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Why is this correct - particle lifetime probability distribution

  1. Sep 18, 2013 #1

    The following problem can be found in van Kampen's "Stochastic Processes in Physics and Chemistry", Third Edition (Exercise I.3.7):

    The probability distribution of lifetimes in a population is P(t). Show that the conditional probability for individuals of age τ is
    P(t|τ) = \frac{P(t)}{\int_τ^{\infty} P(t') dt'} \qquad (t>τ)
    Note that in the case $$P(t)=\gamma e^{-\gamma t}$$ one has $$P(t|\tau)=P(t-\tau),$$ the survival chance is independent of age. Show that this is the only P for which that is true.

    Now, I am only interested in the first part of the problem - the expression for P(t|τ). A solution is given here:
    and I find it correct.

    What I do not understand is: P(t|τ) is a conditional probability density. As such it should be expressed as a fraction of two probability densities. However, the denominator in the problem above is NOT a probability density, but actual probability (that the particle has age >τ).

    Where is the catch?
  2. jcsd
  3. Sep 18, 2013 #2


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    Science Advisor

    no, why?
    After all, you condition on a real probability, namely that t>τ, not on a density.
  4. Sep 18, 2013 #3
    Exactly, right?
    But the definition says otherwise.

    My point is: from the definition of probability density,
    $$P_X(x) dx$$
    is the probability that a random variable X has a value in [x,x+dx]. P(x) by itself is NOT the probability that the value of the random variable X will by precisely x. That is clear.

    But the conditional probability density P_{X|Y}(x|y) has the following interpretation:
    $$ P_{X|Y}(x|y) dx $$ is the probability that X will have a value from the set [x,x+dx], given that Y==y. Precisely equal, and not that Y will be in [y,y+dy].

    And the Bayes' rule for conditional probability density is:
    P_{X|Y}(x|y) = \frac{P_{X,Y}(x,y)}{P_Y(y)}
    where all P-s, in particular P_Y, are probability densities.
    (If the denominator vanishes, the conditional probability is not defined - conditions cannot be met.)
    P_Y(y) is not the probability that Y==y.

    And yet (going back to my original problem) I have P(τ) in the denominator, which is not a probability density, but a crude probability.

    Why isn't there a probability density, that the lifetime of the particle is τ?
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