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$X_1, X_2, \ldots \sim \operatorname{Ber}(p)$, those variables are i.i.d.

Suppose $S_n = X_1 + X_2 + \cdots + X_n$

$$N = \min\{ n \geq 1: S_n = 1\}$$

1) Show that $X_{N+1}, X_{N+2}$ also follow i.i.d $\operatorname{Ber}(p)$;

2) show that $\{X_1, X_2, \ldots , X_N\}$ and $\{X_{N+1}\}$ are independent each other.

I'm a bit of a dummy so a thorough explanation would be much appreciated. Thanks in advance!

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  • $\begingroup$ Probably the reason for the votes to close this question is that it's phrased like a homework problem without expressing any thoughts about it beyond the statement of the problem. $\endgroup$ – Michael Hardy Apr 16 '18 at 17:08
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$\newcommand{\e}{\operatorname{E}}$ The law of total probability says the conditional probability of an event $A,$ given a random variable $N,$ is a random variable whose expected value is the marginal (or "unconditional") probability of $A{:}$ $$ \Pr(A) = \e(\Pr(A \mid N)). $$ We have \begin{align} & \Pr(X_{N+1} = x_1\ \&\ \cdots\ \&\ X_{N+k}= x_k \mid N=n) \\[10pt] = {} & \Pr(X_{n+1} = x_1\ \&\ \cdots\ \&\ X_{n+k} = x_k) \\[10pt] = {} & p^{x_1+\cdots+x_k} (1-p)^{k - (x_1+\cdots+x_k)} \tag 1 \\[10pt] & \text{But that last expression does not depend} \\ & \text{on the value of $n$, so we conclude that} \\[10pt] & \Pr(X_{N+1} = x_1\ \&\ \cdots\ \&\ X_{N+k}= x_k \mid N=n) \\[10pt] = {} & \Pr(X_{N+1} = x_1\ \&\ \cdots\ \&\ X_{N+k}= x_k) \tag 2 \end{align} Since the expressions on lines $(1)$ and $(2)$ are equal to each other, regardless of the value of $k$, we have $$ X_{N+1},X_{N+2},X_{N+3},\ldots\sim\operatorname{i.i.d.} \operatorname{Bernoulli}(p). $$ Maybe I'll be back to address the second question later.

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    $\begingroup$ Thank you so much Michael. I really appreciate it. This helps me understand the problem now. Thanks! $\endgroup$ – Olympia111 Apr 16 '18 at 16:09
  • $\begingroup$ @Olympia111 If this answer meets your needs then it is good thing to accept it. It always feels good for the one who answered. $\endgroup$ – drhab Apr 16 '18 at 17:55

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