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Proving sequence x_t is decreasing

  1. Nov 8, 2012 #1
    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. jcsd
  3. Nov 8, 2012 #2

    Simon Bridge

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    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?
     
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