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In the following passage of A Map and Simple Heuristic to Detect Fragility, Antifragility, and Model Error by Nassim Taleb:

Convexity Effects and Jensen's Inequality

Define a convex function as one with a positive second derivative, but this is a mathematical construct that does not translate well into practice (as it requires twice-differentiability). Recall from Figure 10

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the “flipping” of the exposure from convex in the body of the distribution to severely concave in the tails. So, more practically, convexity over an interval $2 \Delta x$ satisfies the following inequality:

$$\frac{1}{2}\left( f(x+\Delta x) +f(x-\Delta x \right)\geq f(x)$$

Why economics as a discipline made the monstrously consequential mistake of treating estimated parameters as nonstochastic variables and why this leads to fat-tails even while using Gaussian models.

The average of expectations is not the expectation of an average. For $f$ convex across all values of $\{X_i\},$

$$\sum w_i E \Big [f(X_i)\Big ] \geq E\Big [\sum f(w_i X_i)\Big ]$$ For example, take a conventional die (six sides) and consider a payoff equal to the number it lands on. The expected (average) payoff is $\frac{1}{6} \sum_{i=1}^6 i = 3.5 .$ Now consider that we get the squared payoff, $\frac{1}{6} \sum_{i=1}^6 i^2 = \frac{91}{6} \approx15.67,$ while I $\frac{1}{6} \left( \sum_{i=1}^6 i \right)^2= 12.25,$ so, since squaring is a convex function, the average of a square payoff is higher than the square of the average payoff.

The question is:

With the example of the die and the difference between the square of the expectation and the expectation of the squares, he is just calculating the variance. Is this a valid example of Jensen's inequality?

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    $\begingroup$ Yes$$\mathbb{E}[f(X)]\ge f\mathbb{E}[X])$$applied to $f(x)=x^2$ is Jensen's inequality. The previous equation$$\sum w_i \mathbb{E} \Big [f(X_i)\Big ] \geq \mathbb{E}\Big [\sum f(w_i X_i)\Big ]$$is not Jensen's inequality but just the definition of convexity. $\endgroup$
    – Xi'an
    Apr 17, 2018 at 16:52
  • $\begingroup$ @Xi'an I'm going to ask a silly question, probably undeserving of a separate post... I am having problems understanding the notation on the last equation on your comment, which if it is at all imprecise it would be due to the quote on the OP... The expression $\mathbb{E} \Big [f(X_i)\Big ]$ strikes me like conflicting in the way that the expectation operator $\mathbb E$ would apply to a random variable; however, $X_i$ (or perhaps better, $x_i$) seems to refer to a realization of a RV, in which case $f(X_i)$ would not vary, but rather be a single numerical value. What am I missing? $\endgroup$ Mar 4, 2020 at 18:50
  • $\begingroup$ $X_i$ is a random variable and $f(X_i)$ is another random variable. $\endgroup$
    – Xi'an
    Mar 4, 2020 at 20:55
  • $\begingroup$ @Xi'an Right... That's the only way that it makes sense... What is then it the index $i$ in $X_i$ making reference to? $\endgroup$ Mar 4, 2020 at 21:14
  • $\begingroup$ A random variable can be indexed by $i$. $\endgroup$
    – Xi'an
    Mar 4, 2020 at 21:47

2 Answers 2

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The relation to variance is incidental in this example. But you are right, Jensen's inequality tells us that the expected squared payoff is greater than the squared expected payoff. This is a fact that can be used to prove that variance is non-negative.

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Yes, you could say that the fact that the variance is greater than zero in this case can be seen as a demonstration of the Jensen's inequality. I don't think the author meant to link this to variance though. Any convex function will lead to Jensen's inequality. He simply chose a square because it's easy to calculate, in my opinion. He could have chosen an exponent

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