# Proving sequence x_t is decreasing

1. Nov 8, 2012

### peter60185

Hi, I'm sure x_t is a decreasing sequence while y_t is an increasing one. It feels like it should be simple to prove, but I just can't do it. Any suggestions would be great!

Thanks,

Peter

x_t and y_t are defined iteratively by two equations:

1. y_(t+1) = bq x_t + b(1 - q) y _t

2. f[x_(t+1)] = bqf[x_t] + b(1 - q) f[y _t]

IE

2B. x_(t+1) = f^(-1)[bqf[x_t] + b(1 - q) f[y _t]]

Other conditions:

1. f[x] is decreasing and concave
2. x_0 = 1
3. y_0 = 0
4. 0 < q < 1
5. 0 < b < 1
6. f(0) = 1
7. f(1) = 0

2. Nov 8, 2012

### Simon Bridge

Welcome to PF;
You say:
$x_t$ and $y_t$ are defined iteratively by two equations:

1. $y_{t+1} = bq x_t + b(1 - q) y_t$

2. $f(x_{t+1}) = bqf(x_t) + b(1 - q) f(y _t)$

i.e.

2b. $x_{t+1} = f^{-1}(bqf(x_t) + b(1 - q) f(y_t))$

Other conditions:

1. $f(x)$ is decreasing and concave (in what sense "concave"?)
2. $x_0 = 1$
3. $y_0 = 0$
4. $0 < q < 1$
5. $0 < b < 1$
6. $f(0) = 1$
7. $f(1) = 0$

I'd start out by seeing how far I can walk through the iterations by hand.

$y_1=bq$

$f(x_1)=b(1-q)$

... and so on.
Can you rely on $f(x_{t+1})<f(x_t)$ ?
(meaning of "decreasing and concave" see?)

So $f(x_{t>1})<b(1-q)$ and we know $b(1-q)<1$ and $bq<1$ so anything multiplied by them gets smaller.

Perhaps a pattern will emerge?