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Find the expectation of a Geometric distribution using $\mathbb{E}(X)= \sum_{k=1}^\infty P(X \ge k)$.

Okay I know how to find the expectation using the definition of the geometric distribution $P(X=k)= p \cdot(1-p)^{k-1}$ and I figured that $P(X \ge k)=(1-p)^{k-1}$ but I don't know how to show it.

I know the expectation is $\frac{1}{p}$ but I just get $\mathbb E(X)= \frac{1}{p^2}$ using the method specified in the question.

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For |r|<1, the sum of the geometric series $\sum\limits_{k=1}^\infty r^k$ is ${ r\over 1-r}$. So, write $\sum\limits_{k=1}^\infty P[X\ge k]= \sum\limits_{k=1}^\infty (1-p)^{k-1} = {1\over 1-p}\sum\limits_{k=1}^\infty (1-p)^{k },$ and apply the formula with $r=1-p$.

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The answer that is here does not address one aspect of the question:

I figured that $P(X \ge k)=(1-p)^{k-1}$ but I don't know how to show it.[...]

Here is an hint:

$P(X \ge k)=\sum_{i=k}^\infty P(X=i)$

Now, to evaluate the above sum, you need the sum of the geometric series:

For |r|<1, $\sum_{i=k}^\infty r^i=\frac{ r^k }{1-r}$

The rest of the details are there in David's answer...but in case you need to know more about one or more of this, you may want to ping me here...

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    @DilipSarwate That's true. I agree. :)2012-04-20
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A simpler way would be to plug in $q=1-p$ and solve it that way using formula for geometric sequences:

\begin{align*} E(X) &= \sum\limits_{k=1}^\infty kpq^{k-1}\\ &= \frac{p}{q} \sum\limits_{k=1}^\infty kq^{k}\\ &= \frac{p}{q} \frac{q}{(1-q)^2}\\ &= \frac{p}{q} \frac{q}{p^2}\\ &= \frac{p}{q} \frac{q}{p^2}\\ &= \frac{1}{p}. \end{align*}