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Assume $X_1,...,X_n$ are independent, but not identically distributed continuous RVs. I am interested in the record indicators $R_j = \mathbf{1}_{X_j > max(X_1,...,X_{j-1})}$.

According to Nevzorov (Records: Mathematical Theory) these $R_j$ are not independent. But he does neither give a proof or an intuition.

(Note: In the stationary case, $P(R_j = 1) = \frac{1}{j}$ and the $R_j$ are independent.)

Is there a simple counterexample where one can e.g. compute $P(R_j = 1)$ and $P(R_j = 1 | R_i = 1)$ explicitly and see how the independence is violated in the non-stationary case?

Edit: I should have written continuous RVs because otherwise also the iid statement is not necessarily true. I edited above.

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2 Answers 2

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Take $X_i$ uniformly distributed on $\{1,\dots,i\}$. Then $\{R_2=1\}=\{X_2=2\}$ hence $\mathbb P(R_2=1)=1/2$. Moreover, $$\{R_3=1\}=\{X_3>\max\{X_1,X_2\}\}=\{X_3>X_2\}=\{X_3=3\}\cup(\{X_3=2\}\cap\{X_2=1\})$$ hence $\mathbb P(R_3=1)=\mathbb P(X_3=3)+\mathbb P(\{X_3=2\}\cap\{X_2=1\})= \frac 13\left(1+\frac 12\right)=\frac 12$. Finally, $$ \mathbb P(R_3=1,R_2=1)=\mathbb P(X_3>X_2>X_1)=\mathbb P(X_2=2,X_3=3)=\frac 16\neq \frac 12\times\frac 12. $$

In the continuous case, take $X_1,X_2,X_3$ having exponential distribution with respective parameters $\lambda_1,\lambda_2,\lambda_3$. All the probability can be explicitely computed: we find $\mathbb P(R_1)=\mathbb P(X_2>X_1)=\lambda_1/(\lambda_1+\lambda_2)$,
$\mathbb P(R_3=1)=\mathbb P(X_3>X_2>X_1)+\mathbb P(X_3>X_1>X_2)$ and $\mathbb P(R_2\cap R_3)=\mathbb P(X_3>X_2>X_1)$. All these probabilities can be computed and at the end, the equality we want to check is $$ \frac 1{\lambda_2+\lambda_3}= \frac 1{\lambda_1+\lambda_2}\left( \frac 1{\lambda_2+\lambda_3}+ \frac 1{\lambda_1+\lambda_3}\right) $$ which does not always take place by letting for example $\lambda_1$ to infinity.

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To complement the excellent answer by Davide Giraudo, I give a simple intuitive example, and another simple continuous example.

Intuitive argument: Suppose that you are looking at temperature data, which is rising over time. Say $X_i=i+\epsilon_i$, where $\epsilon_i$ are independent standard Gaussians.

We consider whether $R_1$ and $R_2$ are independent. A priori, the strongest 'competitor' for $X_2$ is $X_1$. Now if $X_1$ is not a record - in spite of the positive trend - this means that $\epsilon_1$ was likely low. This (probably) low value of $\epsilon_1$ makes it more likely that $X_2$ will exceed its strongest competitor $X_1$, too. Moreover, $X_2$ is likely to exceed $X_0$, even if it is a bit larger than usual. So, knowing that $R_1=0$ makes it more likely that $R_2=1$.

Here is a similar, simple mathematical argument:

Simple mathematical argument: We take $X_0=0$, and $X_1,X_2$ as independent continuous. $$ \mathbb{P}[R_2=1|R_1=0]=\mathbb{P}[X_2>0|X_1<0]=\mathbb{P}[X_2>0]. $$ The last step follows by independence. On the other hand: $$ \mathbb{P}[R_2=1|R_1=1]=\mathbb{P}[X_2>X_1|X_1>0]<\mathbb{P}[X_2>0|X_1>0]=\mathbb{P}[X_2>0] $$

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