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Let $X_i \sim F_i$, where $F_i = G(F_{i-1})$ where there sequence begins with a known $F_0$ (and all $F_i$ are valid cdfs (monotonic increasing) defined over (0,1)).

I believe from other posts, that it is not enough to show that $\mathbb{E}X_0> \mathbb{E}X_{\infty}$ (where $X_{\infty}$ is the the random variable distributed by $F_i$ as $i \to \infty$, which we can assume is a Dirac mass function) to prove that $\mathbb{E}X_i$ decreases with $i$.

I believe the reason for this is that there could be a converging zig zag behavior, so that the iterations of $F_i$ could make the corresponding expectations go up a little, then down a little, while still converging to a fixed point.

My question is -- are there are properties of the $G$ operator that would allow you to say that there are no zigzags, and thus the above conditions are sufficient for proving the mean decreases with $i$?

Perhaps this is naive, but it would seem to me that most functions that someone like myself would encounter (read: a beginner) would not create zigzag behavior. Since G does not depend on n, it can't be raising a negative number to alternating powers or anything. And we can assume no trig operators.

Is it possible to say something like "$EX_0 > EX_{\infty}$, and the $G$ function only contains such-and-such type operations, therefore $EX_i$ decreases with $i$" ? And if so, what is "such-and-such" ?

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Since $\mathbb{E}(X_i) = \int_0^1 (1-F_i(x)) \mathrm{d} x = \int_0^1 (1-G(F_{i-1}(x))) \mathrm{d} x$, if $G$ is such that $G(0)=0$, $G(1) = 1$ and $0 \le G(x) \le x$, then $\mathbb{E}(X_i) \ge \mathbb{E}(X_{i-1})$. – Sasha Oct 14 '11 at 17:57
@Sasha: Sufficient, but not necessary, correct? – OctaviaQ Oct 14 '11 at 18:08
up vote 1 down vote accepted

It is actually not that hard to find an example of $G(x)$ where the sequence $\mu_k = \mathbb{E}(X_k)$ is not monotonic. Here is one found with little help from Mathematica:

$$ G(x) = x \left( 1 +\frac{2429}{5002} (1-x) -\frac{83}{51} (1-x)^2 + \frac{113}{102} (1-x)^3\right) $$

enter image description here

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