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# Renewal process, generating function question Watch

1. I'm on 3b(i).

I have no idea where to begin.

I'm sure this has something to do with the fact that the interrarival times of a Poisson process are exponentially distributed but I can't see the link.

Part (ii) is fine, and part 3 is really simple, lol.

Part (c) is also a problem. Can someone guide me in the right direction?
2. for 3bi, X_t has poisson distribution with parameter lambda*t so you can use the pmf of a poisson distribution to get P(X_t=k).

Then, so you can try and get this into the form of a power series expansion for 'e to the power of something'.
3. (Original post by ttoby)
for 3bi, X_t has poisson distribution with parameter lambda*t so you can use the pmf of a poisson distribution to get P(X_t=k).

Then, so you can try and get this into the form of a power series expansion for 'e to the power of something'.
Okay, that makes a lot of sense. I think I just confused interarrival times with just the number of arrivals.

For part (c), I'm not sure what's going on.

(c) (i): If the number of arrivals is even, then I understand that we have one renewal every two arrivals, so therefore .

Now if is odd, then since every second arrival is a renewal and our counting process for starts from 1, then we have 2, 4, 6... counting towards the renewal and if we land on say, 7, then we have .

Is this a sufficient explanation?

For (ii), I got:

so

And since P(Even)+P(odd) = 1, I get something close but not quite what I need. What am I doing wrong?

Edit: This does however give me exactly what I need for the last part. Perhaps it's an error?
4. (Original post by wanderlust.xx)
Okay, that makes a lot of sense. I think I just confused interarrival times with just the number of arrivals.

For part (c), I'm not sure what's going on.

(c) (i): If the number of arrivals is even, then I understand that we have one renewal every two arrivals, so therefore .

Now if is odd, then since every second arrival is a renewal and our counting process for starts from 1, then we have 2, 4, 6... counting towards the renewal and if we land on say, 7, then we have .

Is this a sufficient explanation?

For (ii), I got:

so

And since P(Even)+P(odd) = 1, I get something close but not quite what I need. What am I doing wrong?

Edit: This does however give me exactly what I need for the last part. Perhaps it's an error?
For part i, you need to show it in general because if you just give an example then they could say 'well you showed it for 7 but that doesn't necessarily mean it's true for every other odd number' and dock marks. What you could try is an arguement along the lines of 'if then arrivals 2, 4, 6, 8, ..., 2n are counted and there are of these' then something similar for odd numbers.

For ii, I haven't seen the term 'renewal function' before, but assuming you're correct when you say then the rest of your workings do follow so perhaps it's just a typo there.
5. (Original post by ttoby)
For part i, you need to show it in general because if you just give an example then they could say 'well you showed it for 7 but that doesn't necessarily mean it's true for every other odd number' and dock marks. What you could try is an arguement along the lines of 'if then arrivals 2, 4, 6, 8, ..., 2n are counted and there are of these' then something similar for odd numbers.

For ii, I haven't seen the term 'renewal function' before, but assuming you're correct when you say then the rest of your workings do follow so perhaps it's just a typo there.
Yeah it's definitely a typo, I double checked - turns out they don't update papers.

And yeah, I did generalise the case when I wrote it down. Thanks for that.

Updated: April 19, 2012
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