poisson process

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October 7th 2009, 03:01 AM
Pengu

poisson process

For this question:
Eva catches two kinds of fish - salmon and trout. Assume salmons and trouts come independently at arrival times of Poisson process with rate 1 per hour and 2 per hour.

Given that Eva caught 3 salmons in the first hour, what is the expected number of fishes she could have caught for the first two hours?

the answer is:
$ E(N_2^{(s)} + N_2^{(t)} | N_1^{(s)} = 3)$
$= E(N_2^{(s)} - N_1^{(s)}| N_1^{(s)} = 3) + E(N_1^{(s)}|N_1^{(s)} = 3) + E(N_2^{(t)})$
$= 1 + 3 + (2)(2) = 8$

can someone please explain how to get the second line? like is there some sort of formula to get that?
October 7th 2009, 01:35 PM
Laurent

Quote:

Originally Posted by Pengu

For this question:
Eva catches two kinds of fish - salmon and trout. Assume salmons and trouts come independently at arrival times of Poisson process with rate 1 per hour and 2 per hour.

Given that Eva caught 3 salmons in the first hour, what is the expected number of fishes she could have caught for the first two hours?

the answer is:
$ E(N_2^{(s)} + N_2^{(t)} | N_1^{(s)} = 3)$
$= E(N_2^{(s)} - N_1^{(s)}| N_1^{(s)} = 3) + E(N_1^{(s)}|N_1^{(s)} = 3) + E(N_2^{(t)})$
$= 1 + 3 + (2)(2) = 8$

can someone please explain how to get the second line? like is there some sort of formula to get that?

It is rather the third line that requires explanation. Fact is that $N^{(s)}_1$ and $N^{(s)}_2 - N^{(s)}_1$ are independent, hence $E[N^{(s)}_2 - N^{(s)}_1|N^{(s)}_1=n]=E[N^{(s)}_2 - N^{(s)}_1]$ (for any $n$ ). And $N^{(s)}_2 - N^{(s)}_1$ has same distribution as $N^{(s)}_{2-1}=N^{(s)}_1$ .

The aim of the second line is to reduce to independent random variables and thereby enable computation. Indeed $N^{(s)}_2$ and $N^{(s)}_1$ are correlated but, as second line shows, you can rewrite the formula in terms of $N^{(s)}_2-N^{(s)}_1$ and $N^{(s)}_1$ , which are independent.

And of course $N^{(t)}$ and $N^{(s)}$ are independent as well, so that this term is easily computed: $E(N_2^{(t)}|N_1^{(s)}=n)=E(N_2^{(t)})$ .
October 7th 2009, 04:20 PM
Katina88

Your first line is the total number of fish caught - including salmon and trout.

After that you can add this line:

$ E[(N_2^{(s)} - N_1^{(s)}) + N_1^{(s)} + N_2^{(t)} | N_1^{(s)} = 3] $

explanation:
$ E[(N_s^{(s)} - N_2^{(t)})$ this part means the number of salmon caught between the second and first hour..so that's the number of salmon caught in hour 2

$+ N_1^{(s)}$ then you have to add the number of salmon caught in the first hour

$E[N_2^{(t)}$ this is the total number of trout caught in the first and second hour

so to work out that question:

http://www.mathhelpforum.com/math-he...840bcb51-1.gif
$ E[(N_2^{(s)} + N_1^{(s)}) + N_1^{(s)} + N_2^{(t)} | N_1^{(s)} = 3] $
http://www.mathhelpforum.com/math-he...4e4a999e-1.gif
http://www.mathhelpforum.com/math-he...8d6fd9ce-1.gif

the line that you were asking about is now the third line, and basically they just separated the parts in the second line