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$X$ is the random variable that signifies the number of times a coin is thrown such until the appearance of a second "tail". With the probability of an appearance of "tails" on one toss being $p$. Find this expectation.

I know that

$$P\{X=k\}=\binom{k-1}{1}p^2(1-p)^{k-2}$$

I just do not know how to calculate the expectation for this.

David K
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Bozo Vulicevic
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3 Answers3

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Since André has given the complete answer, I will finish mine. This relates to the comment by A.S.

The expectation would be $$ \begin{align} \sum_{k=2}^\infty k\binom{k-1}{1}p^2(1-p)^{k-2} &=2p^2\sum_{k=2}^\infty\binom{k}{2}(1-p)^{k-2}\\ &=2p^2\sum_{k=2}^\infty\binom{k}{k-2}(1-p)^{k-2}\\ &=2p^2\sum_{k=2}^\infty(-1)^{k-2}\binom{-3}{k-2}(1-p)^{k-2}\\ &=2p^2\sum_{k=0}^\infty(-1)^k\binom{-3}{k}(1-p)^k\\ &=2p^2\frac1{(1-(1-p))^3}\\ &=\frac2p \end{align} $$ where $\binom{-3}{k-2}$ is a negative binomial coefficient.

Community
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robjohn
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  • I honestly want you to have the answer too, but the other answer is just as good and done quicker. I hope you can understand. I appreciate your effort very much. :) – Bozo Vulicevic Dec 11 '15 at 21:37
  • The point of my comments is that without showing any of the work you attempted, you are essentially asking us to do your homework for you. This is why I delayed posting the complete solution. Since André already did your homework for you, it seemed pointless not to show an alternate approach. – robjohn Dec 11 '15 at 21:42
  • It's actually not homework in question. – Bozo Vulicevic Dec 11 '15 at 21:44
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    Well, we have no way to know this. It appears so, but this is why this has been abstracted to the idea of wanting context. Context helps us to know where you are having problems and allows us to address this. It also works to prevent people getting their homework done for them. – robjohn Dec 11 '15 at 21:47
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    Thanks, in the future I will take into consideration to give more detail. – Bozo Vulicevic Dec 11 '15 at 21:49
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From your expression for $\Pr(X=k)$ we have $$E(X)=\sum_{k=2}^\infty k(k-1)p^2(1-p)^{k-2}.$$ Let $$f(x)=\frac{1}{1-x}=1+x+x^2+x^3+x^4+\cdots.$$ Take second derivatives. We get $$f''(x)=\frac{2}{(1-x)^3}=\sum_{k=2}^\infty k(k-1)x^{k-2}.$$ Let $x=1-p$ and multiply the result by $p^2$.

Remark: If you already know the mean of the geometric distribution, here is a simple way. Let $U$ be the number of tosses until the first tail, and let $V$ be the number of tosses between the first tail and the second. Then $X=U+V$, where $U$ and $V$ have geometric distribution. The mean of each of $U$ and $V$ is $\frac{1}{p}$, so the mean of $X$ is $\frac{2}{p}$.

André Nicolas
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Let $\mu_{n}$ denote the expectation of the number of times a coin is thrown until the appearance of an $n$-th tail.

Then you are looking for $\mu_{2}$.

Let $E$ denote the event that the first throw is a tail.

Then $\mu_0=0$ and $\mu_{n}=\left(1+\mu_{n-1}\right)P\left(E\right)+\left(1+\mu_{n}\right)P\left(E^{c}\right)$ for $n>0$, so that:

$\mu_{1}=1P\left(E\right)+\left(1+\mu_{1}\right)P\left(E^{c}\right)=1+\mu_{1}\left(1-p\right)$ implying that $\mu_{1}=\frac{1}{p}$.

and:

$\mu_{2}=\left(1+\mu_{1}\right)P\left(E\right)+\left(1+\mu_{2}\right)P\left(E^{c}\right)=\left(1+\frac{1}{p}\right)p+\left(1+\mu_{2}\right)\left(1-p\right)$ implying that $\mu_{2}=\frac{2}{p}$.

With induction it can be shown that $\mu_n=\frac{n}{p}$.

drhab
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