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Distribution function problem

  • Thread starter altegron
  • Start date
  • #1
14
2

Homework Statement



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]

Homework Equations



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

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?
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
250
Hi altegron! :smile:
Consider a population of individuals with a disease. Suppose that [tex]t[/tex] is the number of years since the onset of the disease.

[tex]P(t) = \int_{-\infty}^t p(x) dx[/tex]
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])
 
  • #3
14
2
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])
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.
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
250
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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