How to prove inequality between probabilities of negative-binomial random variable and geometric variable?Let X be a negative binomial random variable of parameters r, p, then X = Y 1 + Y 2 + … + Y r , where Y j , j = 1 , … , r are geometric random.variables of parameter p. Show that: P ( X &gt; k ) ≤ r P ( Y 1 &gt; k / r )

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How to prove inequality between probabilities of negative-binomial random variable and geometric variable?Let X be a negative binomial random variable of parameters r, p, then   X  =      Y    1    +      Y    2    +  …  +      Y    r  , where       Y    j   ,  j  =  1  ,  …  ,  r are geometric random.variables of parameter p. Show that:       P    (  X  &amp;gt;  k  )  ≤  r      P    (      Y    1    &amp;gt;  k      /    r  )

Jean Deleon · Accepted Answer

Step 1The proposed inequality is true in general (continuous as well as discrete, bounded support or unbounded), not just for the nice and well-known examples of NegativeBinomial (sum of Geometric), Binomial (sum of Bernoulli), or Gamma (sum of exponential) etc.Rewrite the left hand side as       P              {        X  &amp;gt;  k            }        =  1  −      P              {        X  ≤  k            }        =  1  −      P              {                  (            ∑          i      =      1        r        Y    i              )        ≤  k            }           .Then, note that the event       (          ∑              i        =        1            r              Y      i        )    ≤  k behaves as if it is a union that contains the intersection of the individual events       ⋂          i      =      1        r        (          Y      i        ≤          k      r        )  . To be convinced of this fact, please consider the events       Y    1    +      Y    2    ≤  2 versus       Y    1    ≤  1  ⋂      Y    2    ≤  1. Step 2Geometrically, one can view this as the simplex solid being larger than the cube. For example, with all       Y    i   being non-negative, the simplex solid       Y    1    +      Y    2    +      Y    3    ≤  3 clearly contains the entirety of the unit cube       Y    1    ≤  1  ⋂      Y    2    ≤  1  ⋂      Y    3    ≤  1.Therefore                               P                                      {                          X        &amp;gt;        k                              }                          =        1        −                                                                              P                                                                      {                                                                    (                                                            ∑                                              i                        =                        1                                            r                                                              Y                      i                                                        )                                ≤                k                                                      }                                                              ⏞                                            larger event                                              ≤        1        −                                                                              P                                                                      {                                                                                        ⋂                                          i                      =                      1                                        r                                                        (                                          Y                      i                                        ≤                                          k                      r                                        )                                                                              }                                                                                  ⏞                                            smaller event                                                            =                  P                                      {                                                [                                                ⋂                          i              =              1                        r                                (                          Y              i                        ≤                          k              r                        )                                                              ]                                      c                                              }                                                                        =                  P                                      {                                                ⋃                          i              =              1                        r                                (                          Y              i                        &amp;gt;                          k              r                        )                                              }                                           .                    Now, the probability of the union is smaller than the sum of the probabilities, by Boole&#039;s inequality, also known as Bonferroni inequality:      P              {                  ⋃              i        =        1            r        (          Y      i        &amp;gt;          k      r        )                            }                    ≤              ∑                  i          =          1                r                    P                              {                            Y        i            &amp;gt;              k        r                              }                    =      r      ⋅              P                              {                            Y        1            &amp;gt;              k        r                              }                    which is the right hand side.Both &quot;simplex versus hypercube&quot; and Bonferroni inequality are very loose, so the overall inequality is very loose.

Let X be a negative binomial random variable of parameters r,p, then X=Y_1+Y_2+…+Y_r, where Y_j , j=1,…,r are geometric random.variables of parameter p. Show that: P(X>k) le rP(Y1>k/r)

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