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 $n \ge 0$, $p_{n+1}=\frac{1}{5} p_n$, 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?
My first step was the convert the probability function from future values by variable substitution. So I assign $k=n+1 \implies n=k-1$:
$\therefore p_k=\frac{1}{5} p_{k-1} \text{ for } k \ge 1\tag{1}$
Now we can calculate the probabilites for some k values:
$k=0: p_0$
$k=1: p_1=\frac{1}{5}p_0$
$k=2: p_2=\frac{1}{5}p_1=(\frac{1}{5})^2 p_0$
$\vdots$
$k=k: p_k=(\frac{1}{5})^k p_0$
Since the probabilities sum to 1:
We can solve for $p_0: 1 = p_0 ( 1 + (\frac{1}{5}) + (\frac{1}{5})^2 + \cdots )=\cfrac{p_0}{1-1/5}$ $\implies p_0=0.8$
Now this is where I get confused. The question asks us to calculate $\Pr[n>1]$, so:
$\Pr[n>1]=1-\Pr[n \le 1]\tag{2}$
Now when calculating equation(2), I thought I would have to transform $n$ to $k$, so Equation (2) becomes:
$\Pr[n>1]=1-\Pr[k \le 2]=1-\Pr[k=0]-\Pr[k=1]-\Pr[k=2]\tag{3}$
But this is not correct. The solution is just replacing exactly n with k with no substitution. In other words, the solution is:
$\Pr[k>1]=1-\Pr[k \le 1]\tag{4}=1-0.8-0.16$
Can someone please explain why this the case? Why don't I have to transform n to k when solving this question? If $n$ is the number of claims during the period, then k is the number of claims per period PLUS ONE. Thanks in advance! Have a great weekend!
Let us keep $k$ out of the game, it interferes with the analysis. Let the probability of no claims be $p_0$. Then the probability of $1$ claim is $p_0\cdot\frac{1}{5}$, the probability of $2$ claims is $p_0\cdot\frac{1}{5^2}$, and so on. Thus $$p_0\left(1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\cdots\right).$$ The sum of the geometric series $1+\frac{1}{5}+\frac{1}{5^2}+\cdots$ is $\frac{5}{4}$, and therefore, as you found, $p_0=\frac{4}{5}$.
The probability of $0$ claims or $1$ claim is $\frac{4}{5}+\frac{4}{5}\cdot\frac{1}{5}$. This is $\frac{24}{25}$, and therefore the probability of $2$ or more claims is $\frac{1}{25}$.
Another way: A coin has probability $\frac{1}{5}$ of landing heads, and $\frac{4}{5}$ of landing tails. Fate decides on the number of claims by tossing the coin until she gets a head. The number of tails she got before the first head is the the number of claims. It is straightforward to verify that this leads to the probability distribution discussed above. So the probability of $2$ or more claims is the probability that the first $2$ tosses result in tails. The probability of this is $\left(\frac{1}{5}\right)^2$.