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Let $X, Y$ be independent geometric r.v.'s with paramaters $\lambda$ and $\mu$. Let $Z = \min(X,Y)$ and show that $Z \sim$ Geometric$(\lambda\mu)$. I have seen 5 posts of the same question on this website and yet I cannot get the right answer. This is how I try to prove it. We have $$P(X = i) = (1 - \lambda)^i\lambda \ \text{and} \ P(Y = i) = (1- \mu)^i\mu$$ where $i = 0, 1, 2, \dotsc$ Then $P(Z \geq i) = P(X \geq i)P(Y \geq i)$ by independent of r.v.'s. Since $$P(X \geq i) =1 - \lambda\sum_{j=0}^{i-1}(1-\lambda)^j = 1 - \lambda\left(\frac{1 - (1-\lambda)^i}{1 - (1-\lambda)}\right) = (1-\lambda)^i,$$ we have $P(Z \geq i) = [(1-\lambda)(1-\mu)]^i.$ So, $$P(Z= i) = P(Z \geq i) - P(Z \geq i + 1)$$ which is equal to $$[(1-\lambda)(1-\mu)]^i - [(1-\lambda)(1-\mu)]^{i+1} =[(1-\lambda)(1-\mu)]^i(1 - [(1-\lambda)(1-\mu)]).$$ Hence $$P(Z = i) = \mu + \lambda - \lambda\mu[1 - \mu - \lambda + \lambda\mu]^i.$$ But then this means $Z \sim$ Geometric$(\mu + \lambda - \lambda\mu)$. In the post: If $X,Y$ are independent and geometric, then $Z=\min(X,Y)$ is also geometric , It says as a hint that $$P(Z > t) = (\lambda\mu)^t.$$ This is not what I acquired however. Can someone please point out my mistake?

EDIT: I have tried using $$P(X = i) = \lambda(1 - \lambda)^{i-1}, \ \ \ \text{for} \ i \in \mathbb{N}.$$ Then $$P(X \geq i) = 1 - \lambda\sum_{j=1}^{i-1}(1-\lambda)^{j-1} = 1- \frac{\lambda}{1 - \lambda}\left(\sum_{j=0}^{i-1}(1-\lambda)^j - 1\right) = (1-\lambda)^{i-1}.$$ Using the same calculation as above I obtain $$P(Z = i) = \mu + \lambda - \lambda\mu[1 - \mu - \lambda + \lambda\mu]^{i=1}.$$ I still cannot conclude that $Z \sim$ Geometric($\lambda\mu$) from this result.

Vicky
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    Both are correct. Your $\lambda$ and $\mu$ are probabilities of success, wheres those parameters in the link are probabilities of failure. So the difference simply stems from how the distribution is quoted. – Sangchul Lee Jun 16 '20 at 11:35
  • In this solution you allow $Z=Y \cap Z=X$. If you require strict inequality, you have to subtract the case $P(X=k, Y=k)$, e.g. for $P(Z=0) = p_1(1-p_2) + p_2(1-p_1) = p_1+p_2 - 2 p_1 p_2$ – Alex Jun 16 '20 at 14:49
  • @SangchulLee But $X$ and $Y$ here denote the number of failures before we reach success. That's why $P(X = i) = (1 - \lambda)^i\lambda$ right? I have tried using $P(X = i) = (1 - \lambda)^{i-1}\lambda$ as well. The result of $P(Z = i)$ I got is the same, but with power $i - 1$ – Vicky Jun 16 '20 at 15:25

2 Answers2

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I tried solving it directly, with $X,Y,Z=n$ is the number of failures before the first success. In this case, $$ P(X \geq n) = 1-P(X<n) = 1- \sum_{k=0}^{n-1}(1-p_1)^kp_1 = (1-p_1)^{n} $$ For each trial, either one or both $X,Y$ are success, so I see three cases: $$ P(Z=k) = P(X=k, Y > k) + P(Y=k, X > k) + P(X=k, Y=k) = P(X=k)P(Y>k) + P(X>k)P(Y=k) + P(X=k)P(Y=k) $$ The last step is by independence. Therefore, $$ P(Z=k) = (1-p_1)^kp_1 (1-p_2)^{k+1} + (1-p_2)^kp_2 (1-p_1)^{k+1} + (1-p_1)^kp_1 (1-p_2)^{k}p_2 = ((1-p_1)(1-p_2))^k[p_1 + p_2 - p_1 p_2] $$

Alex
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$$ \Pr(\min\{X,Y\} \ge i) = \Pr(X\ge i\ \&\ Y\ge i) = \Pr(X\ge i)\Pr(Y\ge i). $$

Michael Hardy
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