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Let $ X_{1},X_{2},X_{3},\ldots $ be i.i.d. geometric random variables, where $ \mathbf{Pr}(X_{i} = k) = p(1 - p)^{k - 1} $.

Define $ \displaystyle Z_{r} \stackrel{\text{def}}{=} \sum_{i=1}^{r} X_{i} $.

What is $ \displaystyle \lim_{k \to \infty} \sum_{r=1}^{\infty} \mathbf{Pr}(Z_{r} = k) $?

Also, define $ \displaystyle F(k) \stackrel{\text{def}}{=} \min_{r} (Z_{r} ~|~ Z_{r} \geq k) $.

What is $ \displaystyle \lim_{k \to \infty} \mathbf{Pr}(F(k)-k=5) $?


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Maybe it would help if you can give some context or your thoughts on the given problem – Belgi Feb 25 '13 at 12:19
@Solver: Are you OK with my answer? – Bravo Feb 27 '13 at 16:16

Part 1: $\{Z_r=k\}$ is the event of the $r^{th}$ success arriving in the $k^{th}$ trial. Thus the distribution of $Z_r$ is as follows.

$$P(Z_r=k)={{k-1} \choose {r-1}} p^{r-1} (1-p)^{k-r} p,\ k=r,r+1,\ldots\\ \sum_{r=1}^{\infty}P(Z_r=k)=\sum_{r=1}^{k}P(Z_r=k)=p\sum_{r=1}^{k}{{k-1} \choose {r-1}} p^{r-1} (1-p)^{k-r}=p\\ \lim_{k\to \infty}\sum_{r=1}^{\infty}P(Z_r=k)=p $$

Part 2: $$\lim_{k\to \infty}P(\min_r (Z_r|Z_r\ge k)-k=5) \\=\lim_{k\to \infty}P(X_1+X_2+\ldots+X_{r^*}=k+5|X_1+X_2+\ldots+X_{r^*-1}\le k)\\ \lim_{k\to \infty}\frac{P(X_1+X_2+\ldots+X_{r^*}=k+5,X_1+X_2+\ldots+X_{r^*-1}\le k)}{P(X_1+X_2+\ldots+X_{r^*-1}\le k)}\\ \lim_{k\to \infty}\frac{P(k+5-X_{r^*}\le k)}{P(X_1+X_2+\ldots+X_{r^*-1}\le k)}\\ \lim_{k\to \infty}\frac{P(X_{r^*}\ge 5)}{P(X_1+X_2+\ldots+X_{r^*-1}\le k)}\\ P(X_{r^*}\ge 5)=1-q^4 $$

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