Entry, Exit & Trade Management

Question 1

(a) Expectancy is the probability-weighted sum of outcomes: $E = pW - (1-p)L.$ (2 — correct construction) With $p=0.40,\ W=2.5,\ L=1.0$: $E = 0.40(2.5) - 0.60(1.0) = 1.0 - 0.6 = +0.40R.$ (2 — value) The system has a positive edge of $0.40R$ per trade.

(b) Each trade multiplies equity by $(1+fW)$ with prob $p$ and $(1-fL)$ with prob $1-p$. Log-growth per trade is the expected log-return: $g(f)=p\ln(1+fW)+(1-p)\ln(1-fL).$ (2 — setup) Differentiate: $g'(f)=\frac{pW}{1+fW}-\frac{(1-p)L}{1-fL}=0.$ (2) Cross-multiplying: $pW(1-fL)=(1-p)L(1+fW).$ Expand: $pW - pWLf = (1-p)L + (1-p)LWf$. $pW-(1-p)L = f\,WL\,[p+(1-p)] = f\,WL.$ So $\boxed{f^\*=\frac{pW-(1-p)L}{WL}=\frac{E}{WL}.}$ (2 — derivation) Numerically: $f^\*=\dfrac{0.40}{2.5\times1.0}=0.16$, i.e. risk 16% of equity per trade. (2)

(c) Compute $g$ at each fraction. A loss of $1R$ costs fraction $f$; note $W=2.5$ means a winner returns $fW$ of equity.

At $f^\*=0.16$: $1+fW = 1+0.16\times2.5 = 1.40$; $1-fL = 1-0.16 = 0.84$. $g=0.4\ln 1.40 + 0.6\ln 0.84 = 0.4(0.33647)+0.6(-0.17435)=0.13459-0.10461=0.02998.$ Multiple over 100 trades: $e^{100\times0.02998}=e^{2.998}\approx 20.0$. (3)

At half-Kelly $f=0.08$: $1+fW=1.20$; $1-fL=0.92$. $g=0.4\ln1.20+0.6\ln0.92=0.4(0.18232)+0.6(-0.08338)=0.07293-0.05003=0.02290.$ Multiple: $e^{100\times0.02290}=e^{2.290}\approx 9.87$. (3)

Deploy half-Kelly: full Kelly has extreme drawdowns (Kelly maximises growth but the equity path is very volatile; a run of losses at 16%/trade produces a punishing drawdown and, with estimation error in $p,W$, can overbet into ruin). Half-Kelly keeps ~75% of the growth with far lower variance — consistent with "cut losers / preserve capital." (1 mark choice + 1 mark reason within the 6)

(d) $g'(f)=\dfrac{pW}{1+fW}-\dfrac{(1-p)L}{1-fL}$. At $f=0$: $g'(0)=pW-(1-p)L=E.$ (3) If $E\le0$ then $g'(0)\le0$: log-growth is non-increasing as you leave zero position size, so any positive $f$ gives $g(f)<g(0)=0$ (equity erodes in the long run). Interpretation: with no positive expectancy, no position size — however clever the sizing — creates a compounding edge; the only optimal size is zero. This is the mathematical statement of "don't trade a negative-expectancy setup." (3)

Question 2

(a) R = 100-96 = \4.00$(initial risk = entry − structure stop). *(1)* Target at RR$1{:}2$:$+2R = +$8$above entry$\Rightarrow$**target$=$108$**. (2)

(b) Chandelier stop $=$ (highest high since entry) $- 3\times2 = \text{HH} - 6$. Track running max of highs:

(4 for table) Break-even ($\ge100$) is first reached on Day 4 (stop $=103>100$). (2)

(c) Pseudocode (assume stops fill at the stop price; check stop against each bar's low; unrealised profit measured off the bar high):

function manage_trade(prices, entry, atr, R):
    stop = entry - R              # initial structure/ATR stop
    hh = entry
    be_armed = false
    for bar in prices:            # bar has .high, .low, .close
        # 1) update highest high
        hh = max(hh, bar.high)
        # 2) arm break-even once +1R touched
        if not be_armed and bar.high >= entry + R:
            be_armed = true
            stop = max(stop, entry)          # move to break-even
        # 3) chandelier trail (only tightens upward)
        if be_armed:
            stop = max(stop, hh - 3*atr)
        # 4) check for stop hit this bar
        if bar.low <= stop:
            exit = stop
            return exit, (exit - entry)/R    # realized R-multiple
    exit = prices.last.close                  # exit at end if never stopped
    return exit, (exit - entry)/R

(5 for correct logic: BE arming, upward-only trail, stop-hit check; 2 for stated assumption on intrabar fill.)

