Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2012 / S9 / 2

Independent identically distributed increments have the same marginal distribution at every step and factorizing joint distributions across steps. With finite mean $\mu$ and variance $s^2$, an additive random walk has mean displacement $n\mu$ and variance $n{s}^2$ after $n$ steps. Zero increment autocovariances follow from independence, but their converse is false. Gaussian increments are an additional assumption, not part of the definition of a random walk.

To investigate a measured price process, first specify the sampling interval and whether the model concerns price differences or log returns. Remove an estimated drift and examine stationarity, increment distributions, autocorrelations and variance growth over nonoverlapping horizons. Also test higher-order statistics, absolute or squared-return correlations, conditional distributions and out-of-sample predictability; those can reveal dependence missed by linear autocorrelation. Check whether different time scales and market regimes give compatible results, and account for bid-ask effects and intraday seasonality at short lags. Finite data cannot prove independence or permanent stationarity. Failure to reject a particular test is weaker than establishing that every joint distribution factorizes; the next example makes that distinction explicit.

Take the driving signs $a_t=\alpha (t)$ to be independent across times, with $\mathbb{E}{a}_t=0$ and $a_t^2=1$. Temporal independence of the drivers is an assumption needed in addition to their stated equal one-time probabilities. For $X_t=\delta y(t)=a_{t-2}{a}_{t-1}+a_t$,As a useful scale check, $X_t$ takes values $-2,0,2$ with probabilities $1/4,1/2,1/4$. Its variance is therefore $\boxed{\mathbb{E}{X}_t^2=2}$, rather than the unit variance of a single driving sign.

Expand $X_t{X}_{t^{\prime }}$ into products of the independent mean-zero signs. Each increment is the sum of two monomials, $a_t$ and $a_{t-2}{a}_{t-1}$. At distinct times, no monomial from one increment has exactly the same set of sign indices as a monomial from the other. Every product therefore contains at least one sign to an odd power, whose average is zero. HenceThis argument covers overlapping driver sets at separations one and two, not just the disjoint case at larger separation. At equal times the value is $2$, so the complete autocovariance is $2{\delta }_{t{t}^{\prime }}$. It establishes absence of second-order serial correlation, not independence of returns.

Expand the three increments into their eight monomial products. Exactly one term has every driving-sign index appearing an even number of times: take $a_{t-2}{a}_{t-1}$ from $X_t$, $a_{t-1}$ from $X_{t-1}$, and $a_{t-2}$ from $X_{t-2}$. Their product is $a_{t-1}^2{a}_{t-2}^2=1$. Every other term leaves an unpaired mean-zero sign. ThusIf the increments were independent, their three zero means would instead give zero. The nonzero third-order moment is therefore a direct demonstration of nonlinear temporal dependence despite vanishing nonzero-lag autocovariances.

The series is predictable in a statistical sense beyond linear return autocorrelation. For example, the observed past feature $Z_t=X_{t-1}{X}_{t-2}$ satisfies $\mathbb{E}[X_t{Z}_t]=1$ and $\mathbb{E}{Z}_t^2=4$, so its least-squares predictor within the family $b{Z}_t$ is $\boxed{{\hat{X}}_t=Z_t/4}$. This is a predictor based on observed increments, not an assumption that the hidden driving signs are directly visible; it need not equal the full conditional expectation.

A finite bound on ordinary peak-to-trough financial drawdown does not follow. Choose a long alternating sequence of driving signs, $a_t=(-1{)}^t$. It produces $X_t=-1+(-1{)}^t$, alternating $-2$ and $0$. The price never recovers during that episode and its cumulative loss grows without bound as the episode length grows. Every finite such driving sequence has positive probability. Thus $\boxed{\text{no finite worst drawdown magnitude or duration exists without a horizon bound}.}$

More precisely, over an $N$-month window and over admissible starting histories, the maximum possible drawdown isOne derivation keeps the last two signs as a four-state dynamic program. Let $d_n(a,b)$ be the smallest cumulative increment after $n$ steps ending with signs $(a,b)$, with $d_0(a,b)=0$ for each permitted initial pair. Then $d_{n+1}(b,c)={\min }_{a=\pm 1}[d_n(a,b)+ab+c]$. Induction gives the negative of the displayed minimum. The bound is attained by increments $(-2,0{)}^k,-2$ for $N=2k+1$, or $(-2,0{)}^{k-1},-2,-2$ for $N=2k$, so it is also an attained peak-to-trough loss over $N$ months, not only an algebraic net displacement bound.

For comparison, the index with increments $a_t=\pm 1$ has $D_{\max }(N)=N$, attained by $N$ successive negative signs; it too has no finite horizon-free bound. If “drawdown” were intended to mean a strictly decreasing consecutive run, the first model permits at most two successive $-2$ increments, for a loss of four over two months, whereas the simple walk permits arbitrarily long such runs. That different run statistic must not be substituted for the usual drawdown definition: zero increments can separate continuing losses without any recovery.

Derivatives can transfer risk, improve price discovery and permit hedging, so their existence is not by itself evidence that they cause crashes. They can also amplify stress through leverage, margin calls, counterparty exposure and forced dynamic hedging into falling markets. A short-gamma hedge can require selling more underlying after a fall, reinforcing feedback when liquidity is thin. The causal issue is the combination of positions, funding, market depth and institutional rules; derivatives are neither necessary nor sufficient for a crash, and correlation between trading volume and a crash does not establish the direction of causation.