Showing that the distribution of record times $(\tau_k)_{k\geq 1}$ doesn't depend on the distribution, $F$, of the records $X_i$

Question

I read that it's possible to show that the distribution of a record time sequence doesn't depend on the distribution of the record sequence itself, but how would one do this?

So $(X_i)$ is an iid sequence with common continuous distribution $F$. Then $X_1$ is the first record, and $\tau_1$ the first record time. From here on, the $(k+1)$st record time is $\tau_{k+1}$, given by $\tau_{k+1}=\min\{i>\tau_k: X_i >M_{\tau_k}\}$, for $k\geq1$, with $X_{k+1}$ being the $(k+1)$st record, and $M_{\tau_k}$ denoting the maximum of $X_{i-1}$'s until time $\tau_k$.

It is supposedly then possible to show that the sequence $(\tau_{k})_{k\geq1}$ doesn't depend on $F$, but I'm unsure how this can be done?

I tried looking at the probability \begin{align*} P(\tau_{k+1} \leq t) & = P(\min\{i>\tau_k: X_i >M_{\tau_k}\}\leq t)\\ & = P(i\leq t, i-1\leq t, \ldots), \end{align*} for some $t$, since if $i\leq t$, then all 'previous' $i$'s must necessarily also be less than or equal to $t$, but I don't know if this is the way to go.

Answer 1

I would show it inductively. For $n=1$ is trivial that $\tau_1$ does not depend on $F$ (simply because $\tau_1=1$.) Next, distribution of $\tau_2$ doesnt depend on $F$:

$$P(\tau_2=k)=P(X_2, \dots, X_{k-1}<X_1<X_k)=\frac{1}{k(k-1)},$$ because $X_i$ are iid and this is simply showing how many ways there is to order variables $X_1, \dots, X_k$ in order such that $X_2, \dots, X_{k-1}<X_1<X_k$ holds. Therefore, distribution of $\tau_2$ is deterministic not depending on $F$.

Now inductively you do the same, $$P(\tau_n=k)=\sum_{i=1}^{k-1} P(\tau_{n-1}=i)P(\tau_n=k\mid \tau_{n-1}=i). $$

Now only thing we need to prove is that $ P(\tau_n=k\mid \tau_{n-1}=i) $ for all $i<k$, doesn't depend on $F$. Then, every element in the sum doestn depend on $F$ which is what we wanted. But

$$ P(\tau_n=k\mid \tau_{n-1}=i)=P[X_k\geq X_i\geq(X_1, \dots, X_{k-1}) \mid X_i\geq(X_1, \dots, X_{i-1})]. $$ This again doesnt depend on $F$, because this is equal to $\frac{1}{k}\cdot (\frac{i-1}{i}\dots\frac{k-2}{k-1})$.

Maybe better explanation: Define $R_i^n$ as the order statistic of $X_i$ in $X_1, \dots, X_n$. I.e. $R_i^n=1$ if $X_i$ is the smallest out of all $X_1, \dots, X_n$.

First observation: $R_i^n$ doesn't depend on $F$. That's because we are looking only for the orders, not on magnitude of each $X_i$.

Second observation: $\tau_k$ can be rewritten using only $R_i^n$. intuitively, thats because we only use $X_i$ is bigger or smaller than $X_j$ in the definition of $\tau$.