What is the formula of the expected value of a geometric random variable?

2 Answers
Nov 19, 2015

If you have a geometric distribution with parameter #=p#, then the expected value or mean of the distribution is ...

Explanation:

expected value #=1/p#

For example, if #p=1/3#, then the expected value is #3#

hope that helped

Jan 11, 2018

#E(X)=1//p#

Explanation:

Where #k# is the number of trials that have elapsed, we see that the number of trials multiplied by the probability of the series ending at that trial is #k(1-p)^(k-1)p#.

Note that #(1-p)^(k-1)p# is the probability of #k# trials having elapsed, where #p# is the probability of the event occurring.

So, the expected value is given by the sum of all the possible trials occurring:

#E(X)=sum_(k=1)^ook(1-p)^(k-1)p#

#color(white)(E(X))=psum_(k=1)^ook(1-p)^(k-1)#

#color(white)(E(X))=p(1+2(1-p)+3(1-p)^2+4(1-p)^3+cdots)#

In my view, the previous step and the following step are the trickiest bits of algebra in this whole process. Pay close attention to how the #k# can be rewritten into the infinite sum of infinite sums starting at ascending values.

#color(white)(E(X))=p(sum_(k=1)^oo(1-p)^(k-1)+sum_(k=2)^oo(1-p)^(k-1)+sum_(k=3)^oo(1-p)^(k-1)+cdots)#

Note that #0lt p lt1#, so we also have that #0 lt 1-p lt 1#. Thus, we can use the sum of the infinite geometric series, i.e., that #sum_(k=1)^oor^(k-1)=1/(1-r)#.

#color(white)(E(X))=p(1/(1-(1-p))+(1-p)/(1-(1-p))+(1-p)^2/(1-(1-p))+cdots)#

#color(white)(E(X))=1+(1-p)+(1-p)^2+cdots#

Which is another geometric series:

#color(white)(E(X))=1/(1-(1-p))#

#color(white)(E(X))=1/p#

So, the expected number of trials is #1//p#.