Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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) $?

Thanks.

share|improve this question
    
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

1 Answer 1

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.