Expected Payoff of the Gaussian Distribution Tail

I wrote this note primarily for myself to understand the equation (6) on the page 5 from Nassim Nicholas Taleb’s paper On the Statistical Differences between Binary Forecasts and Real World Payoffs:

(1)   \begin{equation*}  \int_{K}^{\infty}xg(x)dx = \frac{e^-\frac{K^2}{2}}{\sqrt{2 \pi}} \end{equation*}

where:

  • x is a random variable,
  • g(x) is the probability density function (PDF) of x, which is Gaussian distribution (centered and normalized).

The author assumes familiarity with the equation 1 and, therefore, doesn’t elaborate on it. The steps below show where this equation came from.

More generally, the expected payoff I for all values of x above K, i.e. the tail of the distribution, is:

(2)   \begin{equation*}  I = \int_{K}^{\infty}f(x)g(x)dx \end{equation*}

where f(x) is the payoff function.

Normally, we would use integration by parts for calculating the integral above.

But the author considers a special case where the payoff function f(x) is linear:

(3)   \begin{equation*}  f(x)=x \end{equation*}

So, 2 becomes:

(4)   \begin{equation*}  I = \int_{K}^{\infty}xg(x)dx \end{equation*}

The integral 4 is much easier to solve than more general 2, because the first derivative of the centered and normalized Gaussian PDF has the following property:

(5)   \begin{equation*}  g'(x) = - x g(x) \end{equation*}

We will integrate the equation and apply the second fundamental theorem of calculus to the left side of 5:

(6)   \begin{equation*} \lim_{x \to \infty} g(x) - g(K)= - \int_{K}^{\infty} x g(x) dx \end{equation*}

Notice that the right side became our expected payoff function, therefore:

(7)   \begin{equation*} g(K) - \lim_{x \to \infty}g(x) = I \end{equation*}

Since the probability approaches 0 as x approaches infinity, the expected payoff of all x above K equals the probability of K:

(8)   \begin{equation*} I = g(K) = \frac{e^-\frac{K^2}{2}}{\sqrt{2 \pi}} \end{equation*}

Thanks for sharing

Nenad Noveljic

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