Finding the Probability function of the remainder.Let X have a geometric distribution with f ( x ) = p ( 1 − p ) x, x = 0 , 1 , 2 , …. Find the probability function of R, the remainder when X is divided by 4.I am stuck on the above practice question: If I say R = 0 , 1 / 4 , 2 / 4 , …… and then use the continuous case equivalent of geometric distribution to find the probability function of R, which is exponential distribution, will I be correct?

Question

Finding the Probability function of the remainder.Let X have a geometric distribution with   f  (  x  )  =  p  (  1  −  p  )  x,   x  =  0  ,  1  ,  2  ,  …. Find the probability function of R, the remainder when X is divided by 4.I am stuck on the above practice question: If I say   R  =  0  ,  1      /    4  ,  2      /    4  ,  …… and then use the continuous case equivalent of geometric distribution to find the probability function of R, which is exponential distribution, will I be correct?

Dominique Mayer · Accepted Answer

Step 1I&#039;m not sure what you mean by &quot;using the continuous case equivalent of geometric distribution to find the probability function of R&quot;. However, here&#039;s how I would approach the problem:When you divide X by R, there are three possible remainders: 0,1,2, and 3. You will get a remainder of 0 for multiples of 4, that is, when   X  =  0  ,  4  ,  8  ,  …. You will get a remainder of 1 for numbers one more than the multiples of 4, such as 1,5,9,…. Hopefully, you get the idea by now.With that in mind, the probabilities are as follows: clearly,   P  (  R  =  r  )  =  0 for all numbers r besides 0,1,2,3. For 0,1,2,3, we have:                    P        (        R        =        0        )                            =        p        (        1        −        p                  )          0                +        p        (        1        −        p                  )          4                +        p        (        1        −        p                  )          8                +        ⋯                                          =                  ∑                      k            =            0                    ∞                p        (        1        −        p                  )                      4            k                                              P        (        R        =        1        )                            =        p        (        1        −        p                  )          1                +        p        (        1        −        p                  )          5                +        p        (        1        −        p                  )          9                +        ⋯                                          =                  ∑                      k            =            0                    ∞                p        (        1        −        p                  )                      4            k            +            1                                              P        (        R        =        2        )                            =        p        (        1        −        p                  )          2                +        p        (        1        −        p                  )          6                +        p        (        1        −        p                  )          1                0        +        ⋯                                          =                  ∑                      k            =            0                    ∞                p        (        1        −        p                  )                      4            k            +            2                                              P        (        R        =        3        )                            =        p        (        1        −        p                  )          3                +        p        (        1        −        p                  )          7                +        p        (        1        −        p                  )          1                1        +        ⋯                                          =                  ∑                      k            =            0                    ∞                p        (        1        −        p                  )                      4            k            +            3                              Step 2From there, you could use the formula for the sum of a geometric series. Alternatively, you could simply make the following observations:  P  (  R  =  3  )  =  (  1  −  p  )  P  (  R  =  2  )    P  (  R  =  2  )  =  (  1  −  p  )  P  (  R  =  1  )    P  (  R  =  1  )  =  (  1  −  p  )  P  (  R  =  0  )    P  (  R  =  0  )  +  P  (  R  =  1  )  +  P  (  R  =  2  )  +  P  (  R  =  3  )  =  1And solve the above system of equations to find the required probabilities.

Let X have a geometric distribution with f(x)=p(1-p)x, x=0,1,2,… Find the probability function of R, the remainder when X is divided by 4.

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