The minimum of two independent geometric random variablesLet X ∼ ‬ G ( p 1 ), Y ∼ ‬ G ( p 2 ), X and Y are independent. Prove that the minimum is also geometric, meaning: min ( X , Y ) ∼ G ( 1 − ( 1 − p 1 ) ( 1 − p 2 ) ).Instructions: first calculate the probability P ( min ( X , Y ) &gt; k ) and compare it to the parallel probability in (of?) a geometric random variable.

Question

The minimum of two independent geometric random variablesLet   X  ∼  ‬  G  (      p    1    ),   Y  ∼  ‬  G  (      p    2    ), X and Y are independent. Prove that the minimum is also geometric, meaning:   min  (  X  ,  Y  )  ∼  G  (  1  −  (  1  −      p    1    )  (  1  −      p    2    )  ).Instructions: first calculate the probability   P  (  min  (  X  ,  Y  )  &amp;gt;  k  ) and compare it to the parallel probability in (of?) a geometric random variable.

antidootnw · Accepted Answer

Step 1Let X and Y be independent random variables having geometric distributions with probability parameters       p    1   and       p    2   respectively. Then if Z is the random variable min(X,Y) then Z has a geometric distribution with probability parameter   1  −  (  1  −      p    1    )  (  1  −      p    2    ).There are essentially two ways to see this:First, the method outlined by the hint in your homework - Note that the cdf of X is   1  −  (  1  −      p    1        )    k   and the cdf of Y is   1  −  (  1  −      p    2        )    k  , so the probability that   X  &amp;gt;  k is   (  1  −      p    1        )    k   and the probability that   Y  &amp;gt;  k is   (  1  −      p    2        )    k   and so the probability that both are greater than k is             [      (      1      −              p        1            )      (      1      −              p        2            )      ]        k  . But the probability that both are greater than k is the same as the probability that the minimum of the two is greater than k. From this we can get the cdf of Z as   1  −            [      (      1      −              p        1            )      (      1      −              p        2            )      ]        k  , and we can note that this is the cdf of a geometric random variable with probability parameter   1  −  (  1  −      p    1    )  (  1  −      p    2    ).Step 2Second, and more intuitively to me, we can go back to the definition of a geometric random variable with probability parameter p : the number of Bernoulli trials with probability p needed to get one success. So min(X,Y) is the number of trials of simultaneously running a Bernoulli experiment with probability       p    1   and one with probability       p    2   before one or the other experiments succeeds. The probability of one of the two experiments succeeding at any step is just   1  −  (  1  −      p    1    )  (  1  −      p    2    ), so Z is a geometric random variable with probability parameter   1  −  (  1  −      p    1    )  (  1  −      p    2    )

Let X∼‬G(p_1), Y∼‬G(p_2), X and Y are independent. Prove that the minimum is also geometric, meaning: min(X,Y)∼G(1-(1-p_1)(1-p_2)).

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