Tuesday 11 August 2026

Parity

$$C-P=S_0-K{e}^{-rT}$$

The proof is a payoff identity plus the law of one price. For every value of $S_T$,

$$(S_T-K{)}^{+}-(K-S_T{)}^{+}=S_T-K$$

which you check on the two cases $S_T\ge K$ and $S_T<K$. The left side is the maturity value of long one call, short one put; the right side is the maturity value of one share, short $K$ zero-coupon bonds each worth $e^{-rT}$ today. Two portfolios with identical payoffs in every state must have identical prices, or you short the dearer, buy the cheaper, and pocket the difference with no exposure. The step worth stating out loud is that last one: parity is not algebra, it is the no-arbitrage assumption applied to an algebraic identity.

The violation

$$S_0-K{e}^{-rT}=100-100e^{-0.05}=4.8771$$

$$C-P=10.45-5.00=5.45$$

The synthetic forward embedded in the options is rich by $5.45-4.8771=0.5729$ in present value, so sell the synthetic and buy the real thing.

The trade

Short one call, long one put, long one share. Today's cash flow is

$$10.45-5.00-100=-94.55$$

so borrow $94.55 at $r$; the net outlay is zero. At maturity the stock-plus-options position is

$$S_T+(K-S_T{)}^{+}-(S_T-K{)}^{+}=S_T+(K-S_T)=K=100$$

for every $S_T$ — that is the whole point of the parity identity, and it is why the position carries no market risk. Repaying the loan costs $94.55e^{0.05}=99.3977$, leaving

$$100-99.3977=0.6023.$$

Answer: a riskless $0.60 per unit at maturity, equivalently $0.5729$ today, and indeed $0.5729e^{0.05}=0.6023$.

Note. The profit is the parity gap, nothing more. A common slip is to also claim the position benefits if the stock rallies; it does not, and the fact that the maturity value is exactly $K$ in every state is what makes the profit riskless rather than merely likely.

1. Variance of a correlated average

The covariance matrix of the ${\widehat{f}}_i$ has $v$ on the diagonal and $\rho v$ off it, so

$$\mathrm{V}\mathrm{a}\mathrm{r}\left({\sum }_{i=1}^B{\widehat{f}}_i\right)=Bv+B(B-1)\rho v$$

and dividing by $B^2$,

$$\mathrm{V}\mathrm{a}\mathrm{r}(\overline{f})=\rho v+\frac{(1-\rho )v}{B}.$$

Since $𝔼[\overline{f}]=m$, the bias-variance decomposition gives

$$𝔼[(\overline{f}-y{)}^2]=b^2+\rho v+\frac{(1-\rho )v}{B}.$$

2. Many members

$$\mathrm{V}\mathrm{a}\mathrm{r}(\overline{f})⟶\rho v,𝔼[(\overline{f}-y{)}^2]⟶b^2+\rho v$$

Averaging drives out only the independent slice $(1-\rho )v$ of the variance. Two terms survive any amount of averaging: the squared bias, and the shared variance $\rho v$.

3. Reading it as a design rule

Bagging pays when $v$ is large, $\rho$ is small and $b$ is small — which is exactly the profile of a deep, fully grown decision tree: nearly unbiased, high variance, and decorrelated from its siblings by bootstrap resampling. It does essentially nothing for a high-bias learner, because $b^2$ is untouched: averaging a hundred depth-1 stumps leaves the same systematic error as one.

The formula also explains why a random forest subsamples features at each split rather than only bootstrapping rows. Bootstrapping alone leaves trees strongly correlated, so $\rho v$ dominates and the ensemble stops improving; forcing splits to consider different features lowers $\rho$, which lowers the floor itself. Note the tradeoff that the formula does not show: restricting features usually raises each tree's own $v$ and $b$ a little, so the gain is real but not free.