Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What is $f(t)=X_{t+1}$, if $X_{t+1}=(1-p)(1-X_{t})+pX_{t}$ and $X_{0},p \in [0,1]$?

And what are general methods for finding functions defined by such recurrent equations?

share|improve this question
    
I'm not sure that the ~s were for, so I got rid of them. Also, it's probably better to use *words* to get italics, rather than math mode. I thought it looked kind of funky so I put it in a block quote. –  Dylan Moreland Jul 20 '12 at 23:14

2 Answers 2

up vote 4 down vote accepted

I will assume that $t$ ranges over, say, the non-negative integers.

For your particular example, we have $$\begin{align*}X_{t+2}&=(1-p)(1-X_{t+1})+pX_{t+1},\\ X_{t+1}&=(1-p)(1-X_t)+pX_t.\end{align*}$$

Subtract and simplify. We get $$X_{t+2}-X_{t+1}=(2p-1)(X_{t+1}-X_t).$$ So if $Y_t=X_{t+1}-X_t$, we find that $Y_{t+1}=(2p-1)Y_t$, which is easily solved, and therefore $X_t$ is $X_0$ plus the sum of a geometric series.

Remark: There are a number of good general methods for solving linear recurrences with constant coefficients. The generating functions approach is quite useful. Or else in our case we can rewrite our recurrence as $$X_{t+1}=(2p-1)X_t + 1-p.$$ First find a general solution of the homogeneous recurrence $X_{t+1}=(2p-1)X_t$. This is easy. Then find a particular solution of our non-homogeneous recurrence. Easy, by guessing that a constant might work. Then the general solution of our recurrence is the general solution of the homogeneous recurrence, plus our particular solution.

share|improve this answer
    
@TMM: Thank you, fixed. Someday I may write a typo-free answer. –  André Nicolas Jul 20 '12 at 23:30
1  
Surely with your 80k worth of contributions to this site, you will be forgiven for the occasional typo :) –  TMM Jul 20 '12 at 23:32

Well, you have $X_{t+1}$ equal to a weighted average between $X_{t}$ and $1 - X_{t}$. So there's a fixed point where these are equal, i.e., at $X=1/2$. To find the behavior away from that fixed point, define $Y_{t} = X_{t} - 1/2$: then $$ \begin{eqnarray} Y_{t+1}&=&X_{t+1}-\frac{1}{2} \\ &=& \left(1-p\right)\left(1 - X_{t}\right) + pX_{t} - \frac{1}{2}\\ &=& \left(1 - p\right)\left(\frac{1}{2} - Y_t\right) + p\left(Y_{t} + \frac{1}{2}\right) -\frac{1}{2}\\ &=& (2p-1)Y_{t}. \end{eqnarray} $$ So $Y_{t}$ is multiplied by a constant at each step, giving $$ X_{t} = Y_{t} + \frac{1}{2} = \left(2p-1\right)^{t} Y_{0} + \frac{1}{2} = \left(2p-1\right)^{t} \left(X_{0} - \frac{1}{2}\right) + \frac{1}{2}. $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.