Alternate T-Rex Solution

[Anonymous]

Hi Everyone,
I was just looking at the Nov 2001 exam and answer key posted by the SOA, and I don't agree with the answer they gave for the T-Rex question of c) .5

My solution which I think is totally valid gives an answer of a) .3. Let me know what you guys think. Maybe I'm way off here, but I don't think so.

Alternate solution

T-Rex starts with 10,000 calories. He burns calories at a uniform rate of 10,000 per day. So 2.5 days later you would assume he had burned 25,000 calories.
Now since he started with 10,000 calories, that leaves T-Rex with 15,000 calories to make up.

To make up these calories he has to eat scientists. Since, each scientist is worth 10,000 calories, then he would have to eat two scientist in the 2.5 days to make up the 15,000 calories. Thus the probability that he dies, is if he eats either zero or one scientist in the 2.5 days.

Now since we have a poisson distribution lambda equal to one, lambda*time = 2.5
So the probability
he eats zero scientists in 2.5 days is
exp(- 2.5);
he eats one scientist in 2.5 days is
2.5exp(-2.5)
which gives a probability of him dying equal to .287 answer a) .3 not c) .5?

This leads me to the second part of this problem. Question 11, What is the expected calories eaten in the next 2.5 days.
Well one would assume, it would be zero if he dies, and 25,000 if he lives, probability .7, thus the expected value would be 25,000*(.7)= 17,500
Which would be the closest answer a) 17,800. NOT b.

What do you guys think?

[Anonymous]

ur answer underestimates the prob. of dying because the trex has to eat at least one scientist before his initial 10k calories are gone.

[Anonymous]

in other words, ur criterion of eating 2 scientists in 2.5 days is too lax.

[Anonymous]

Sorry Roger Lee, could you elaborate a little more. I don't quite understand.

[Actuary321]

The probablity that the first waiting time > 1.0 days plus probability that time to second is > 2.0 days.

sorry can't elaborate beyond that.

and when I say can't, I mean not able, rather than not willing.

[Anonymous]

your probability of the trex surviving 2.5 days is too high .. because ur definition of death is not stringent enough.

for instance, your answer says that the trex survives as long as he eats 2 scientists in 2.5 days.

which includes the trex eating the 2 scientists on day 2. however, he wouldn't have survived past day 1 to do that.

[Anonymous]

I see what your saying now. The time when the T-Rex eats the scientists is important.

Man this is so annoying. (Getting questions wrong)

[AOnly]

This is how far I got:

Let T_1 and T_2 the arrival time of two scientists. It is not known which comes first, although their arrival times are generated by a poisson process with rate 1.

A necessary condition for the dinosaur to survive:
1. min{T_1,T_2} < 1

Assuming this occurs, a second meal must be consumed before time 2. Otherwise 20000 calories will have been consumed and only 20000 created (incl. original surplus) However if a second meal is eaten, survival to 2.5 is guaranteed. Thus the condition for survival is

2. min{T_1,T_2} < 1
max{T_1,T_2} < 2

If you plot this region on the (T_1,T_2) plane, you get an "L-shaped region" with corners (0,0) (2,0) (2,1) (1,1) (1,2) (0,2)

The probability of this region can then be computed by integration. The joint density of (T_1=t,T_2=s) is exp(-t)exp(-s), assuming the arrivals are independent.

Of course, this is highly unlikely. I think the gruesome demise of your fellow scientist might influence your future behavior, but independence is the assumption unless stated otherwise.

When I integrate this, I get .6935, which would give me the wrong answer. Any ideas?

This vaguely reminds me of those joint status problems from Chapter 9 of Bowers. You have the first to arrive and the last to arrive random variables. I suspect you can rig this up using the appropriate q formulas from multiple lives theory. I'm working on that approach now.

x=status of first scientist
y=status of second scientist
q_{xy}=probability one has died before time 1
2q_{overline{xy}}=probability both have died before time 2


<font size=-1>[ This Message was edited by: AOnly on 2001-11-09 12:40 ]</font>

<font size=-1>[ This Message was edited by: AOnly on 2001-11-09 13:21 ]</font>

[Macroman]

AOnly:

You are making this much too hard.

1) There are not just two scientists, but infinitely many. They are expected to arrive at the rate of 1 per day, but the possibility exists (although remote) that 500,000 might show up in one day. Similarily, there is some possibility that there might be no scientists for a year. The Poisson distribution will do these things to you.

2) The solution by Math Guy is much too neat and accurate not to go with.

[Flying Squirrel]

I don't believe your assumption that arrival (meal) times are distributed as a poisson is correct. The question only states that Rex eats scientists at a poisson rate of one per day. If meal times were distributed like a poisson, then your way may work. However, it seems to me that you are dealing with discrete probabilities which, of course, cannot be integrated. The solution on the other thread does seem correct, that it is P(0)+P(1)*P(0).

[AOnly]

I have not seen this other solution to which you refer, but it is important to me that if a method is correct (even if too hard) then it should give the correct answer. I still like to see a problem from all angles. I absolutely hated these problems when I studied the material. Long term ruin, I liked, but these short term ones were a combinatorial pain.

I agree there are not just two scientists. But the arrival times of the first two determine the probability of survival.

If you think about the sample space being ALL interrarival times (T_1,T_2,....), and S being the set of all such points that results in survival, then P(S) is the measure of this set. There may be more than two scientist, but only T_1 and T_2 determine survival to time 2.5.

I do think meal times are poisson distributed. Presumably the only doubt as to T-Rex's survival is his ability to find food. Adn this is controlled by a poisson process.

For the Poisson process, there is a theorem that states that if you are told only that if the second arrival occurs at (or is that before?) t=2, then the arrival time of the first is Uniformly distributed on [0,1]. I think this is why the SOA said 0.5 is the probability. But i don't think this probability is the solution to the problem.


The other important issue is that if I made a mistake, that I know how to find it.

I am going to run a simulation study and see what the results look like. Did I hear somebody whisper (obsessive-compulsive?). I'll ignore it :wink:

1. Simulate time of first meal by taking a uniform RV U_1 and computing T_1=-log[U_1].

2. Compute # of calories remaining just after the meal

3. Compute U_2 and T_2=-log[U_2]. See if this occurs before # of calories expire.