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In my lecture notes there is written an equation

$$\mathbb{E}[X_{t+1} \mid X_t =x ] \leq (1-\delta) x,$$

(how this equation is derived, does not really matter here).

Then the next part is:

"Taking the expectation on both sides,

$$\mathbb{E}[\mathbb{E}[X_{t+1} \mid X_t ] \mid X_0 = s_0] \leq (1-\delta) \mathbb{E}[X_t \mid X_0 = s_0]. \quad (1)$$

By the definition of a conditional expectation, we have

$$\mathbb{E}[X_{t+1} \mid X_0 = s_0]= \mathbb{E}[\mathbb{E}[X_{t+1} \mid X_t ] \mid X_0 = s_0], \quad (2)$$

so

$$ \mathbb{E}[X_{t+1} \mid X_0=s_0] \leq (1-\delta) \mathbb{E}[X_t \mid X_0 = s_0]." \quad (3)$$

So I am really confused about this derivation and need some help further explaining these steps. The step from (2) to (3), however, is obvious, since it is only putting (1) into (2). I am a beginner in working with conditional expectations, so this is probably why I have no idea how to tackle such a problem. Thank you very much for your help.

Edit: Maybe one sentence about what confuses me the most: Taking the conditional expectation (conditioned on $X_0=s_0$) seems to be legit, but why does the conditioning on $X_t=x$ (in the first equation) disappear?

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That first (un-numbered) inequality is valid for every possible value $x$ of $X_t$. If so, then the inequality is valid for $X_t$ itself: $$ E(X_{t+1}|X_t) \le (1-\delta)X_t\tag0$$ That's how the conditioning on $X_t=x$ disappeared; it was replaced by the above statement about $X_t$, omitted from your lecture notes, which I've labeled ($0$).

Line ($1$) is a straightforward application of $E(\cdots|X_0=s_0)$ to both sides of ($0$). Line ($2$) follows from the def of conditional expectation because $\sigma(X_0)$ is a coarser $\sigma$-algebra than $\sigma(X_t)$ (the "tower" property of conditional expectation).

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  • $\begingroup$ Thank you very much. This explains everything and is now clear to me. Also thanks for mentioning the name of this tower property, now I know what I have to learn/repeat in order to understand similar derivations, so I'll have a look at the properties of conditional expectation. Thanks a lot. $\endgroup$ Commented Feb 21, 2015 at 8:56
  • $\begingroup$ I posted a question concerning this power property here (math.stackexchange.com/questions/1158742/…) and got the answer that this is not correct. I must misunderstand something and I would really appreciate if you could help me here! $\endgroup$ Commented Feb 22, 2015 at 9:00
  • $\begingroup$ The remarks under your second question are correct: the tower property holds if $\sigma(X_0)$ is contained in $\sigma(X_t)$, which I was assuming to be true in your example. Can you provide additional information about the what the $\{X_t\}$ represent? $\endgroup$ Commented Feb 22, 2015 at 16:15
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    $\begingroup$ This is useful information because $X_0$ is constant, so $\sigma(X_0)$ is a trivial sigma-field (it consists of the empty set and the entire space), hence it is indeed contained within $\sigma(X_t)$ for every $t$, so we're good to apply the tower property. $\endgroup$ Commented Feb 23, 2015 at 17:45
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    $\begingroup$ The intuition is that conditional expectation is a coarsening operation on a random variable. Conditioning on a smaller (coarser) sigma-algebra forces the result to be more discretized (in the extreme case, the result is constant), while conditioning on a finer sigma algebra allows more of the original random variable to be "retained" (in the extreme case, $X$ is unchanged). In general if $\mathcal G$ and $\mathcal F$ are sigma-algebras with $\mathcal G\subset\mathcal F$, then the coarser sigma-algebra prevails: $E(E(X|\mathcal G)|\mathcal F) = E(E(X|\mathcal F)|\mathcal G) = E(X|\mathcal G)$. $\endgroup$ Commented Feb 24, 2015 at 18:29

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