Evaluating min/max probability with Geometric random variablesSuppose that X 1 , … , X n are independent, geometric 1 ( p ) random variables. Evaluate P ( min ( X 1 , … , X n ) &gt; l ) and P ( max ( X 1 , … , X n ) ≤ l )The textbook solution is as follows: P ( min ( X 1 , … , X n ) &gt; l ) = P ( ⋂ k = 1 n { X k &gt; l } ) = ∏ k = 1 n P ( X k &gt; l ) = ∏ k = 1 n p l = p l n and P ( max ( X 1 , … , X n ) ≤ l ) = P ( ⋂ k = 1 n { X k ≤ l } ) = ∏ k = 1 n P ( X k ≤ l ) = ∏ k = 1 n ( 1 − p ) l = ( 1 − p l ) n I'm having trouble understanding pretty much the entire solution. If I were to summarize what's particularly puzzling me:1) How was the first line of each solution derived (i.e. P ( min ( X 1 , … , X n &gt; l ) = P ( ∩ k = 1 n { X k &gt; l } ) and the corresponding equation for max)?2) The PMF for Geometric 1 ( p ) is given as p X ( k ) = ( 1 − p ) p k − 1 . How is P ( X k &gt; l ) = p l ? The same applies for the max case as well.

Question

Evaluating min/max probability with Geometric random variablesSuppose that       X    1    ,  …  ,      X    n   are independent,       geometric    1    (  p  ) random variables. Evaluate   P  (  min  (      X    1    ,  …  ,      X    n    )  &amp;gt;  l  ) and   P  (  max  (      X    1    ,  …  ,      X    n    )  ≤  l  )The textbook solution is as follows:                    P        (        min        (                  X          1                ,        …        ,                  X          n                )        &amp;gt;        l        )                            =        P                  (                      ⋂                          k              =              1                        n                    {                      X            k                    &amp;gt;          l          }          )                                                                 =                  ∏                      k            =            1                    n                P        (                  X          k                &amp;gt;        l        )                                                         =                  ∏                      k            =            1                    n                          p          l                                                                 =                  p                      l            n                              and                     P        (        max        (                  X          1                ,        …        ,                  X          n                )        ≤        l        )                            =        P                  (                      ⋂                          k              =              1                        n                    {                      X            k                    ≤          l          }          )                                                                 =                  ∏                      k            =            1                    n                P        (                  X          k                ≤        l        )                                                         =                  ∏                      k            =            1                    n                (        1        −        p                  )          l                                                                 =        (        1        −                  p          l                          )          n                    I&#039;m having trouble understanding pretty much the entire solution. If I were to summarize what&#039;s particularly puzzling me:1) How was the first line of each solution derived (i.e.   P  (  min  (      X    1    ,  …  ,      X    n    &amp;gt;  l  )  =  P  (      ∩          k      =      1        n    {      X    k    &amp;gt;  l  }  ) and the corresponding equation for max)?2) The PMF for       Geometric    1    (  p  ) is given as       p    X    (  k  )  =  (  1  −  p  )      p          k      −      1      . How is   P  (      X    k    &amp;gt;  l  )  =      p    l  ? The same applies for the max case as well.

encaselatqr · Accepted Answer

Step 11) This is because the minimum of a set of numbers is greater than l if and only if all the numbers are greater than l. So   min  (      X    1    ,  …  ,      X    n    )  &amp;gt;  l if and only if       X    1    &amp;gt;  l and       X    2    &amp;gt;  l and ... and       X    n    &amp;gt;  l. Note that the event &quot;      X    1    &amp;gt;  l and       X    2    &amp;gt;  l and ... and       X    n    &amp;gt;  l&quot; is written as       ⋂          k      =      1              n        {      X    k    &amp;gt;  l  }. Thus   P  (  min  (      X    1    ,  …  ,      X    n    )  &amp;gt;  l  )  =  P      (          ⋂              k        =        1                    n              {          X      k        &amp;gt;    l    }    )    ..The max case is similar reasoning; it&#039;s because the maximum of a set of numbers is less than or equal to l if and only if all the numbers are less than or equal to l.Note that what I said above holds true regardless of the distribution of the       X    k  &#039;s.Step 22) Note that if X has PMF   (  1  −  p  )      p          k      −      1       for   k  ≥  1, then for any integer   l  ≥  1, we have                    P        (        X        ≤        l        )                            =                  ∑                      k            =            1                                l                          P        (        X        =        k        )                                          =                  ∑                      k            =            1                                l                          (        1        −        p        )                  p                      k            −            1                                                            =        (        1        −        p        )        ×                              1            −                          p                              l                                                          1            −            p                                  (        using the                   geometric series formula                )                                          =        1        −                  p          l                .            (And so   P  (  X  &amp;gt;  l  )  =  1  −  P  (  X  ≤  l  )  =      p    l  .)

Suppose that X_1,…,X_n are independent, geometric_1(p) random variables. Evaluate P(min(X_1,…,X_n)>l) and P(max(X_1,…,X_n) le l).

Answered question

Answer & Explanation

New Questions in High school geometry