Finding the probability distribution of the sum of geometric distributions?X and Y are both geometric distributions with success p. What it the P r ( X + Y = n )Would I use a convolution with a sum for this? Do I need to define a third random variable?

Question

Finding the probability distribution of the sum of geometric distributions?X and Y are both geometric distributions with success p. What it the   P  r  (  X  +  Y  =  n  )Would I use a convolution with a sum for this? Do I need to define a third random variable?

asijikisi67 · Accepted Answer

Explanation:We have                                   Pr        (        X        +        Y        =        n        )                            =                                       ∑                      i            =            1                                n            −            1                          Pr        (        X        =        i        )        ⋅        Pr        (        Y        =        n        −        i        ∣        X        =        i        )                            =                                       ∑                      i            =            1                                n            −            1                          Pr        (        X        =        i        )        ⋅        Pr        (        Y        =        n        −        i        )                            =                                       ∑                      i            =            1                                n            −            1                          p        (        1        −        p                  )                      i            −            1                          ⋅        p        (        1        −        p                  )                      n            −            i            −            1                                              =                                       ∑                      i            =            1                                n            −            1                                    p          2                (        1        −        p                  )                      n            −            2                                              =                            (        n        −        1        )                  p          2                (        1        −        p                  )                      n            −            2

Alan Sherman · Accepted Answer

Step 1From the definition of Geometric Random Variables:If X,Y are the independent counts of trials, from sequences of Bernoulli trials, until first success (with rate p), then   X  +  Y would be the count of trials from a sequence of Bernoulli trials until the second success. Step 2We would then be seeking the probability of obtaining exactly one success somewhere among   n  −  1 trials, then a second success.      P    (  X  +  Y  =  n  )  =  (  n  −  1  )      p    2    (  1  −  p      )          n      −      2                     1              n      ∈      {      2      ,      3      ,      …      }

X and Y are both geometric distributions with success p. What it the Pr(X+Y=n). Would I use a convolution with a sum for this? Do I need to define a third random variable?

Answered question

Answer & Explanation

New Questions in High school geometry