Given a payoff function $F(X)$ of a random variable $X$, and a Taylor expansion of $F(X)$ around $X=a$, then the expecation of $F(X)$ can be written as

$$ E[F(X)] = F(a) + E[ O((X-a))] $$

Under what conditions can I write $$ E[F(X)] = F(a) + O(E[(X-a)]) \,\,? $$

And if there is no condition under which the two are equivalent, any ideas how I could quantify

$$ E[ O((X-a))] - O(E[(X-a)]) \,\, ? $$

I suspect convexity will play a role, but not clear yet to me how.

  • $\begingroup$ Writing $F(X) = F(a) + O(X-a)$ is the first question is not straightforward if $X$ is stochastic. A first way intuitive way to approach the problem would be to write the taylor expansion of $F(X)$ (assumption 1: it's sufficiently smooth for such an expansion to exist) around the mean of $X$, $\Bbb{E}[X]=a$ (assumption 2), to obtain e.g. "at order 2": $$\Bbb{E}[F(X)] \approx F(\Bbb{E}[X]) + \frac{1}{2} \Bbb{V}[X] F''(\Bbb{E}[X])$$. Hence the role of convexity (or see Jensen's inequality). You could go further in the developments and add other terms. $\endgroup$ – Quantuple Mar 13 at 8:18
  • $\begingroup$ @Quantuple. If you don't mind, could you explain where your equation comes from. I'm not questioning it but rather just wondering how you get it. Thanks. $\endgroup$ – mark leeds Mar 13 at 23:31
  • $\begingroup$ Act like $f$ was a deterministic function and consider its Taylor expansion of around $X=\Bbb{E}[X]$. This gives (at order 2) $$f(X) = f(\Bbb{E}[X]) + f'(\Bbb{E}[X])(X-\Bbb{E}[X]) + \frac{1}{2}f''(\Bbb{E}[X])(X-\Bbb{E}[X])^2 $$ Now take the expectation on both sides and use linearity of expectation for the RHS. The second term on the RHS drops, the third is related to the variance. $\endgroup$ – Quantuple Mar 13 at 23:40
  • $\begingroup$ @Quantuple Thanks, yes indeed the convexity correction is important. The issue I am dealing with is that I would like to somehow bound the error terms in a Taylore expansion as above. And it would be easier to bound terms $(E(X))^n$ instead of terms of the form $E(X^n)$. Need to think about this more, no easy answer probably. $\endgroup$ – ilovevolatility Mar 17 at 16:15

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