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    Please see attached screenshot!

    What happens between 'starting from x=1' to 'using y=x-1'
    I know its times by (1-n) and divide by (1-n) - but I don't know why the 'to the power of x' goes to x-1 and how the x within the sum goes to (y+1)..

    All help appreciated..
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    (Original post by JKITFC)
    Please see attached screenshot!

    What happens between 'starting from x=1' to 'using y=x-1'
    I know its times by (1-n) and divide by (1-n) - but I don't know why the 'to the power of x' goes to x-1 and how the x within the sum goes to (y+1)..

    All help appreciated..
    It's just using the rule of indices:

    a^n = a^{n-1} \times a.

    In this case: (1-\pi)^x = (1-\pi)^{x-1} \times (1-\pi).

    Since (1-\pi) is just a constant, you can pull it out of the sum:

    \displaystyle

\begin{equation*}\sum x(1-\pi)^x = \sum x (1-\pi)^{x-1} (1-\pi) = (1-\pi)\sum x(1-\pi)^{x-1}\end{equation*}
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    (Original post by Zacken)
    It's just using the rule of indices:

    a^n = a^{n-1} \times a.

    In this case: (1-\pi)^x = (1-\pi)^{x-1} \times (1-\pi).

    Since (1-\pi) is just a constant, you can pull it out of the sum:

    \displaystyle

\begin{equation*}\sum x(1-\pi)^x = \sum x (1-\pi)^{x-1} (1-\pi) = (1-\pi)\sum x(1-\pi)^{x-1}\end{equation*}
    Perfect - thankyou
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    (Original post by JKITFC)
    Perfect - thankyou
    And for the next part, think about it just like a "integration substitution"

    The sub y=x-1 means that when x=1, we have y = 1-1 = 0 (that's the bottom limit sorted) and the top limit is y = \infty -1 = \infty so that stays the same.

    Then, the y = x-1 means that the (1-\pi)^{x-1} = (1-\pi)^y and the x = y + 1 by re-arranging the sub y=x-1 by adding 1 to both sides of the equation.
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    Thanks - that part clicked just after I posted!

    I get the fact that the second part of the addition is a PMF so equals one, and the fact the first part = E(X).
    How does (Y+1) go to y(1-n) - is it becasue the (1-n) outside the sum multiplies with the (y+1) inside?
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    (Original post by JKITFC)
    Thanks - that part clicked just after I posted!

    I get the fact that the second part of the addition is a PMF so equals one, and the fact the first part = E(X).
    How does (Y+1) go to y(1-n) - is it becasue the (1-n) outside the sum multiplies with the (y+1) inside?
    If you're asking about (1-\pi)[E(X) + 1], then yeah - it's just multipling out, just like a(b+c) = ab + bc; we have: (1-\pi)[E(X) + 1] = (1-\pi)E(X) +(1-\pi)1
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    (Original post by Zacken)
    If you're asking about (1-\pi)[E(X) + 1], then yeah - it's just multipling out, just like a(b+c) = ab + bc; we have: (1-\pi)[E(X) + 1] = (1-\pi)E(X) +(1-\pi)1
    I mean from the very top line of the second screenshot.
    How does (Y+1) go to y(1-n) - so how do they make the (y+1) at the bottom of the first screenshot into the E(X) in the top line of the second screenshot?
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    (Original post by JKITFC)
    I mean from the very top line of the second screenshot.
    Oh, I think I understand what you mean.

    \displaystyle 

\begin{equation*}(y+1)(1-\pi)^y \pi  = (y)(1-\pi)^y \pi + (1)(1-\pi)^y \pi \end{equation*}

    By expanding the (y+1)(a) = ay + a where a = (1-\pi)^y \pi.

    So:

    \displaystyle

\begin{align*}(1-\pi)\sum(y+1)(1-\pi)^y y &= (1-\pi) \sum \big( (y)(1-\pi)^y \pi + (1)(1-\pi)^y \pi \big) \\ & = (1-\pi)\bigg[ \sum y(1-\pi)^y \pi +  \sum (1-\pi)^y \pi \bigg]\end{equation}
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    (Original post by Zacken)
    Oh, I think I understand what you mean.

    \displaystyle 

\begin{equation*}(y+1)(1-\pi)^y \pi  = (y)(1-\pi)^y \pi + (1)(1-\pi)^y \pi \end{equation*}

    By expanding the (y+1)(a) = ay + a where a = (1-\pi)^y \pi.

    So:

    \displaystyle

\begin{align*}(1-\pi)\sum(y+1)(1-\pi)^y y &= (1-\pi) \sum \big( (y)(1-\pi)^y \pi + (1)(1-\pi)^y \pi \big) \\ & = (1-\pi)\bigg[ \sum y(1-\pi)^y \pi +  \sum (1-\pi)^y \pi \bigg]\end{equation}
    Thankyou - exactly what I was looking for - would give rep but wont let me!
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    (Original post by JKITFC)
    Thankyou - exactly what I was looking for - would give rep but wont let me!
    No problem.
 
 
 
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