11
$\begingroup$

Let $X_1, X_2, \cdots$ be jointly continuous and independently distributed with marginal pdf $f(x)$, where each $X_i$ represents annual rainfall at a given location. How can I find the distribution for the number of years until the first year's rainfall $(X_1)$ is exceeded for the first time?

I assume this is a fairly straight forward problem, but I am failing to see what the most logical first step would be. How would you go about thinking of the first step?

$\endgroup$
2
  • $\begingroup$ "independent" and also "identical"? The "identical" changes the complexity of this problem considerably. $\endgroup$
    – Him
    Commented Jun 10, 2023 at 13:48
  • $\begingroup$ @Him - the fact that all the $X_i$ have the same marginal PDF and are independent implies identical. $\endgroup$
    – jbowman
    Commented Jun 15, 2023 at 15:58

2 Answers 2

9
$\begingroup$

Let us assume that annual rainfall can be considered a continuous variable (which requires the probability of no rainfall in a year to be equal to zero.) The following steps get us to our goal:

  1. The first year's rainfall has cumulative distribution function $F(X_1)$, and, as is well-known, $F(X_1) \sim \text{Uniform}(0,1)$ (the Probability Integral Transform)

  2. The probability that any given successive year's rainfall exceeds the first year's rainfall is $1-F(X_1)$, label it $p$. (This is because the probability of any given successive year's rainfall being $\leq$ year 1's rainfall is $F(X_1)$, so the probability of exceeding it is just $1-F(X_1)$.) Since $F(X_1) \sim \text{Uniform}(0,1)$, $p$ is too; $1 - $ a $\text{Uniform}(0,1)$ variate is also a $\text{Uniform}(0,1)$ variate.

  3. The number of years ($k$) until the first exceedence of the first year's rainfall, conditional upon $p$, has a Geometric distribution with probability parameter $p$: $p(k | p) = (1-p)^{k-1}p$.

To remove the conditioning upon $p$, we integrate the Geometric distribution with respect to the Uniform distribution on $p$, which "disappears" in the expression below because it is equal to $1$ everywhere:

$$p(k) = \int_0^1(1-p)^{k-1}p\text{d}p$$

which is the $\text{Beta}(2,k)$ function. Expanding this function leads to:

$$p(k) = {1 \over k(k+1)}, \, k \geq 1$$

A quick check in R that this (plausibly) does sum to $1$:

> sum(beta(2,1:100000))
[1] 0.99999
$\endgroup$
6
  • $\begingroup$ So $p(k)=\frac{1}{k(k+1)}$ is the distribution, where $k \ge 0$, correct? Also, how do you know $F(X_1) \sim U(0,1)$? $\endgroup$
    – Ron Snow
    Commented Nov 28, 2019 at 21:13
  • $\begingroup$ Also, I don't understand your (2) entirely. How do you make those claims? $\endgroup$
    – Ron Snow
    Commented Nov 28, 2019 at 21:15
  • 1
    $\begingroup$ I've edited my answer to address your questions. $\endgroup$
    – jbowman
    Commented Nov 28, 2019 at 21:20
  • $\begingroup$ Awesome, thanks for adding the link and a bit of an explanation in #2. I did not know about that 1 - $U(0,1) \sim U(0,1)$. That seems like a nice property to know! $\endgroup$
    – Ron Snow
    Commented Nov 29, 2019 at 3:25
  • $\begingroup$ There is an unstated premise in this analysis that all years' chances of exceeding the first years' rainfall are i.i.d. $\endgroup$
    – Him
    Commented Jun 10, 2023 at 13:45
8
$\begingroup$

jbowman's answer is correct, though there is an alternative approach to the same answer

By symmetry/exchangeability and assuming a continuous distribution so no ties, the probability that $X_1$ is the largest of $\{X_1,X_2,\cdots X_n\}$ is $\frac1n$ and the probability that $X_1$ is the largest of $\{X_1,X_2,\cdots X_n, X_{n+1}\}$ is $\frac1{n+1}$.

This means the probability $X_{n+1}$ is the first value to exceed $X_1$ is $\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}.$

Slightly counter-intuitively, this implies that the expected number of years until $X_1$ is exceeded is therefore infinite. Suppose $X_{n+1}$ is the first to exceed $X_1$, which takes $n$ years to happen. Then the expected number of years is $\sum\limits_1^\infty n \times \frac1{n(n+1)} = \sum\limits_1^\infty \frac1{n+1}$ and this is infinite.

$\endgroup$
11
  • 1
    $\begingroup$ Elegant! (+1) I think you could also have done it with the second "probability that $X_1$ is the..." replaced by "probability that $X_{n+1}$ is the largest...", then just multiplied the two probabilities, since they are clearly independent. $\endgroup$
    – jbowman
    Commented Dec 1, 2019 at 1:59
  • 1
    $\begingroup$ @jbowman that works too - both working allows independence to be proved formally, and settle any doubts of somebody who thinks perhaps $X_1$ being high (if it is the highest of the first $n$) makes it less likely $X_{n+1}$ will exceed it $\endgroup$
    – Henry
    Commented Dec 1, 2019 at 2:08
  • 1
    $\begingroup$ @Edison Try writing $X_1 = \max\limits_{1 \le i \le n} (X_i)$ to say $X_1$ is the largest of the first $n$ $X_i$s. Then the following three equations are true as the first says either $X_1 \gt X_{n+1}$ or $X_1 < X_{n+1}$ and the others are rearrangements with the third being what we want $\endgroup$
    – Henry
    Commented Dec 3, 2019 at 1:42
  • 1
    $\begingroup$ $\mathbb P(X_1 = \max\limits_{1 \le i \le n} (X_i)) = \mathbb P(X_1 = \max\limits_{1 \le i \le n} (X_i)\gt X_{n+1}) +\mathbb P(X_1 = \max\limits_{1 \le i \le n} (X_i)\lt X_{n+1})$ $\mathbb P(X_1 = \max\limits_{1 \le i \le n} (X_i)) = \mathbb P(X_1 = \max\limits_{1 \le i \le n+1} (X_i)) +\mathbb P(X_{n+1} \gt X_1 = \max\limits_{1 \le i \le n} (X_i))$ $\mathbb P(X_{n+1} \gt X_1 = \max\limits_{1 \le i \le n} (X_i)) = \mathbb P(X_1 = \max\limits_{1 \le i \le n} (X_i)) -\mathbb P(X_1 = \max\limits_{1 \le i \le n+1} (X_i))$ $\endgroup$
    – Henry
    Commented Dec 3, 2019 at 1:43
  • 1
    $\begingroup$ @xabush I have added to the final paragraph $\endgroup$
    – Henry
    Commented Jun 10, 2023 at 13:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.