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I'm looking at this problem

A $500$-year flood is one that occurs once in every $500$ years.

a) What is the probability of having at least $3$ such floods in $500$ years?

b) What is the probability that a flood will occur within the next $100$ years?

c) What is the expected number of years until the next flood?

Attempt:

a) The window size is $500$ years, $\lambda=1$ and the number of floods is a Poisson variable $X$. So the answer would be $1-P(X=0)-P(X=1)-P(X=2)$, where $P(X=n)$ is the Poisson pmf.

c) I can consider a window size of $1$ year, $\lambda=\frac{1}{500}$ and consider the time to next flood as the exponential random variable $T$, so that $E[T]=1/\lambda=500$.

Are the above two correct?

Finally for part b), one approach is to decide on a window size, let's say $1$ year, set $\lambda$ appropriately (in this case $\frac{1}{500}$), let $T$ be the time to next flood (exponential RV), and find $$P(T=1)+P(T=2)+P(T=3)+\ldots+P(T=100)$$

The other approach is to set the window to $100$ years, set $\lambda=\frac{100}{500}=0.2$, model number of floods as a Poisson RV $X$ and find $1-P(X=0)$.

Numerically, these approaches give slightly different answers ($0.1811$ vs $0.1813$). Which of the two approaches is better for this purpose (i.e. which gives a more accurate answer)?

Also, in the first approach to part b), instead of a one-year window, I could've taken a half-year window and set $\lambda=\frac{1}{1000}$ and summed from $P(T=1)$ to $P(T=200)$. And of course there's no limit to how small I can set that window. What is the recommended granularity of time windows for problems like these?

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  • $\begingroup$ I would write > in the first two questions, and < in the final one - given that climate change means that all estimates based on historical rates will underestimate future rates. In other words, what we consider a 500 year flood today, will not be a 500 year flood tomorrow. $\endgroup$ – Mooks Oct 18 at 12:14
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Your treatment of (a) and (c) are correct. In (b), the second approach is the correct one because in your first approach, you treat exponential RVs as if they're discrete. You should've done the following ($\lambda=1/500$): $$P(T\leq100)=\int_0^{100}\lambda e^{-\lambda t}dt=1-e^{-100\lambda}=1-e^{-0.2}$$ which gives the same answer as your Poisson approach. By the way, for (b), if you discretize at infinitely small steps, you get this integral.

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  • $\begingroup$ Ah you're right. Since Poisson and exponential distributions are closely related, and since Poisson is associated with discrete values, I had this notion that exponential should also be treated as discrete. Thanks! $\endgroup$ – Shirish Kulhari Oct 13 at 13:48

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