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Thread: Coin tossing, Expected values N and N^2

  1. #1
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    Coin tossing, Expected values N and N^2

    Here's the problem I'm working with, what I'm stumped on is below:
    A biased coin has probability p of coming up Heads. The coin is tossed repeatedly. Let N be the number of tosses until Heads appears for the first time. By conditioning on the outcome of the first toss, find the expected values of $\displaystyle N$ and $\displaystyle N^2$. Use this to find the variance of $\displaystyle N$.

    So, I know how to compute $\displaystyle E[N]$ & $\displaystyle E[N^2]$ here without conditioning on the first toss (i.e. I know what the answers are), but I'm a little stumped on how to get $\displaystyle E[N^2]$ via that route. I know I can utilize $\displaystyle E[N^2]=E[E[N^2|Y]]$, but I'm not sure how to solve $\displaystyle E[N^2|Y]$. I think a little nudge in the right direction is all I need! Thanks
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  2. #2
    Grand Panjandrum
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    Re: Coin tossing, Expected values N and N^2

    Quote Originally Posted by puggles View Post
    Here's the problem I'm working with, what I'm stumped on is below:
    A biased coin has probability p of coming up Heads. The coin is tossed repeatedly. Let N be the number of tosses until Heads appears for the first time. By conditioning on the outcome of the first toss, find the expected values of $\displaystyle N$ and $\displaystyle N^2$. Use this to find the variance of $\displaystyle N$.

    So, I know how to compute $\displaystyle E[N]$ & $\displaystyle E[N^2]$ here without conditioning on the first toss (i.e. I know what the answers are), but I'm a little stumped on how to get $\displaystyle E[N^2]$ via that route. I know I can utilize $\displaystyle E[N^2]=E[E[N^2|Y]]$, but I'm not sure how to solve $\displaystyle E[N^2|Y]$. I think a little nudge in the right direction is all I need! Thanks
    $\displaystyle E(N^2|Y=y)=\sum_n n^2 p(n|y)$

    CB
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  3. #3
    MHF Contributor chisigma's Avatar
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    Re: Coin tossing, Expected values N and N^2

    Quote Originally Posted by puggles View Post
    Here's the problem I'm working with, what I'm stumped on is below:
    A biased coin has probability p of coming up Heads. The coin is tossed repeatedly. Let N be the number of tosses until Heads appears for the first time. By conditioning on the outcome of the first toss, find the expected values of $\displaystyle N$ and $\displaystyle N^2$. Use this to find the variance of $\displaystyle N$.

    So, I know how to compute $\displaystyle E[N]$ & $\displaystyle E[N^2]$ here without conditioning on the first toss (i.e. I know what the answers are), but I'm a little stumped on how to get $\displaystyle E[N^2]$ via that route. I know I can utilize $\displaystyle E[N^2]=E[E[N^2|Y]]$, but I'm not sure how to solve $\displaystyle E[N^2|Y]$. I think a little nudge in the right direction is all I need! Thanks
    The probability that a head appears for the first time at the k-th toss is...

    $\displaystyle p_{k}= \frac{1}{2^{k}}$ (1)

    ... and from (1) You derive...

    $\displaystyle E \{n\} = \sum_{k=1}^{\infty} \frac{k}{2^{k}}$ (2)

    $\displaystyle E \{n^{2}\} = \sum_{k=1}^{\infty} \frac{k^{2}}{2^{k}}$ (3)

    For the series (2) You can consider that for $\displaystyle |x|<1$ is...

    $\displaystyle \sum_{k=0}^{\infty} x^{k}= \frac{1}{1-x} \implies \sum_{k=1}^{\infty} k\ x^{k-1}= \frac{1}{(1-x)^{2}} \implies\sum_{k=1}^{\infty} k\ x^{k}= \frac{x}{(1-x)^{2}} $ (4)

    ... and inserting in (4) $\displaystyle x=\frac{1}{2}$ You obtain...

    $\displaystyle E \{n\}= 2$ (5)

    For the series (3) You can consider that for $\displaystyle |x|<1$ is...

    $\displaystyle \sum_{k=2}^{\infty} k\ (k-1)\ x^{k-2}= \frac{2}{(1-x)^{3}} \implies \sum_{k=2}^{\infty} k\ (k-1)\ x^{k}= \frac{2\ x^{2}}{(1-x)^{3}} $ (6)

    ... and combining (4), (6) and a bit of patience You arrive to write...

    $\displaystyle \sum_{k=1}^{\infty} k^{2}\ x^{k} = \frac{x\ (1+x)}{(1-x)^{3}}$ (7)

    ... ... and inserting in (7) $\displaystyle x=\frac{1}{2}$ You obtain...

    $\displaystyle E \{n^{2}\}= 6$ (8)

    The computation of variance using (5) and (8) is left to You...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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    Re: Coin tossing, Expected values N and N^2

    Hmmm, I don't think that's how I'm supposed to approach the problem here (we've already worked out this problem in my class doing the above method). I solved for $\displaystyle E[N]$ by conditioning on the first toss:
    let Y = 0 If first toss is tails & 1 if first toss is heads
    then $\displaystyle E[N|Y=1] = 1 $ because the first toss would give you heads
    and $\displaystyle E[N|Y=0] = 1 + E[N] $

    then: $\displaystyle E[N]=E[N|Y=0]P(Y=0) + E[N|Y=1]P(Y=1) = 1/p$

    I was hoping that $\displaystyle E[N^2]=E[N^2|Y=0]P(Y=0) + E[N^2|Y=1]P(Y=1) $
    would be just as easy, but like I said earlier, $\displaystyle E[N^2|Y]$ doesn't jump out right away like the earlier work I did.
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  5. #5
    Grand Panjandrum
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    Re: Coin tossing, Expected values N and N^2

    OK time to be explicit: The conditional probabilities are:

    $\displaystyle Pr(N=1|Y=1)=1$

    $\displaystyle Pr(N > 1|Y=1)=0$

    $\displaystyle Pr(N=n|Y=0)=Pr(N=n-1);\ \ n>1$

    $\displaystyle Pr(N=1|Y=0)=0$.

    $\displaystyle E(N^2|Y=0)= \sum_{n=1}^{\infty}n^2 Pr(n|Y=0)$

    $\displaystyle E(N^2|Y=1)=1$

    CB
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  6. #6
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    Re: Coin tossing, Expected values N and N^2

    Thanks for the clarification!
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