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I'm looking for the Expectation of the discrete random variable X, E[X], with pmf: $$p(x)=(\frac 16)^{x+1}, x=0,1,2,3...$$

so what I tried is as follows... $$E[X]= \sum_{0}^\infty xp(x) =$$

so then $$=\sum_0^\infty x(\frac 16)(\frac 16)^x = \frac 16\sum_{0}^\infty x(\frac 16)^x $$

which is $$ \frac 16 [0(\frac 16)^0+1(\frac 16)^1+2(\frac16)^2+3(\frac 16)^3+...] $$

so if we were to use the rule that: $$ \sum_0^\infty ar^x = \frac {a}{(1-r)} $$ when |r|<1.

Then it seems like the difference between that formula and my problem is the increasing coefficient on the (1/6)^x...

My math book (which doesn't really say anything more about it)... states that "there is a general increasing geometric series relation which is $$1 + 2r + 3r^2 + 4r^3+...= \frac {1}{(1-r)^2} $$

is that what I need to know? and if so, can someone please show me why? Thanks!

Adam
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2 Answers2

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This "general relation" follows from calculus where you learn about Maclaurin series

In particular, it talks about how since

$$\sum_{n=0}^\infty x^n = {1\over 1-x},\quad |x|<1$$

We can take derivatives of both sides and get

$$\sum_{n=0}^\infty{d\over dx}(x^n)={d\over dx}\left(\sum_{n=0}^\infty x^n\right)={d\over dx}\left({1\over 1-x}\right)$$

therefore

$$\sum_{n=0}^\infty nx^{n-1}={1\over (1-x)^2}$$

In your case you use $x$ instead of $n$ and ${1\over 6}$ instead of $x$, but it amounts to the same thing, just using different letters.

So you are trying to solve

$$\sum_{n=0}^\infty n\left({1\over 6}\right)^{n+1}={1\over 36}\sum_{n=0}^\infty n\left({1\over 6}\right)^{n-1}={1\over 36}\cdot{1\over \left(1-{1\over 6}\right)^2}={1\over (6-1)^2}={1\over 25}$$

Addendum

I almost forgot, you should be careful to renormalize your mass function. Note that

$$\sum_{n=0}^\infty\left({1\over 6}\right)^{n+1}={1\over 6}\cdot {1\over 1-{1\over 6}}={1\over 5}\ne 1.$$

This is a problem, because all the probabilities have to add up to $1$. So either there's a missing piece which has the rest of the probability, or you really mean that the pmf is

$$P(X=x)=5\cdot \left({1\over 6}\right)^{x+1}$$

in which case the final answer changes from ${1\over 25}$ to ${1\over 5}$. Either way, something has to change, as without a true probability measure, the answer doesn't make sense in context.

Adam Hughes
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  • So then if we take $$\sum_o^\infty nx^{n-1}$$ and have n=x and x=(1/6)... we get, almost, my initial problem of $$\sum_0^\infty x(1/6)^{x−1}=\frac {1}{(1−x)^2}$$ which is really nearly my problem.. if we multiplied both sides by $$(\frac 16)^2$$ – Adam Aug 31 '14 at 22:04
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    @Adam reread my answer, it has changed since that was what I wrote. (notice the "edited" thing at the bottom of posts, when they happen it means the op has changed something. Often just minor grammar or typos, sometimes substantial changes are made). – Adam Hughes Aug 31 '14 at 22:11
  • thanks buddy, you rock. I'm pretty new to series and it looks like I'm going to need to do some research to find a better understanding of them in general.. And are you referring to renormalizing? Thanks – Adam Aug 31 '14 at 22:12
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    @Adam renormalizing means "making sure the probabilities actually add up to 1 (=100%)" Since your probabilities only add up to ${1\over 5}$ there's an issue with them, since probabilities cannot work that way. Welcome to Math Stack Exchange, and enjoy your time here. As a matter of courtesy, once you are able, you should always be sure to "accept" and "upvote" the answer you think helped you the most (you can only accept one per question, but you can upvote as many as you like.) It also helps ensure the people of the site know you consider your problem "dealt with." – Adam Hughes Aug 31 '14 at 22:14
  • @Adam_Hughes Thanks for the welcoming. I understand what you're saying, probability must sum to 1. Does it normally make me wait a certain amount of time before I get to accept and upvote? – Adam Aug 31 '14 at 22:27
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    @Adam You need I think 15 reputation to upvote--privileges on this site are awarded based on "reptutation," not "time." You can click "accept" or the little check mark next to an answer to accept it, but you need 15 privilege to upvote, which you will get soon enough (and the site will notify you when you gain this privilege). Try accepting now, I am looking at the FAQ and I don't see a reputation requirement to accept.

    You are correct about the probabilities needing to add up to 1.

     – Adam Hughes Aug 31 '14 at 22:36
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Try multiplying out $(1+r+r^2+\cdots)^2$. The coefficient of $r^k$ will be the number of ways of adding two nonnegative integers to get $k$, i.e. $k+1$.

Andrew Dudzik
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  • Ah I see what you're saying...

    well, I multiplied them anyway.. and I got ... 1 + 2r +3r^2...

    so are you saying that.. if I looked at the term with a k in the exponent, then the coefficient is 1 more than that

     – Adam Aug 31 '14 at 22:18