(d) Moving to break-even too early converts many trades that would have retraced then continued into scratch (0R) exits and truncates would-be winners at the low end, cutting the average win $W$. Because expectancy $E=pW-(1-p)L$, shrinking $W$ (and only slightly raising effective $p$ of non-losses) can lower $E$ even though loss frequency feels reduced — the "safety" is paid for by giving up right-tail R-multiples that the whole edge depends on. "Cut losers quickly" applies to real adverse structure breaks, not to normal noise near entry. (4)

Question 3

(a) Portion $x$ books $+1R$. Portion $(1-x)$ resolves to $+4R$ (prob $q$) or $0R$ (prob $1-q$): $\bar R(x)=x(1)+(1-x)\big[q(4)+(1-q)(0)\big]=x+(1-x)\,4q.$ (4) With $q=0.35$: $4q=1.4$, so $\bar R(x)=x+1.4(1-x)=1.4-0.4x$.

(b) $\bar R(x)=1.4-0.4x$ is linear; slope $=-0.4<0$. (3) Hence expectancy is maximised at the lower endpoint $x=0$ (hold the full position for the runner), giving $\bar R(0)=1.4R$; scaling everything at $+1R$ gives $\bar R(1)=1.0R$. (2) Interpretation: scaling out feels good (locks profit, boosts hit rate) but here it reduces expectancy because the runner has a positive edge worth more than the booked $1R$; the psychological comfort costs return. (1)

(c) Outcomes of the whole trade as a function of the runner's resolution:

• With prob $q$: $R = x + (1-x)4 = 4-3x$.
• With prob $1-q$: $R = x + 0 = x$.

Mean $\bar R(x)=1.4-0.4x$ (from (a)). Variance: $\mathrm{Var}=q(1-q)\big[(4-3x)-x\big]^2=q(1-q)(4-4x)^2=16q(1-q)(1-x)^2.$ (2 — derivation) With $q(1-q)=0.35\times0.65=0.2275$:

• $x=0$: mean $=1.40$, Var $=16(0.2275)(1)=3.64$, sd $=1.908$; ratio $=1.40/1.908=0.734$.
• $x=1$: mean $=1.00$, Var $=0$, sd $=0$; ratio undefined/$\infty$ (a certain $+1R$). (2)

Argument: the full-runner ($x=0$) has higher expectancy but lower per-trade Sharpe (0.734) and real downside dispersion. Scaling fully out ($x=1$) yields a certain $+1R$ — zero variance — so a strongly risk-averse trader (minimising ruin/drawdown, or with poor edge estimates) may prefer larger $x$ despite lower mean, trading ~0.4R of expectancy for elimination of variance. The rational choice is an interior compromise if a smoothness/drawdown constraint is added; on pure expectancy alone, $x=0$ wins. (2)

[
  {"claim":"Expectancy E = 0.40R for Q1a","code":"p=Rational(4,10);W=Rational(5,2);L=1;E=p*W-(1-p)*L;result=(E==Rational(4,10))"},
  {"claim":"Kelly fraction f* = 0.16","code":"p=Rational(4,10);W=Rational(5,2);L=1;E=p*W-(1-p)*L;f=E/(W*L);result=(f==Rational(16,100))"},
  {"claim":"100-trade full-Kelly multiple ~20.0","code":"import sympy as sp;g=sp.Float('0.4')*sp.log(sp.Float('1.40'))+sp.Float('0.6')*sp.log(sp.Float('0.84'));m=sp.exp(100*g);result=(abs(m-20.0)<0.5)"},
  {"claim":"Half-Kelly multiple ~9.87","code":"import sympy as sp;g=sp.Float('0.4')*sp.log(sp.Float('1.20'))+sp.Float('0.6')*sp.log(sp.Float('0.92'));m=sp.exp(100*g);result=(abs(m-9.87)<0.3)"},
  {"claim":"Chandelier stop reaches break-even (>=100) on Day4","code":"highs=[102,105,104,109,108];hh=100;days=[];\nfor i,h in enumerate(highs,1):\n    hh=max(hh,h);days.append((i,hh-6))\nfirst=[d for d,s in days if s>=100][0];result=(first==4)"},
  {"claim":"Scaling expectancy R-bar(0)=1.4 and R-bar(1)=1.0","code":"q=Rational(35,100);f=lambda x:x+(1-x)*4*q;result=(f(0)==Rational(14,10) and f(1)==1)"},
  {"claim":"Variance at x=0 is 3.64","code":"q=Rational(35,100);V=16*q*(1-q)*(1-0)**2;result=(V==Rational(364,100))"}
]