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Let $X,Y$ be two independent r.v's with geometric distribution: parameters $\lambda, \mu $, respectively. Let $Z = \min(X, Y) $. Show $Z$ is geometric with parameter $\lambda \mu $.

$$Attempt $$

By hypothesis, I know $P^X( A \in X) = \sum_{j \in A} \lambda^j (1-\lambda) $ and $P^Y(B \in Y) = \sum_{j \in B} \mu^j(1 - \mu) $. Next, I have

$$ P^Z(Z \in C) = \sum_{j \in C} P(Z = j ) = \sum_{j \in C} P( \min(X,Y) = j) $$

We know $X,Y \geq \min(X,Y) $, therefore, $P( \min(X,Y) =j ) = P(X \geq j, Y \geq j ) $ and since $X,Y$ are independent, we have $P(X \geq j, Y \geq j ) = P(X \geq j) P(Y \geq j )$. Here is where I am stuck. Am I on the right track ?

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    Note, if the minimum is equal to $j$, then at least one of the two variables must equal to $j$. $$\begin{align}P(\min(X,Y)=j) & = P((X=j\cap Y\geq j)\cup(X>j\cap Y=j)) \[1ex] & = P(X=j)P(Y\geq j)+P(X>j)P(Y=j)\end{align}$$ – Graham Kemp Mar 26 '15 at 09:39
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    I found several earlier posts of exactly the same question. Judging by the content, it's hard to decide which one should be deemed the original and the others duplicate. In chronological order: 90782, 845706, 1040620, 1056296, and 1169142. – Lee David Chung Lin Mar 21 '18 at 01:19

2 Answers2

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Use distribution function:

$$P(\min(X,Y) \le x) = 1 - P(\min(X,Y) \ge x) = 1 - P(X \ge x, Y\ge x) = 1 - P(X \ge x) P(Y \ge x)$$

Now compute that and recognise the distribution function of a geometric distribution

Ant
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    Im confused, inst it what I have just done ? –  Mar 26 '15 at 08:55
  • @Anonaki no, you started with a general event $Z$ and wrote that $P(\min(X,Y) = j) = P(X\ge j, Y\ge j) $ which is not true – Ant Mar 26 '15 at 08:59
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    Why is not true? isnt $X,Y \geq \min(X,Y)$ ? –  Mar 26 '15 at 09:00
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    @Anonaki yes, that is true. But it doesn't imply what you've written.. Try to think about why and construct s counterexample :-) – Ant Mar 26 '15 at 09:02
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    What is $P( \min(X,Y) \leq x ) $. the $x$ confuses me. Can you explain ? –  Mar 26 '15 at 09:04
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    @Anonaki its the point in which we are calculating the distribution function $F$ :$F(x) = P(X \in (-\infty, x]) = P(X \le x)$ – Ant Mar 26 '15 at 09:06
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    but geometric rv is discrete i think –  Mar 26 '15 at 09:07
  • @Anonaki indeed. So what? Distribution functions exists for every probability distribution. Search for "distribution function geometric" to actually look at it and understand what it is – Ant Mar 26 '15 at 09:08
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    I am so confused. I learned that the distribution of geometric r.v is $P(X = j) = p^j(1-p) $ where $p$ is parameter. IT doesnt look like your distribution. –  Mar 26 '15 at 09:11
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    @anonaki we are using different terms :-) I am referring to the cumulative distribution function, or cdf:-) – Ant Mar 26 '15 at 09:14
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Given that $X,Y$ are two independent, geometic distributed random variables with parameters $\lambda, \mu$ (respectively), then each is a count failures before a success event (say $S_X,S_Y$) occurs in a series of Bernoulli trials with success probability $\lambda, \mu$.

Thus $Z:=\min(X,Y)$ measures the count of failed trials before either success event first occurs.   As these events are independent, the unioned success event has probability:

$$P(S_X\cup S_Y) = \lambda +\mu-\lambda \mu$$

$Z$ is thus a geometric distributed random variable with parameter: $\lambda +\mu-\lambda \mu$

Graham Kemp
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