Question

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

Answer 1

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

Answer 2

`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`.