| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Given the life table function Q(x), P(X > x) Probability that a
person survives at least x years. What is the probability that having
reached x years the person survives at least y further years?
Tony.
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 910.1 | BEING::POSTPISCHIL | Always mount a scratch monkey. | Thu Jul 28 1988 09:48 | 16 | |
Re .0:
P(x+y)/P(x) -- also known as Bayes' Rule: The probability that A
occurs given that B occurs is equal to the probability of both A and B
happening divided by the probability that B happens. (In other words,
it is the proportion of space that "A and B" takes up in the space "B
happens".)
In this case, A is "survives at least y further years" which is
equivalent to "survives at least x+y years". B is "survives at least x
years". "A and B" is simply A, since "surviving x+y years and
surviving x years" is equivalent to "surviving x+y years". So the
answer is P(A and B)/P(B) = P(A)/P(B).
-- edp
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| 910.2 | B is x years | VIVIAN::MILTON | I'm thinking about it! | Thu Jul 28 1988 12:42 | 13 |
> B is "survives at least x years".
B is the more specific - given exactly x years, the prob that survives
a further y years:-
P(X>=x+y|x=x)
does that make a difference.
I have a problem in that I think Q(x) is a continuous random variable
and as such P(X=x)=0 or am I confusing things?
Tony.
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| 910.3 | BEING::POSTPISCHIL | Always mount a scratch monkey. | Thu Jul 28 1988 17:15 | 6 | |
Re .2:
P(X >= x+y | X >= x) = P(X >= x+y) / P(X >= x).
-- edp
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