Trading Psychology

Question 1

(a) [8 marks]

• Bankroll after $n$ trades: $B_n = B_0\prod (1+f R)^{\text{win}}(1-f)^{\text{loss}}$. Taking log and dividing by $n$, the per-trade expected log return is the expectation of the log multiplier: $g(f)=p\ln(1+fR)+q\ln(1-f)$. (3)
• Maximise: $g'(f)=\dfrac{pR}{1+fR}-\dfrac{q}{1-f}=0$. (2)
• Cross-multiply: $pR(1-f)=q(1+fR)\Rightarrow pR-pRf = q+qfR$. Collect: $pR-q = f R(p+q)=fR$. Hence $f^\*=\dfrac{pR-q}{R}$. (2)
• Half-Kelly rationale: variance of log-growth scales roughly with $f^2$ while growth is near-flat around the optimum (concave, zero first derivative at $f^\*$); half-Kelly captures ~75% of growth with ~half the volatility → reduces drawdown-induced fear/greed deviations and keeps the trader in the process (4.8.10). (1)

(b) [8 marks] With $p=0.55,q=0.45,R=1$:

• $f^\* = \dfrac{0.55\cdot1-0.45}{1}=0.10$. (2)
• $g(f^\*)=0.55\ln(1.10)+0.45\ln(0.90)=0.55(0.095310)+0.45(-0.105361)=0.052421-0.047412=0.005008$. (3)
• Half-Kelly $f=0.05$: $g=0.55\ln(1.05)+0.45\ln(0.95)=0.55(0.048790)+0.45(-0.051293)=0.026835-0.023082=0.003753$. (2)
• Comment: half-Kelly gives ≈75% of the growth rate ($0.003753/0.005008\approx0.75$) with substantially lower drawdown variance — favourable risk trade-off. (1)

(c) [8 marks] $q=0.45$.

• $P(\ge5 \text{ consecutive losses}) = q^5 = 0.45^5 = 0.0184528\approx 0.01845$. (3)
• Expected trades between starts of such runs $\approx 1/q^5 = 1/0.0184528 \approx 54.2$ trades. (3)
• Plan use: because a 5-loss streak is expected roughly every ~54 trades, it is a normal statistical event, not a signal of broken edge. Pre-committing a max-consecutive-loss stop and a cooling-off rule prevents interpreting variance as failure → blocks tilt (4.8.5) and revenge trading (4.8.4). (2)

Question 2

(a) [8 marks]

• For $N$ i.i.d. standard normals, the maximum concentrates near the level $x$ where $N(1-\Phi(x))\approx 1$, i.e. $1-\Phi(x)\approx 1/N$. (3)
• Using tail bound $1-\Phi(x)\approx \phi(x)/x \sim e^{-x^2/2}$, set $e^{-x^2/2}\approx 1/N \Rightarrow x^2/2\approx \ln N \Rightarrow x\approx\sqrt{2\ln N}$. (3)
• $N=1000$: $\sqrt{2\ln 1000}=\sqrt{2(6.90776)}=\sqrt{13.8155}=3.717$. So the best of 1000 no-edge strategies shows Sharpe ≈3.7 by chance. (1)
• FOMO trap (4.8.3): chasing the top backtest = selecting noise; disciplined process rejects the strategy unless it survives multiple-testing correction/out-of-sample. (1)

(b) [8 marks] Sample pseudo-code (mark for correct structure):

def walk_forward(data, param_grid, folds, tau):
    flags = []
    for train, test in split_folds(data, folds):   # chronological, no leakage
        best_p = argmax(param_grid, key=lambda p: sharpe(backtest(train, p)))
        is_sharpe  = sharpe(backtest(train, best_p))
        oos_sharpe = sharpe(backtest(test,  best_p))
        flags.append(is_sharpe - oos_sharpe > tau)  # overfit flag
    overfit = mean(flags) > 0.5
    return overfit

