Expected value and linearity

Every price in finance is an expected value of something. An option, a bond, a bet on a dice game: once you can write down what it pays in each outcome and how likely each outcome is, the fair price is a weighted average. This lesson is about computing that average, and about one property of it that makes hard questions easy.

The weighted average

A random variable $X$ has a probability mass function $p(x)$ if it is discrete or a density $f(x)$ if it is continuous. Its expected value (or mean, $\mu$) weights every value by its probability:

The same recipe prices any payoff $g(X)$: weight $g(x)$ instead of $x$.

A fair six-sided die pays $E[X] = (1+2+3+4+5+6)/6 = 3.5$. If a game pays the roll, its fair price is $3.50. Pay less and you profit on average; pay more and you lose.

Pricing a choice: the re-roll game

Interviewers love to add a decision. Roll a die; you may re-roll once and must keep the second roll. What is the game worth? Work backwards. A fresh roll is worth 3.5, so keep anything above 3.5 and re-roll 1, 2 or 3:

If each re-roll costs $1, a re-roll is worth $3.5 - 1 = 2.5$, so you only re-roll 1s and 2s: $\tfrac{2}{6}\cdot 2.5 + \tfrac{1}{6}(3+4+5+6) = 23/6 \approx 3.83$.

With unlimited $1 re-rolls the game is worth some value $V$, and the rule is to keep any face above $V - 1$. Keeping 4, 5, 6 gives $V = 2.5 + \tfrac12(V-1)$, so $V = 4$. Keeping 3 or more gives the same $V = 4$, because a 3 is exactly break-even.

Linearity: the trick that solves half the questions

Expectation is linear: $E[aX + b] = aE[X] + b$. More surprisingly, it adds across any random variables, even dependent ones:

So when a question asks for the expected number of somethings, write the count as a sum of indicators, $I_i = 1$ if thing $i$ happens and 0 otherwise. Each indicator has $E[I_i] = P(\text{thing } i)$, and you add those up. You never need the joint distribution.

Example. Ten boys and ten girls stand in a random line. What is the expected number of adjacent boy–girl pairs?

There are 19 neighbouring slots. Slot $i$ is mixed with probability

so the answer is $19 \times \tfrac{10}{19} = 10$. The slots are dependent (a mixed slot changes who is left for the next one), and it doesn’t matter.

In code

import numpy as np
rng = np.random.default_rng(1)

# Fair value of "roll, optionally re-roll once" by simulation
first = rng.integers(1, 7, 1_000_000)
second = rng.integers(1, 7, 1_000_000)
payoff = np.where(first <= 3, second, first)
print(payoff.mean())                     # ≈ 4.25

# Linearity: expected mixed neighbours in a line of 10 boys + 10 girls
line = np.array([0] * 10 + [1] * 10)
mixed = []
for _ in range(100_000):
    p = rng.permutation(line)
    mixed.append(np.sum(p[1:] != p[:-1]))
print(np.mean(mixed))                    # ≈ 10.0