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I have a monotonically increasing series of values $X, x\in[0,1] \ \forall i\in1,2,...,60$ the weighted mean of which is defined by $$\bar{x}=\frac{x_1+\sum_{i=2}^{60} i(x_i-x_{i-1})}{x_{60}}.$$ Given an input $\alpha$, I would like to figure out how to shift the values of $X$ to $X^{'}$ such that the weighted mean of $X^{'}$ is equal to $\alpha$.

For a "real world" example of this, imagine I am a member of a support team who is given a number of new accounts to support each month. We have a curve based on past experience that defines the percentage of accounts starting in any given month that will have left after $y$ months over a 5 year period (a persistence curve, if you will). The weighted mean I described above gives the average time any one account will continue receiving support from my team. If we expect the average time in which accounts will stay with us to increase from, say, 12 months to 18 months, how should I adjust this curve to shift the weighted mean to 18 from 12 while keeping the general shape of the curve the same?

Thanks in advance!

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Generalizing a little,

$\overline{x} =\dfrac{x_1+\sum_{i=2}^{n} i(x_i-x_{i-1})}{x_{n}}\\ $

So, replacing $x_i$ by $x_i+c$, we get

$\begin{array}\\ \overline{x+c} &=\dfrac{x_1+c+\sum_{i=2}^{n} i((x_i+c)-(x_{i-1}+c))}{x_{n}+c}\\ &=\dfrac{x_1+c+\sum_{i=2}^{n} i((x_i+c)-(x_{i-1}+c))}{x_n}\dfrac{x_n}{x_{n}+c}\\ &=\left(\dfrac{x_1+\sum_{i=2}^{n} i(x_i-x_{i-1})}{x_{n}}+\dfrac{c}{x_{n}}\right)\dfrac{x_n}{x_{n}+c}\\ &=\left(\overline{x}+\dfrac{c}{x_{n}}\right)\dfrac{x_n}{x_{n}+c}\\ &=\left(\overline{x}x_n+c\right)\dfrac{1}{x_{n}+c}\\ \end{array} $

and if this $=a$, then $a =\left(\overline{x}x_n+c\right)\dfrac{1}{x_{n}+c} $ or $a(x_n+c) =\overline{x}x_n+c $ or $c(a-1) =\overline{x}x_n-ax_n =x_n(\overline{x}-a) $ or $c =\dfrac{x_n(\overline{x}-a)}{a-1} $.

Note that if you want $a=1$, this requires $\overline{x} = 1$, so no change is needed.

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    $\begingroup$ Thanks for the answer. However, I am having a little trouble with this and was hoping you could help me clarify. Consider the sample sequence $x^T=[0,0.4,.7,.9,1]$. The weighted mean of this sequence is $3$. If I want to change it to $4$, your answer directs me to find $c=(1)*(4-3)=1$. Then I add this to each $x$ to get $y^T=[1,1.4,1.7,1.9,2]$. However, the weighted mean of this sequence is $2$. Am I doing something incorrectly here? $\endgroup$ – Jon May 3 '16 at 0:27
  • $\begingroup$ I see my error. I am trying to fix it. Thanks. $\endgroup$ – marty cohen May 3 '16 at 1:52

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