What is the distribution function for a population with a disease?

[altegron]

Homework Statement

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

<br /> C(t)=\left\{\begin{array}{cc}A(t),&amp; t &lt; 0\\<br /> B(t), &amp; t \geq 0\end{array}\right<br />

Homework Equations

P(t) = \int_{-\inf}^t p(x) dx

The Attempt at a Solution

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

<br /> \frac{-(kt+1) \cdot e^{-kt} \cdot c}{k^2}<br />

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?

 

[tiny-tim]

Hi altegron!

Consider a population of individuals with a disease. Suppose that t is the number of years since the onset of the disease.
…
P(t) = \int_{-\infty}^t p(x) dx

No, nobody died before the onset of the disease!

(and that bar would be |_a^b\ \big|_a^b\ \Big|_a^b\ \bigg|_a^b\ \Bigg|_a^b)

 

[altegron]

Hi altegron!


No, nobody died before the onset of the disease!

(and that bar would be |_a^b\ \big|_a^b\ \Big|_a^b\ \bigg|_a^b\ \Bigg|_a^b)

Thanks for the response,

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

<br /> \lim_{x\rightarrow\infty} \int_0^x cte^{-kt} dt = 1<br />

Which is easier written as

<br /> \int_0^{\infty} cte^{-kt} dt = 1<br />

Which is:

<br /> \frac{-(kt+1) \cdot e^{-kt} \cdot c}{k^2} \bigg|_0^{\infty} = 1 <br />

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

<br /> \frac{(k \cdot (0)+1) \cdot e^{-k \cdot (0)} \cdot c}{k^2} = 1 <br />

<br /> \frac{(0+1) \cdot e^{0} \cdot c}{k^2} = 1 <br />

<br /> \frac{c}{k^2} = 1 <br />

So, finally:

<br /> c = k^2<br />

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

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

For B(t) 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:

<br /> B(t) = 1 - \frac{(kt+1) \cdot e^{-kt} \cdot c}{k^2}<br />

And that looks correct on my calculator when I graph it with k=1.

 

[tiny-tim]

Looks good!

(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]" )