A geometrically distributed random variable where the first success probability is differentYour first success probability p a is higher on the first trial, than on the remaining trials (where it is p b , constantly). How does this impact the mean of the general experiment (1/p in the general, fixed geometric case) and the standard deviation ( 1 − p ) / p 2 in the general, fixed geometric case)?I think I got the mean: p a + ( 1 − p a ) ( 1 / p b + 1 )

Question

A geometrically distributed random variable where the first success probability is differentYour first success probability       p    a   is higher on the first trial, than on the remaining trials (where it is       p    b  , constantly). How does this impact the mean of the general experiment (1/p in the general, fixed geometric case) and the standard deviation       (    1    −    p    )          /              p      2       in the general, fixed geometric case)?I think I got the mean:      p    a    +  (  1  −      p    a    )  (  1      /        p    b    +  1  )

Tristan Pittman · Accepted Answer

Step 1It seems you&#039;ve already figured out how to adjust the expected value: Consider the experiment as one trial with probability       p    a  , and with probability   1  −      p    a   a standard series of Bernoulli trials with parameter       p    b   plus one extra trial.Since the variance can be written as the difference of two expectation values, you can do the same thing for the variance. From   ⟨  n  ⟩  =  1      /    p and       σ    2    =  (  1  −  p  )      /        p    2    =  ⟨      n    2    ⟩  −  ⟨  n      ⟩    2   in the unmodified case, we get   ⟨      n    2    ⟩  =  (  1  −  p  )      /        p    2    +  1      /        p    2    =  (  2  −  p  )      /        p    2  .Step 2Thus in the modified case, we have                     ⟨                  n          ′                ⟩                            =                  p          a                +        (        1        −                  p          a                )        ⟨        n        +        1        ⟩                                          =                  p          a                +        (        1        −                  p          a                )        (        1                  /                          p          b                +        1        )                                          =        1        +        (        1        −                  p          a                )                  /                          p          b                        ,                            ⟨                  n          ′                −        1                  ⟩          2                                    =                              (            ⟨                          n              ′                        ⟩            −            1            )                    2                                                  =        (        1        −                  p          a                          )          2                          /                          p          b          2                        ,                            ⟨        (                  n          ′                −        1                  )          2                ⟩                            =                  p          a                ⋅        0        +        (        1        −                  p          a                )        ⟨                  n          2                ⟩                                          =        (        1        −                  p          a                )        (        2        −                  p          b                )                  /                          p          b          2                        ,            and thus                               σ                      ′            2                                              =        ⟨        (                  n          ′                −        1                  )          2                ⟩        −        ⟨                  n          ′                −        1                  ⟩          2                                                  =        (        1        −                  p          a                )        (        2        −                  p          b                )                  /                          p          b          2                −        (        1        −                  p          a                          )          2                          /                          p          b          2                                                  =                              1            −                          p              a                                            p            b            2                          (        1        +                  p          a                −                  p          b                )                .            To check the result, note that it yields the correct values 0 for       p    a    =  1,   (  1  −      p    b    )      /        p    b    2   for       p    a    =  0 and   (  1  −      p    a    )      p    a   for       p    b    =  1

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