Thayer Watkins

Evaluating the Expected Value of Functions
by Means of the Characteristic Function
Let p(z) be the probability distribution function of the variable z. The expected value of a function of z, say g(z) is defined as:

E{g} = ∫-∞g(z)p(z)dz

Sometimes the probability distribution p(z) is not known but its characteristic function is known. The probabililty distribution p(z)can be found from its characteristic function Φ(ω) by means of the inverse formula:

p(z) = (1/2π)∫-∞exp(-iωz)φ(ω)dω

Instead of carrying out the inversion of the characteristic function and using the result to compute the expected value consider the substitution of the inversion formula into the expression for the expected value; i.e.,

E{g} = ∫-∞g(z)p(z)dz

= (1/2π)∫-∞-∞g(z)exp(-iωz)φ(ω)dωdz

Reversing the integration operation gives:

E{g} = ∫-∞Γ(ω)φ(ω)dω
where Γ(ω) = (1/2π)∫-∞g(z)exp(-iωz)dz

Thus the expected value of g can be computed as the integral of the product of the characteristic function of the probability distribution and the characteristic function of the function g.


Consider a call option, the right to purchase a security at a specified price, the exercise price, within a specified period of time. Let S be the market price of the security and X the exercise price. At the time of expiration of the call option the value C0 is equal to u(S-X) where the function u(z) is equal to zero if z<0 and z if z≥0.

The value of the call option when there is t time before expiration is:

Ct(S,X,r) = e-rtC0(S0-X)

where S0 is the security price on the expiration day of the call option and r is the risk-free interest rate. The expected value of the option with t time until expiration is then:

E{Ct(St,X,r)} = e-rtE{C0(S0-X)}

The distribution of S0 will depend upon the value of the price t time periods before. The relationship is

0 = St(1+z)
where z is proportional change in price and is a random variable with distribution p(z).

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