I am working on a question as follows:

Let $\xi_1,\xi_2,\ldots$ be i.i.d. random elements with distribution $\mu$ in some measurable space $(S,\mathcal{S})$. Fix $A∈\mathcal{S}$ with $\mu A>0$,and put $\tau=\inf\{k:\xi_k\in A\}$. Show that $\xi_\tau$ has distribution $\mu[\cdot|A]=\mu(\cdot\cap A)/\mu A$.

I'm disturbed by fact that some $\xi_i^{-1}A$ is not measurable for some $i$ (e.g. $\xi_i$, $i<\tau$) and measurable for the other (here $\xi_\tau$), while $\xi_i$'s are i.i.d.. Could someone give an example of two random variables with the stated property? I'd appreciate any thoughts on this.

  • 1
    $\begingroup$ I a not sure I understand: by definition of a random variable, the set $\xi_i^{-1}A$ is measurable. The random variables are defined on a probability space $(\Omega,\mathcal A,\mathbb P)$ and take their values on $S$ endowed with the $\sigma$-algebra $\mathcal S$. $\endgroup$ – Davide Giraudo Oct 20 '17 at 8:26
  • $\begingroup$ But in that case why do we need to define $\tau$, since all $\xi_i\in A$? $\endgroup$ – Chiying Wang Oct 20 '17 at 8:39
  • $\begingroup$ We can imagine the simplest non-degenerated case: $S=\{0;1\}$ and $\xi_i$ takes the value $0$ with probability $1/2$ and the value $1$ with probability $1/2$. Let $A:=\{0\}$. $\endgroup$ – Davide Giraudo Oct 20 '17 at 8:51
  • $\begingroup$ Thanks, I think I must misunderstood the notation here. Because sometimes in the textbook $\xi\in A$ means $\{\omega\in\Omega:\xi(\omega)\in A\}$. $\endgroup$ – Chiying Wang Oct 20 '17 at 9:00
  • $\begingroup$ If you prefer here, the random variable $\tau$ is defined by $\tau(\omega)=\inf\{k: \xi_k(\omega)\in A\}$. $\endgroup$ – Davide Giraudo Oct 20 '17 at 9:02

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