• Chronological split (no lookahead) — enforces realism/consistency (4.8.1). (2)
• Params chosen in-sample only, scored out-of-sample — separates fitting from validation. (2)
• $\tau$ threshold operationalises "degradation" objectively → removes discretionary/emotional override. (2)
• Majority-of-folds decision → robust to single-fold luck. (2)

(c) [4 marks]

• Paper trading validates mechanics/slippage/logic (necessary) but P&L is not real, so it does not test loss-aversion, fear, greed and tilt under real financial/emotional stress (not sufficient). Named psychological variable: emotional response to real capital loss / fear of realising losses. (4)

Question 3

(a) [8 marks]

• Proposed weights (ordering justified by severity of rule-breach): risk control first. $w_s=0.35$ (position-size breach can cause ruin), $w_r=0.30$ (revenge entries are highest-EV-destroying), $w_a=0.25$ (plan adherence), $w_j=0.10$ (journaling supports learning but is lowest immediate risk). Sum $=1.00$. (3)
• Per-trade score $D_i = w_a a + w_s s + w_r r + w_j j \in[0,1]$. (2)
• Rolling mean $\bar D = \frac{1}{20}\sum_{i=n-19}^{n} D_i$. (1)
• Lock-out threshold: halt trading if $\bar D < 0.80$ (i.e. discipline slipping) until a review is logged — enforces cool-down. Any justified threshold in $[0.7,0.85]$ accepted. (2)

(b) [8 marks]

• Assumptions: per-trade P&L $X_i$ are i.i.d. with finite mean $\mu=\mathbb{E}[X]$ and finite variance $\sigma^2$; decisions depend only on the process (entry rules), so outcomes don't alter the rule set → i.i.d. preserved. (3)
• Sample mean $\bar X_n=\frac1n\sum X_i$. WLLN: for any $\varepsilon>0$, $P(|\bar X_n-\mu|>\varepsilon)\to 0$ as $n\to\infty$. (3)
• (Chebyshev proof) $P(|\bar X_n-\mu|>\varepsilon)\le \dfrac{\operatorname{Var}(\bar X_n)}{\varepsilon^2}=\dfrac{\sigma^2}{n\varepsilon^2}\to 0$. (2)
• Interpretation: process-focused discipline keeps the distribution stationary; realised P&L then converges to the true edge, which is why traders must judge process not short-run outcome (4.8.10). Emotional deviation breaks the i.i.d. assumption and invalidates the guarantee.

[
 {"claim":"Kelly fraction f* = 0.10 for p=0.55,R=1","code":"p=Rational(55,100);q=1-p;R=1;fstar=(p*R-q)/R;result=(fstar==Rational(1,10))"},
 {"claim":"Full-Kelly growth rate ~0.005008","code":"import sympy as sp;p=sp.Rational(55,100);q=sp.Rational(45,100);f=sp.Rational(1,10);g=p*sp.log(1+f)+q*sp.log(1-f);result=abs(float(g)-0.005008)<1e-5"},
 {"claim":"Half-Kelly growth ~0.003753 and about 75% of full","code":"import sympy as sp;p=sp.Rational(55,100);q=sp.Rational(45,100);gf=p*sp.log(sp.Rational(105,100))+q*sp.log(sp.Rational(95,100));gk=p*sp.log(sp.Rational(11,10))+q*sp.log(sp.Rational(9,10));result=abs(float(gf)-0.003753)<1e-5 and abs(float(gf/gk)-0.75)<0.02"},
 {"claim":"P(>=5 losses)=0.45**5 ~0.01845 and expected gap ~54.2","code":"q=0.45;p5=q**5;gap=1/p5;result=abs(p5-0.0184528)<1e-6 and abs(gap-54.19)<0.5"},
 {"claim":"sqrt(2 ln 1000) ~ 3.717","code":"import sympy as sp;val=sp.sqrt(2*sp.log(1000));result=abs(float(val)-3.717)<1e-3"}
]