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As it is written here, using the Taylor's expansion, one can write

$$ \mathbb{E}^xf(X_t) \approx f(x)+t Af(x) $$ from $$ Af(x) := \lim_{t \downarrow 0} \frac{\mathbb{E}^x(f(X_t))-f(x)}{t} $$ where $A$ is the infinitesimal generator and $\mathbb{E}^x$ is the expected value.

Can anybody explain how the first approximation was achieved? It is just simply a cross multiplication in the limits? Thanks.

Denis
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1 Answers1

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$$Af(x) = \lim_{t \downarrow 0} \frac{\mathbb{E}^x(f(X_t))-f(x)}{t}$$ implies that $$\lim_{t \downarrow 0} \mathbb{E}^x(f(X_t))-f(x) - tAf(x) = \lim_{t \downarrow 0} t \left[ \frac{\mathbb{E}^x(f(X_t))-f(x)}{t} - Af(x) \right] = 0$$ which tells you that $\mathbb{E}^x(f(X_t)) \approx f(x) + tAf(x)$ for small $t \geq 0$.

Rhys Steele
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  • How $\lim_{t \downarrow 0} \mathbb{E}^x(f(X_t))-f(x) - tAf(x)$ was derived from the definition? multiplied both sides by t? And why could it be assumed that the term in the big bracket is zero? – Kevin Nov 19 '23 at 16:06
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    We want to look at $\lim_{t \downarrow 0} \mathbb{E}^x(f(X_t))-f(x) - tAf(x)$ because we want to see that $\mathbb{E}^x[f(X_t)] \approx f(x) + t Af(x)$; it isn't derived from the definition of $A$. The definition of $A$ is used to see that the limit as $t \to 0$ of the term in the big brackets is $0$ (the term in the brackets itself isn't $0$, it's important there's a limit there) – Rhys Steele Nov 19 '23 at 16:10
  • Thank you for the fast response. How did you conclude $\mathbb{E}^x(f(X_t)) \approx f(x) + tAf(x)$ from $\lim_{t \downarrow 0} \mathbb{E}^x(f(X_t))-f(x) - tAf(x) = 0$ – Kevin Nov 19 '23 at 17:10
  • This isn’t a precise statement since $\approx$ isn’t precisely defined. The heuristic statement being made is just the meaning of the limit though – Rhys Steele Nov 19 '23 at 17:55
  • so why $\lim_{t \downarrow 0} \mathbb{E}^x(f(X_t)) \approx \mathbb{E}^x(f(X_t))$? – Kevin Nov 19 '23 at 18:00
  • What you’ve written doesn’t make sense? On the left hand side you send $t$ to 0 but on the right it’s fixed (and unspecified) – Rhys Steele Nov 19 '23 at 18:07
  • I think i confused every thing. I dont understand why $\lim_{t \downarrow 0} \mathbb{E}^x(f(X_t))-f(x) - tAf(x) = \lim_{t \downarrow 0} t \left[ \frac{\mathbb{E}^x(f(X_t))-f(x)}{t} - Af(x) \right] = 0$ tells us that $\mathbb{E}^x(f(X_t)) \approx f(x) + tAf(x)$ – Kevin Nov 19 '23 at 18:22
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    The statement I made is that for small t $\mathbb{E}^x(f(X_t)) \approx f(x) + tAf(x)$ where you should notice that the symbol $\approx$ has no precise meaning (though I will say what I mean by it in this comment for clarity). The limiting statement then means that for any 'amount of error' $\varepsilon$ there is a $\delta$ such that $0<t< \delta$ ('for small t') $|\mathbb{E}^x(f(X_t))-f(x) - tAf(x)| < \varepsilon$ (which is what should be understood as $\mathbb{E}^x(f(X_t))\approx f(x) + tAf(x)$). – Rhys Steele Nov 19 '23 at 18:27