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

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

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Answer 1 · 2015-11-19T02:30:13

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

Explanation:

expected value $= \frac{1}{p}$ $=1/p$

For example, if $p = \frac{1}{3}$ $p=1/3$ , then the expected value is $3$ $3$

hope that helped

Answer 2 · 2018-01-11T23:57:07

$E (X) = 1 / p$ $E(X)=1//p$

Explanation:

Where $k$ $k$ 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 $k {(1 - p)}^{k - 1} p$ $k(1-p)^(k-1)p$ .

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

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

$E (X) = \infty \sum k = 1 k {(1 - p)}^{k - 1} p$ $E(X)=sum_(k=1)^ook(1-p)^(k-1)p$

$E (X) = p \infty \sum k = 1 k {(1 - p)}^{k - 1}$ $color(white)(E(X))=psum_(k=1)^ook(1-p)^(k-1)$

$E (X) = p (1 + 2 (1 - p) + 3 {(1 - p)}^{2} + 4 {(1 - p)}^{3} + \dots)$ $color(white)(E(X))=p(1+2(1-p)+3(1-p)^2+4(1-p)^3+cdots)$

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 $k$ $k$ can be rewritten into the infinite sum of infinite sums starting at ascending values.

$E (X) = p (\infty \sum k = 1 {(1 - p)}^{k - 1} + \infty \sum k = 2 {(1 - p)}^{k - 1} + \infty \sum k = 3 {(1 - p)}^{k - 1} + \dots)$ $color(white)(E(X))=p(sum_(k=1)^oo(1-p)^(k-1)+sum_(k=2)^oo(1-p)^(k-1)+sum_(k=3)^oo(1-p)^(k-1)+cdots)$

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

$E (X) = p (\frac{1}{1 - (1 - p)} + \frac{1 - p}{1 - (1 - p)} + \frac{{(1 - p)}^{2}}{1 - (1 - p)} + \dots)$ $color(white)(E(X))=p(1/(1-(1-p))+(1-p)/(1-(1-p))+(1-p)^2/(1-(1-p))+cdots)$

$E (X) = 1 + (1 - p) + {(1 - p)}^{2} + \dots$ $color(white)(E(X))=1+(1-p)+(1-p)^2+cdots$

Which is another geometric series:

$E (X) = \frac{1}{1 - (1 - p)}$ $color(white)(E(X))=1/(1-(1-p))$

$E (X) = \frac{1}{p}$ $color(white)(E(X))=1/p$

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

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

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