, because failures are supposed instantaneous events".
Also, how many times a component will fail in a interval of time of length T, given the probability of failure?
Im confused. Thanks for your time. |
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| David Winsemius |
| Posted: Sun Mar 11, 2007 10:12 pm |
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Alejandro113 <alejandro113@gmail.com> wrote in
news:21360617.1173392035385.JavaMail.jakarta@nitrogen.mathforum.org:
Quote: Hi, I`m a bit confused:
Giver the probability density funcion that a component will faill "f"
(the cumulative distribution is F), an interval of time T, and a
sub-interval R. Failures occur in an instant of time (have no
duration), they are simply events in the interval. So, what`s wrong
with this: " There are T/R sub intervals of lenght R in the interval
T. So T/R*(1-F(R)) is the mean number of intervals of length R in wich
the component doesn`t fail.
That does not look correct. R is presumably the length of the "sub-
interval" you are constructing, whereas a cumulative distribution
function would take as its argument a particular time point rather than
an interval. The two values of F(.) that will be interesting are
F(end_time) and F(start_time). No such animal as F(end - start).
Quote: So R*(T/R)*(1-F(R)) is the mean time when
no failures are observed. But that doesn`t make sense, because
failures are supposed instantaneous events".
Now you are tying yourself in knots. Intervals have positive measure
whereas the measure of a finite set of instants would be zero. The marvel
of calculus is that you can start with a density that has measure zero at
every point and then construct a sum that is non-zero. F(.) is just the
sum (integral) from the lowest end of the range of the function to any
particular point.
Quote: Also, how many times a component will fail in a interval of time of
length T, given the probability of failure?
That does not parse as an English sentence. I am guessing you want a way
of estimating the number of events from the non-decreasing CDF. Take t_1
and t_2 (greater than t_1) to be the end points of your interval. If N is
number of items at risk (assuming no censoring), the number of estimated
events in the interval would be:
N*(F(t_2)-F(t_1))
--
David Winsemius |
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