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Math Help - independant random variables ?

  1. #1
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    independant random variables ?

    Suppose that X and Y are independent random variables with the same geometric distribution,  P(X = k) = P(Y = k) = pq^(k-1) where q = 1 - p

    Find
    P( X = k | X + Y = n + 1) n = 2,3,....
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  2. #2
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    Quote Originally Posted by abrahamtim View Post
    Suppose that X and Y are independent random variables with the same geometric distribution,  P(X = k) = P(Y = k) = pq^(k-1) where q = 1 - p

    Find
    P( X = k | X + Y = n + 1) n = 2,3,....
    I suggest you first work with concrete cases. eg. n = 2 and k = 0, 1, 2, 3. etc. Then see if you can generalise.
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  3. #3
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    Hello,

    Let's call f the function : f(x)=pq^{x-1}

    We can notice that \displaystyle p\frac{f(x)}{f(y)}=f(x-y)

    From Bayes Formula, it appears that
    \displaystyle \begin{aligned} P(X=k|X+Y=n+1)&=\frac{P(X=k,X+Y=n+1)}{P(X+Y=n+1)} \\<br />
&=\frac{P(X=k,Y=n-k+1)}{P(X+Y=n+1)} \\<br />
&=\frac{f(k)f(n-k+1)}{P(X+Y=n+1)} \end{aligned}
    (by independence for the last line)

    But \{X+Y=n+1\} is the disjoint union \displaystyle \bigcup_{j=1}^n \{X=j,Y=n-j+1\}

    Hence
    \displaystyle \begin{aligned}P(X+Y=n+1)&=\sum_{j=1}^n P(X=j,Y=n-j+1) \\<br />
&=\sum_{j=1}^n f(j)f(n-j+1) \end{aligned}


    Back to the formula...
    \displaystyle \begin{aligned}P(X=k|X+Y=n+1)&=\frac{f(k)f(n-k+1)}{\sum_{j=1}^n f(j)f(n-j+1)} \\<br />
&=\frac{1}{\sum_{j=1}^n \frac{f(j)}{f(n-k+1)}\cdot\frac{f(n-j+1)}{f(k)}} \\<br />
&=\frac{1}{\sum_{j=1}^n \frac{pf(j-n+k-1)}{pf(k-n+j-1)}} \\<br />
&=\frac 1n \end{aligned}

    Ain't that nice ?
    Last edited by Moo; March 8th 2011 at 03:45 PM.
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