Probability function of Y = m a x { X , m } for m positive integer when X is geometric distributionX has geometric distribution so f X ( x ) = p ( 1 − p ) x . I wrote this: f Y ( y ) = P ( Y = y ) = P ( m a x { X , m } = y )So if y = m it means that X &lt; m so f Y ( y ) = P ( X &lt; m ) = 1 − P ( X ≥ m ) = 1 − ( 1 − p ) m and if y = m + 1 , m + 2 , . . . then f Y ( y ) = p ( 1 − p ) y Is it right?

Question

Probability function of   Y  =  m  a  x  {  X  ,  m  } for m positive integer when X is geometric distributionX  has geometric distribution so       f    X    (  x  )  =  p  (  1  −  p      )    x  . I wrote this:      f    Y    (  y  )  =  P  (  Y  =  y  )  =  P  (  m  a  x  {  X  ,  m  }  =  y  )So if   y  =  m it means that   X  &amp;lt;  m so       f    Y    (  y  )  =  P  (  X  &amp;lt;  m  )  =  1  −  P  (  X  ≥  m  )  =  1  −  (  1  −  p      )    m   and if   y  =  m  +  1  ,  m  +  2  ,  .  .  . then       f    Y    (  y  )  =  p  (  1  −  p      )    y  Is it right?

Audrey Rosales · Accepted Answer

Step 1Assuming as correct your       f    X    (  x  ), that is X is a geometric rv counting the failures before the first success, your pmf       f    Y    (  y  ) does not sum up to 1...Observe that   Y  =  m when   X  ≤  m and this happens with probability   1  −  (  1  −  p      )          m      +      1       thus      P    [  Y  =  y  ]  =      {                            1          −          (          1          −          p                      )                          m              +              1                                ,                                      if                           y              =              m                                                                         p          (          1          −          p                      )            y                    ,                                      if                           y              =              m              +              1              ,              m              +              2              ,              m              +              3              ,              …                                                                         0          ,                          elsewhere                         Step 2As you can see, this pmf works, being   1  −  (  1  −  p      )          m      +      1        +  p      ∑          y      =      m      +      1              ∞        (  1  −  p      )    y    =  1  −  (  1  −  p      )          m      +      1        +  p            (      1      −      p              )                  m          +          1                            1      −      (      1      −      p      )        =  1

Probability function of Y=max{X,m} for m positive integer when X is geometric distribution

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Answer & Explanation

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