I'm stuck with this probability exercise

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Gudjon.Smith

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The expected number of moles in a park depends on the time since the last maintenance. The expected number is given by ct^2 ( c times t square) as a function of time since the last maintenance. The constant c is given by c=4,6 moles per year. The maintenances are done exactly every 5 years. Calculate the expectation and variance of the number of moles from the point of view of a person who has no idea when the last maintenance occurred.


Solution so far:
First of all the expected number of moles increases quadratically.
I calculated the expectation of the number of moles: E[X]= integral[0,5] ct^2 dt = 191,67
Now this is the part where I'm stuck. First of all I don't know if I calculated the expectation correctly. Second of all the variance=E[X^2]-E^2[X] but I have no idea how to calculate E[X^2]. Any help?
 
The first step is to determine what the distribution of time of last maintenance is. Can you do that?
 
Why did you integrate? From what I read and understand the expected number of moles as a function of time is ct^2. There is no need to add up the number of moles, the given formula does that for you. Next question is why are you finding the number of moles after 5 years. Is 5 years the average time period for someone who randomly asks how many moles are there?
 
Romsek said:
The first step is to determine what the distribution of time of last maintenance is. Can you do that?
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Do you mean like determine the number of moles for t=0, t=1, t=2, t=3, t=4 and t=5 by using ct^2 and replacing t?
 
Jomo said:
Why did you integrate? From what I read and understand the expected number of moles as a function of time is ct^2. There is no need to add up the number of moles, the given formula does that for you. Next question is why are you finding the number of moles after 5 years. Is 5 years the average time period for someone who randomly asks how many moles are there?
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Honestly now that you mention it, no 5 years isn't the average time period for someone who randomly asks how many moles are there. I've been struggling with this exercise for a little while now so I asked my teacher for a hint and he told me at some point I would have to integrate. Since I've been struggling to find a good starting point to solve the exercise I thought I would integrate as my teacher suggested I would have to do at some point in the exercise to maybe figure out how to solve it eventually. But it's wrong and I'm back to the starting point...
 
Last edited:
First thing is to identify the distribution of the last date maintenance occurred from the point of view of someone w/no knowledge beforehand.

The best case is that it took place right as the person was considering it. The worst case is that it took place 5 yrs ago. We have no reason to expect that any time within these limits is any more probably than another so it should be clear that this gives us a uniform distribution between 0 and 5 yrs.

That is

[MATH]p_T(t) = \begin{cases} \dfrac{1}{5} &0 \leq t \leq 5\\0&\text{else} \end{cases}[/MATH]
Now. The expectation of a function of a random variable [MATH]T[/MATH] is given as

[MATH]E[f(T)] = \displaystyle \int ~f(t) p(t) ~dt[/MATH]
In this case that equates to

[MATH]E[c t^2] = \displaystyle \int \limits_0^5 c t^2 \cdot \dfrac 1 5~dt[/MATH]
The variance is found as follows

[MATH]Var[f(T)] = E[f(T)^2] - \left(E[f(T)]\right)^2[/MATH]
and is calculated in a similar fashion.

I leave you to your integrals.
 
@Romsek I am a bit confused here. If ct^2 was the number of moles at time t then I understand why you integrate. However ct0^2 is the number of moles from t=0 to t=t0 so why integrate?
 
Jomo said:
@Romsek I am a bit confused here. If ct^2 was the number of moles at time t then I understand why you integrate. However ct0^2 is the number of moles from t=0 to t=t0 so why integrate?
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I don't see anything about [MATH]c t^2[/MATH] being the number of moles from [MATH]t_0 \to t[/MATH]
I just see it as the instantaneous number of moles at time [MATH]t[/MATH]
There is a questionable use of the word "expected" in the problem statement. That might mean something other than I think.
 
First of all thank you very much for your help so far. I've spoken to my teacher and here is what he told me:

"Here we give the expectation of the number of moles as a function of time since the last maintenance. This time is, itself, a random variable: we therefore speak of a conditional expectation! You have to think about the law followed by the number of moles (in combination with the expectation) to know the variance. In this exercise, you will also need the probabilistic law for the time since the last maintenance."

I really hope that I translated this correctly into english, let me know if something doesn't make any sense, probably a translation mistake...but anyway given what he told me I'm pretty sure that @Romsek 's way of solving the exercise is correct.
 
The problem right now is if you do not show us your work we can't find your error.
 
Jomo said:
The problem right now is if you do not show us your work we can't find your error.
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I found my mistake: for the variance instead of calculating 1/5 * (ct^2)^2 I calculated (1/5)^2 * ct^2 and by doing so I found a negative variance, my bad
 
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