Expected Payoff of the Gaussian Distribution Tail Part 2

Consider the following equation in Nassim Nicholas Taleb’s paper On the Statistical Differences between Binary Forecasts and Real World Payoffs

(1) $\begin{equation*} \lim_{K \to \infty}\frac{\int_{K}^{\infty}xf(x)dx}{K \int_{K}^{\infty}f(x)dx} = \frac{I_1}{I_2} = 1 \end{equation*}$

where:

I1 is the expected payoff for all the values above K.
I2 is the payoff of K multiplied by the probability of exceeding K.
x is the impact function g(x)=x
f(x) is the probability density function of centered and normalized Gaussian distribution.

In other words, I1 and I2 are close to each other only for thin tailed distributions. Conflating I1 and I2 leads to underestimating the expected payoff, especially with fat tailed distributions.

I wrote this note first and foremost to understand the Taleb’s equation 1.

The proof will be done for the Gaussian distribution probability density function (centered and normalized):

(2) $\begin{equation*} f(x) = \frac{e^-\frac{x^2}{2}}{\sqrt{2 \pi}} \end{equation*}$

Since for a thin tailed distribution

(3) $\begin{equation*} \lim_{x \to \infty}f(x) = 0 \end{equation*}$

1 is an indeterminate form.

After applying Hôpital’s rule and the first fundamental theorem of calculus to 1:

(4) $\begin{equation*} \begin{aligned} \frac{I1}{I2} =& \lim_{K \to \infty}\frac{\left[ xf(x) \right]_{K}^{\infty} }{K \left[ f(x)\right]_{K}^{\infty} } \\=& \lim_{K \to \infty}\frac{ \sqrt{2 \pi} \left[ \frac {x} { e^\frac{x^2}{2} } \right]_{K}^{\infty} }{ \sqrt{2 \pi} K \left[ e^-\frac{x^2}{2} \right]_{K}^{\infty} } \\ =& \lim_{K \to \infty}\frac{ \left[ \frac {1} { x e^\frac{x^2}{2} } \right]_{K}^{\infty} }{ K \left[ e^-\frac{x^2}{2} \right]_{K}^{\infty} } \\ =& \lim_{K \to \infty}\frac{ \left[ \frac {1} { x e^\frac{x^2}{2} } \right]_{K}^{\infty} }{ K \left[ e^-\frac{x^2}{2} \right]_{K}^{\infty} } \\ =& \lim_{K \to \infty}\frac{ \lim_{x \to \infty}\frac {1} { x e^\frac{x^2}{2} }-\frac{1}{Ke^\frac{K^2}{2}} }{ K ( \lim_{x \to \infty}e^-\frac{x^2}{2}-e^-\frac{K^2}{2} ) } \\ =& \lim_{K \to \infty}\frac{ \frac{1}{Ke^\frac{K^2}{2}} }{ K e^-\frac{K^2}{2} } \\ =& \lim_{K \to \infty}\frac{ \frac{1}{Ke^\frac{K^2}{2}} }{ \frac{K'}{ \left( e^\frac{K^2}{2} \right)^{'} } } \\ =& \lim_{K \to \infty}\frac{ \frac{1}{Ke^\frac{K^2}{2}} }{ \frac{1}{Ke^\frac{K^2}{2}} } \\ =& 1 \end{aligned} \end{equation*}$

Thanks for sharing

Expected Payoff of the Gaussian Distribution Tail Part 2

Nenad Noveljic

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