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Distribution function problem

  1. Mar 8, 2010 #1
    1. The problem statement, all variables and given/known data

    Consider a population of individuals with a disease. Suppose that [tex]t[/tex] is the number of years since the onset of the disease. The death density function, [tex]f(t) = cte^{-kt}[/tex], approximates the fraction of the sick individuals who die in the time interval [t, t+Δt] as follows:
    Fraction who die: [tex]f(t)\Delta t = cte^{-kt} \Delta t[/tex]
    where [tex]c[/tex] and [tex]k[/tex] are positive constants whose values depend on the particular disease.

    (a) Find the value of c in terms of k.

    (b) Express the cumulative death distribution function in the form below. Your answer will be in terms of k.

    [tex]
    C(t)=\left\{\begin{array}{cc}A(t),& t < 0\\
    B(t), & t \geq 0\end{array}\right
    [/tex]

    2. Relevant equations

    [tex]P(t) = \int_{-\inf}^t p(x) dx[/tex]

    3. The attempt at a solution

    To solve part a, I know that [tex]\lim_{t\rightarrow\infty} P(t) = 1[/tex]. So c and k must have values so that it equals one. So I integrate [tex]f(t)[/tex] with the relevant equations to get:

    [tex]
    \frac{-(kt+1) \cdot e^{-kt} \cdot c}{k^2}
    [/tex]

    from negative infinity to t.

    The problem is that this diverges to negative infinity and doesn't give me a meaningful answer. So what do I do?


    Also, how do I get the the bar with a superscript and subscript for 'from a to b' in tex?
     
  2. jcsd
  3. Mar 8, 2010 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi altegron! :smile:
    No, nobody died before the onset of the disease! :wink:

    (and that bar would be [tex]|_a^b\ \big|_a^b\ \Big|_a^b\ \bigg|_a^b\ \Bigg|_a^b[/tex])
     
  4. Mar 8, 2010 #3
    Thanks for the response,

    That makes sense -- I guess that means I should integrate from 0 to infinity then?

    [tex]
    \lim_{x\rightarrow\infty} \int_0^x cte^{-kt} dt = 1
    [/tex]

    Which is easier written as

    [tex]
    \int_0^{\infty} cte^{-kt} dt = 1
    [/tex]

    Which is:

    [tex]
    \frac{-(kt+1) \cdot e^{-kt} \cdot c}{k^2} \bigg|_0^{\infty} = 1
    [/tex]

    The infinity term goes to zero and the zero term (negated for subtraction) is:

    [tex]
    \frac{(k \cdot (0)+1) \cdot e^{-k \cdot (0)} \cdot c}{k^2} = 1
    [/tex]

    [tex]
    \frac{(0+1) \cdot e^{0} \cdot c}{k^2} = 1
    [/tex]

    [tex]
    \frac{c}{k^2} = 1
    [/tex]

    So, finally:

    [tex]
    c = k^2
    [/tex]

    That seems plausible but I don't know how to check it.

    Then for part b, I know [tex]A(t) = 0[/tex] because the disease hasn't started yet.

    For [tex]B(t)[/tex] we use the integral we've already computed, from zero to t years. We know the value at zero is 1 (actually, -1, but it is being subtracted), so we have:

    [tex]
    B(t) = 1 - \frac{(kt+1) \cdot e^{-kt} \cdot c}{k^2}
    [/tex]

    And that looks correct on my calculator when I graph it with k=1.
     
  5. Mar 8, 2010 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Looks good! :biggrin:

    (btw, that | is really only used for "value at", with only a subscript …

    for integrals, use [] … LaTeX will automatically size them to fit if you type "\left[" and "\right]" :wink:)
     
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