Paradoxul St. Petersburg

X \sim (\begin{matrix} 2^{0} & 2^{1} & 2^{2} & \dots & 2^{n} & \dots \\ p & q p & q^{2} p & \dots & q^{n} p & \dots \end{matrix})

Media

E [X] = \infty

Avem $p = \frac{1}{6} q = \frac{5}{6}$

Bernoulli expected value

\begin{aligned} BEV [X] & = \prod_{n \geq 1} 2^{n q^{n} p} = 2^{\sum_{n \geq 1} p n q^{n}} \\ = 2^{\sum_{n \geq 1} p q \frac{d}{d q} (q^{n})} \\ = 2^{p q \frac{d}{d q} (\frac{1}{1 - q}) =} \\ = 2^{p q \cdot (1 - q)^{- 2}} \\ = 2^{\frac{q}{p}} = 2^{5} \end{aligned}

Dacă avem $p = q = \frac{1}{2}$ , atunci:

E [X] = \infty BEV [X] = 2

Experimentul 1

X \sim (\begin{matrix} 0 & 10000 \\ 0.05 & 0.95 \end{matrix})

Media

E [X] = 9500

$w = 3000$ , $E [X + w] = 12500$
Asigurare $a = 800$ . Câștigul este, în mod sigur $3000 + 10000 - 800 = 13000 - 800 = 12200 < 12500$ .

BEV [X] = 3000^{\frac{1}{20}} \cdot 13000^{\frac{19}{20}} = 12081

Experimentul 2

X_{i} \sim (\begin{matrix} \frac{1}{2} & \frac{21}{20} & \frac{3}{2} \\ \frac{1}{6} & \frac{4}{6} & \frac{1}{6} \end{matrix})

Media este

\begin{aligned} E [X_{i}] & = \frac{1}{2} \cdot \frac{1}{6} + \frac{21}{20} \cdot \frac{4}{6} + \frac{3}{2} \cdot \frac{1}{6} \\ = \frac{31}{30} = 1.0 (3) \end{aligned}

Evenimentele sunt independente

X_{i} ⊥ ⊥ X_{j} \forall i \neq j

Avem suma inițială $w_{0} = 1$

\begin{aligned} w_{1} = X_{1} w_{0} \\ w_{2} = X_{2} X_{1} w_{1} \\ ⋮ \\ w_{300} = \prod_{i = 1}^{300} X_{i} \cdot w_{0} \end{aligned}

Media fiind im))
gross = (1 - f) + f * M # rebalance to fraction f each round
W *= gross

E [w_{300}] = \prod_{i = 1}^{300} E [X_{i}] \cdot w_{0} = [1.0 (3)]^{300} \cdot w_{0} = 18713 \cdot w_{0}