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I am always confused how to change the index of summation.

$$\sum_{y=1}^\infty (1-\theta)^{y-1}\theta$$

The above is supposed to be a geometric sum and sum up to $1-(1-\theta)^x$? But how?

Jiew Meng
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3 Answers3

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Is your upper bound of summation $x$? If yes, set $n=y-1$. $y$ takes the values $1,2,3,\dots x$, so $n$ will take the values $0,1,2,\dots x-1$. So, $$\sum_{y=1}^x(1-\theta)^{y-1}\theta=\sum_{n=0}^{x-1}(1-\theta)^n\theta=\theta\sum_{n=0}^{x-1}(1-\theta)^n=\theta\frac{(1-\theta)^x-1}{(1-\theta)-1}=1-(1-\theta)^x.$$

detnvvp
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  • How did u get from the 3rd expression to the 4th? I think its something to do with the upper limit of the sum being $x-1$ instead of $x$ ... but how? – Jiew Meng Oct 15 '13 at 23:46
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    This comes from the geometric series: the sum of $a^n$ for $n=0,\dots k-1$ is $\frac{a^k-1}{a-1}$. – detnvvp Oct 15 '13 at 23:48
  • Oh ... ic ... the formula I had is just $\sum_{k=0}^\infty z^k = \frac{1}{1-z}$. When I try to work out $\sum_{n=0}^{x-1} z^n = \frac{1}{1-z} - (z^x)$? Reason being: $\sum_{n=0}^{x-1} z^n = \sum_{n=0}^{x} z^n - z^x$? – Jiew Meng Oct 15 '13 at 23:51
  • It's easier to see this from the identity $$a^k-1=(a-1)(a^{k-1}+a^{k-2}+\dots+a+1).$$Then, you divide with $a-1$. – detnvvp Oct 15 '13 at 23:52
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$$\sum_{y=1}^x(1-\theta)^{y-1}\theta$$ more precisely $$\sum_{y=1}^{y=x}(1-\theta)^{y-1}\theta \tag{1}$$ Now the equation of lower bound $$y=1 $$ can be transformed to $$y-1=0$$ and the equation of the upper bound $$y=x$$ to $$y-1=x-1$$ by subtracting $1$ from the equation. We get $$\sum_{y-1=0}^{y-1=x-1}(1-\theta)^{y-1}\theta \tag{2}$$ In this expression we substitute $y-1$ by $y$ to get $$\sum_{y=0}^{y=x-1}(1-\theta)^{y}\theta$$ This substitution is only correct if all old $y$ are substituted by new $y$. This is more clear if in $(2)$ you first substitute $y-1$ by $z$

$$\sum_{z=0}^{z=x-1}(1-\theta)^{z}\theta$$ check if all $y$ have disappeared and now substitute $z$ by $y$ $$\sum_{y=0}^{y=x-1}(1-\theta)^{y}\theta$$

A sligthly other way to process $(1)$. We want to change $y-1$ to $z$ so we set $$y-1=z$$ and from this we get $$y=z+1$$ Substituting this in $(1)$ we get

$$\sum_{z+1=1}^{z+1=x}(1-\theta)^{(z+1)-1}\theta $$

and this can be transformed to

$$\sum_{z+1=1}^{z+1=x}(1-\theta)^{(z+1)-1}\theta $$ or better

$$\sum_{z=0}^{z=x-1}(1-\theta)^{z}\theta $$ You can change the variable name from $z$ to $y$ and get $$\sum_{y=0}^{y=x-1}(1-\theta)^{y}\theta$$

miracle173
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You can see change of variable more clearly if you write the summed terms out:

$$\begin{align*} \sum_{y=1}^x (1-\theta)^{y-1}\theta =& (1-\theta)^{1-1}\theta + (1-\theta)^{2-1}\theta+(1-\theta)^{3-1}\theta + \cdots + (1-\theta)^{x-1}\theta\\ =& (1-\theta)^0\theta + (1-\theta)^1\theta+(1-\theta)^2\theta + \cdots + (1-\theta)^{x-1}\theta\\ =& \sum_{z=0}^{x-1} (1-\theta)^z\theta \end{align*}$$ The remaining is simply a geometric sum, as you have noted.

peterwhy
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