MHB Probability more than one claim filed?

[schinb65]

In modeling the number of claims filed by an individual under an automobile policy
during a three-year period, an actuary makes the simplifying assumption that for all
integers n ≥ 0, $p_n+1 = \frac{1}{5} p_n$ , where $p_n$ represents the probability that the policyholder files $n$ claims during the period.
Under this assumption, what is the probability that a policyholder files more than one
claim during the period?

So my question is the solution which I have attached. How do we go from
$\sum_{k=0}^{\infty}\frac{1}{5}^kp_0$ to the next step $\frac{p_0}{1-\frac{1}{5}}$

Thank you.

 

[I like Serena]

Re: Probability more then one claim filed?

In modeling the number of claims filed by an individual under an automobile policy
during a three-year period, an actuary makes the simplifying assumption that for all
integers n ≥ 0, $p_n+1 = \frac{1}{5} p_n$ , where $p_n$ represents the probability that the policyholder files $n$ claims during the period.
Under this assumption, what is the probability that a policyholder files more than one
claim during the period?

So my question is the solution which I have attached. How do we go from
$\sum_{k=0}^{\infty}\frac{1}{5}^kp_0$ to the next step $\frac{p_0}{1-\frac{1}{5}}$

Thank you.

Welcome to MHB, schinb65! :)

What you have is a so called geometric series:

$a+ar+ar^2+ar^3+ar^4+\cdots = \displaystyle\sum_{k=0}^\infty ar^k = \dfrac{a}{1-r}$​

In your case you have $a=p_0$ and $r=\frac 1 5$, so:

$p_0 + \frac 1 5 p_0 + (\frac 1 5)^2 p_0 + (\frac 1 5)^3 p_0 + (\frac 1 5)^4 p_0 + ... = \displaystyle\sum_{k=0}^{\infty} \left(\frac{1}{5}\right)^k p_0 = \dfrac{p_0}{1-\frac 1 5}$​

(Btw, I suspect you intended $p_{n+1} = \frac{1}{5} p_n$.)

 

[Jameson]

Re: Probability more then one claim filed?

Welcome to MHB, schinb65! :)

What you have is a so called geometric series:

$a+ar+ar^2+ar^3+ar^4+\cdots = \displaystyle\sum_{k=0}^\infty ar^k = \dfrac{a}{1-r}$​

In your case you have $a=p_0$ and $r=\frac 1 5$, so:

$p_0 + \frac 1 5 p_0 + (\frac 1 5)^2 p_0 + (\frac 1 5)^3 p_0 + (\frac 1 5)^4 p_0 + ... = \displaystyle\sum_{k=0}^{\infty} \left(\frac{1}{5}\right)^k p_0 = \dfrac{p_0}{1-\frac 1 5}$​

(Btw, I suspect you intended $p_{n+1} = \frac{1}{5} p_n$.)

Hi ILikeSerena, :)

I can't spot an error in this but I don't see where you used the fact that we are calculating $P[n>1]$. I would say we need to solve it this way: $P[n>1]=1-P[n=0]-P[n=1]$. Is that what you did in fact?

 

[I like Serena]

Re: Probability more then one claim filed?

Hi ILikeSerena, :)

I can't spot an error in this but I don't see where you used the fact that we are calculating $P[n>1]$. I would say we need to solve it this way: $P[n>1]=1-P[n=0]-P[n=1]$. Is that what you did in fact?

Hi Jameson ;)

The attachment in the OP contains the full solution including what you just mentioned.
I only clarified the step in it that schinb65 asked about.

 

[Jameson]

Re: Probability more then one claim filed?

Hi Jameson ;)

The attachment in the OP contains the full solution including what you just mentioned.
I only clarified the step in it that schinb65 asked about.

(Headbang) Doh! Sorry about that. I just saw the full solution just before reading you reply. I'll try to be more observant in the future, haha. :)