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View Full Version : Course 3, May 2001 #28

boognish
11-04-2001, 09:01 PM
This seems like it should be an easy problem!! Any insight into how to approach beyond the solution the SOA gives?

Thanks.

phdmom
11-04-2001, 10:57 PM
Maybe this is more intuitive....

The probability that an individual dies within one year is Pr(T<1) = integral from 0 to 1 of f(t)dt. You're given that the conditional distribution of t (given the force of mortality is m) is f(t)=me^(-mt), so to find the distribution of t, you need to multiply by the distribution of m and integrate over m.

Therefore f(t) = integral from 0 to 2 of (1/2)me^(-mt)dm, and so Pr(T<1) is the double integral. To me it's easier to set up this way (integral from 0 to 1, integral from 0 to 2 (1/2)me^(-mt) dm dt), then switch the order of integration to avoid int. by parts.

Whew, I don't know if that was any clearer than the answer key. I can think much better when I can write the symbols.

Now, what are your thoughts on #8? :wink:

boognish
11-05-2001, 07:35 AM
How would the answer change if it was uniformly distributed b/w (0,10)? What the second integral just be integrated from 1-10, then multiply the result by 1/10?

Thanks.

phdmom
11-05-2001, 08:20 AM
Yes, that's right.

Anonymous
11-05-2001, 01:43 PM
Double integrals seem like a little bit of overkill here.

Think of the probability px as the expected value of (e to the negative mu power), where mu is a random variable.

The expected value of (e to the negative mu) is the integral (from 0 to 2) of (e to the negative mu times 1/2 d(mu)).

Therefore you have an easy single integral, producing .4323324.

Note that if you make the incorrect assumption that mu = 1, you obtain Answer E.

The required answer, the complement of that, is then .5676676.

boognish
11-05-2001, 02:09 PM
Professor,

Are you forgetting a mu in that first integral? This seems like May 2000 #18, although the solution they give for that question seems to make more sense to me. Your integrating f(t) from (0,2)times (1/2)d(mu). Thats s(x), then just solve for F(1).

I don't really understand the theory behind the statistics of the question, but if its a variant of this type of question, I think i can work the mechanics of it.

Anonymous
11-05-2001, 02:38 PM
Candyland:

I don't see the connection between this problem and #18 on May, 2000. In that 2000 problem, the force of mortality is given to be a specific constant, while in #28 on May, 2001, the force of interest is a uniformly distributed random variable.

boognish
11-05-2001, 03:52 PM
Professor,

I meant May 2000 #17!!

Sorry, does that make sense now?

Anonymous
11-05-2001, 04:58 PM
Candyland:

You are exactly right---those two questions are virtually identical. The 2000 problem involves the probability of living half a year, while the 2001 problem involves the probability of living a whole year. In each case, the constant force of mortality is assumed to be a uniformly distributed random variable.

However, I still do not see what you mean by my having left out a "mu" in my solution to the 2001 question. There is no "mu" (or in this case, "theta") multiplier, nor should there be, in the SOA solution to the 2000 question either.

Good luck tomorrow!

boognish
11-05-2001, 05:02 PM
Professor:

I setup the integral in this way:

s(x)= 1/2 (integral from 0 to 2) mu exp^[-mu(t)]dt

then solved for 1-s(1)

so in the integral that mu is the one i spoke of.

Am I missing something here?

Anonymous
11-05-2001, 06:36 PM
Candyland:

Your set-up of the integral can't be correct, as you have s(x) on the left hand side, but no "x" on the right-hand side, making your s(x) a constant (not a function of x), which is impossible. Your s(0), for example, would not be 1, as it must be.

The confusion may be that the variable of integration must be "mu", not "t", just as it is "theta" in the 2000 problem. "Mu" is the variable for which the distribution is given; "t" has no relevance.

Go back and re-think the solution I posted at 13:43; I think it'll make sense now.

boognish
11-05-2001, 06:50 PM
I meant s(t) on the left hand side, good observation.

I rethought your approach and it does seem to be much easier. Thanks.