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One statistics course of mine has the following recurrence.

$$p(n)=\frac{0.16p(n-1)+0.8p(n-2)+0.72p(n-3)}{n}$$

Assuming I know $p(0),$ is there a general formula for getting a solution?

Adrian Keister
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Preston Lui
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    Since the formula for $p(n)$ involves $3$ previous values, i.e., $p(n-1)$, $p(n-2)$ and $p(n-3)$, wouldn't you need to also know at least $p(1)$ and $p(2)$ to get any additional values from the recurrence formula, much less a general solution? – John Omielan Dec 09 '19 at 17:16

1 Answers1

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You can solve it only asymptotically. Notice that no matter what starting values are $p(n)$ tends to $0$ as $n$ tends to infinity.

Therefore you can write

$$p_\sim(n)=\frac{1.68p_\sim(n-1)}{n}$$

which is giving

$$p_\sim(n)=\frac{c}{n!}1.68^{n}$$

where $c$ depends on the initial conditions.