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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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2 Answers 2

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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The calculation showing that $P\{X\geq k\} = (1-p)^{k-1}$ has appeared many times on this site, and almost every time it has been pointed out that rather than "doing the math" and summing the series, it is simpler to look at the basic problem: the event $\{X\geq k\}$ occurs if and only if the first $k-1$ independent trials ended in failure, and this has probability $(1-p)^{k-1}$. No muss, no fuss, no remembering the sum of the geometric series. –  Dilip Sarwate Apr 20 '12 at 20:02
@DilipSarwate That's true. I agree. :) –  user21436 Apr 20 '12 at 20:19

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