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I need a little help for the following exercise:

Let $X_1, X_2,\ldots$ be sequence of iid rv with values in $\{1,2,3 \}$ and $p(i):=P(X=i) \gt 0$ for $i \in \{1,2,3\}$. Define an another rv $R:=\inf\{n\in \mathbb{N}:X_n\in\{2,3\} \}$

a) Show that $P(R<\infty)$ and determine the distribution of R

b) Show R and $X_n$ are independent

At a) I startet with:

$ P(X_1,\ldots,X_n\notin\{2,3\}) = (1-P(X_1=1))^{n}$

from here I concluded that $P(X_1,\ldots,X_n\notin\{2,3\} = 1)=0$, if $n \rightarrow \infty$ but i cannot find an argument that $P(R <\infty) = 1$

thanks

jed
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3 Answers3

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Your notation here is really rough... to the point that I can't understand what you are getting at with the last part. In fact, one of your statements is entirely wrong as written. But, let me try to give you a push in the right direction.

Note that the event $\{R<\infty\}$ is precisely the event "there is some $n$ such that $X_n\neq 1$".

Let $E_n$ denote the event $\{R>n\}$. Note that $E_1,E_2,\ldots$ is a decreasing set of events; that is, $E_1\supseteq E_2\supseteq E_3\supseteq\cdots$. And furthermore, $E:=\{R=\infty\}$ is precisely $\bigcap_{n=1}^{\infty}E_n$. In particular, for any $N\in\mathbb{N}$, this tells us that $$\tag{1} P(E)=P\left(\bigcap_{n=1}^{\infty}E_n\right)\leq P\left(\bigcap_{n=1}^{N}E_n\right)=P(E_N). $$ But, we have $$\tag{2} P(E_N)=P(X_i=1\text{ for }i=1,2,\ldots,N)=\left(P(X_1=1)\right)^N. $$ Because $P(X\in\{2,3\})>0$, we know that $P(X_1=1)<1$; so, letting $N\rightarrow\infty$ in (2) and plugging the result in to (1) shows that $P(E)=P(R=\infty)=0$. Hence $P(R<\infty)=1$.

As for the distribution, note that again $$ P(R\geq n)=P(X_1,\ldots,X_{n-1}=1)=P(X_1=1)^{n-1}, $$ and use the fact that for an $\mathbb{N}$-valued variable we have $P(R=n)=P(R\geq n)-P(R\geq n+1)$.

Nick Peterson
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It looks as if the symbols are getting in the way of solving the problem. We are repeatedly tossing a three-sided coin, that has $1$, $2$, and $3$ marked on its sides. Think of getting a $1$ as a failure, and getting a $2$ or a $3$ as a success.

Then $R=k$ if we have $k-1$ failures in a row, followed by a success.

Let $p=p(1)$. This is the probability of a failure. Then $1-p=p(2)+p(3)$ is th probability of a success. Note that $p\ne 0$ and $1-p\ne 0$.

The probability $\Pr(R=k)$ of $k-1$ failures in a row followed by success is $p^{k-1}(1-p)$.

Sum $\Pr(R=k)$ from $k=1$ to $\infty$. We have a geometric series, with sum $1$.

Now you can attack the independence question. For this, we need to consider $\Pr((R=k)\land (X_n=i)$.

André Nicolas
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$R$ follows Geometric distribution:

$$ R= \begin{array}{&&} 1 \ \text{w.p.} \ p_2+p_3\\ 2 \ \text{w.p.} \ p_1(p_2+p_3)\\ \ldots\\ n \ \text{w.p.} \ p_1^{n-1}(p_2+p_3) \end{array} $$ Hence $P(R< \infty)=1-P(R=\infty)=1-(p_2+p_3)\lim_{n \to \infty}p_1^{n-1}=1$

Alex
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