44
4 Stochastic Processes
4.7 Digression on Expectation Values
In (4.33) we calculate the present value of a call option c by back-propagating
the payoff function max(S − K , 0). A complementary view is based on the notion
of averaging the payoff function, weighted by the probability distribution function
(S, t), which is the same as the expectation value of the payoff function in the
“state” (S, t).
We note that this is conceptually analogous to calculating the expectation value
|φ|H |φ of an operator H in the state |φ in quantum mechanics. Let us assume
that H is the Hamiltonian. Expanding both H and φ into eigenfunctions ψ i of H we
have H =
i E i |ψ i ψ i | and |φ =
j c k |ψ j with, in general complex, expansion
coefficients c j . Here E i are the eigenvalues of H and c j the expansion coefficients
of |φ. For the expectation value we find
|φ|H |φ =
k
c
∗
k ψ k |
i
E i |ψ i ψ i |
j
c j |ψ j
(4.46)
=
k
i
j
c
∗
k c j |ψ k |ψ i |ψ i |ψ j =
i
|c i |
2 E i
where we used that the eigenfunctions |ψ i |ψ j = δ i j are orthogonal. We observe
that the expectation value |φ|H |φ can be calculated as the eigenvalues E i weighed
by the probability |c i |
2 of finding the state |φ in eigenstate |ψ j .
Calculating the odds of in a game of throwing dice is conceptually similar of
betting on stocks exceeding a certain strike price. Let us work out what is more
likely, either rolling seven eyes or rolling a double-digit value—10, 11, or 12. Let
us assume that the payoff for rolling seven eyes is V 1 = 13 $ and for rolling doubledigit eyes it is V 2 = 10 $. We can work out the expected payoff V by weighing the
payoff with the probabilities of rolling either outcome. Since each dice has six faces,
the probability of rolling a specific one is 1/6. Since the dices roll independently
the probability of any throw with two dices is 1/36. Now we only have to calculate
the number of possible combinations that result in seven to calculate the probability
p 1 . There are six different combitations: 1 + 6, 2 + 5, . . . , 6 + 1, which results in
a probability of p 1 = 6/36 = 1/6 to roll seven eyes. Rolling double-digit eyes is
also possible with 6 different combinations: 4 + 6, 5 + 5, 6 + 4 all result in 10 eyes,
and 5 + 6, 6 + 5, and 6 + 6 result in 11 or 12 eyes. Even for these combinations the
probability is 6/36 or p 2 = 1/6. If I bet on rolling seven eyes I receive 13 $ if I win
and have to pay 10 $ to my opponent if I lose. Therefore the expectation value of
my payoff is given by V = p 1 V 1 − p 2 V 2 = 3/6 $. Thus, on average I will have an
advantage of half a dollar per game.
The notion of propagating a source function to a different point—the payoff
function back in time in the financial context—plays a central role in optics, where
we use Huygen’s principle [6] to find the intensity on a distant screen cause by light
shining through an aperture. Here the aperture function that is unity where light shines
through and zero otherwise plays the role of the source. In optics, the point-spread
4 Stochastic Processes
4.7 Digression on Expectation Values
In (4.33) we calculate the present value of a call option c by back-propagating
the payoff function max(S − K , 0). A complementary view is based on the notion
of averaging the payoff function, weighted by the probability distribution function
(S, t), which is the same as the expectation value of the payoff function in the
“state” (S, t).
We note that this is conceptually analogous to calculating the expectation value
|φ|H |φ of an operator H in the state |φ in quantum mechanics. Let us assume
that H is the Hamiltonian. Expanding both H and φ into eigenfunctions ψ i of H we
have H =
i E i |ψ i ψ i | and |φ =
j c k |ψ j with, in general complex, expansion
coefficients c j . Here E i are the eigenvalues of H and c j the expansion coefficients
of |φ. For the expectation value we find
|φ|H |φ =
k
c
∗
k ψ k |
i
E i |ψ i ψ i |
j
c j |ψ j
(4.46)
=
k
i
j
c
∗
k c j |ψ k |ψ i |ψ i |ψ j =
i
|c i |
2 E i
where we used that the eigenfunctions |ψ i |ψ j = δ i j are orthogonal. We observe
that the expectation value |φ|H |φ can be calculated as the eigenvalues E i weighed
by the probability |c i |
2 of finding the state |φ in eigenstate |ψ j .
Calculating the odds of in a game of throwing dice is conceptually similar of
betting on stocks exceeding a certain strike price. Let us work out what is more
likely, either rolling seven eyes or rolling a double-digit value—10, 11, or 12. Let
us assume that the payoff for rolling seven eyes is V 1 = 13 $ and for rolling doubledigit eyes it is V 2 = 10 $. We can work out the expected payoff V by weighing the
payoff with the probabilities of rolling either outcome. Since each dice has six faces,
the probability of rolling a specific one is 1/6. Since the dices roll independently
the probability of any throw with two dices is 1/36. Now we only have to calculate
the number of possible combinations that result in seven to calculate the probability
p 1 . There are six different combitations: 1 + 6, 2 + 5, . . . , 6 + 1, which results in
a probability of p 1 = 6/36 = 1/6 to roll seven eyes. Rolling double-digit eyes is
also possible with 6 different combinations: 4 + 6, 5 + 5, 6 + 4 all result in 10 eyes,
and 5 + 6, 6 + 5, and 6 + 6 result in 11 or 12 eyes. Even for these combinations the
probability is 6/36 or p 2 = 1/6. If I bet on rolling seven eyes I receive 13 $ if I win
and have to pay 10 $ to my opponent if I lose. Therefore the expectation value of
my payoff is given by V = p 1 V 1 − p 2 V 2 = 3/6 $. Thus, on average I will have an
advantage of half a dollar per game.
The notion of propagating a source function to a different point—the payoff
function back in time in the financial context—plays a central role in optics, where
we use Huygen’s principle [6] to find the intensity on a distant screen cause by light
shining through an aperture. Here the aperture function that is unity where light shines
through and zero otherwise plays the role of the source. In optics, the point-spread