Entry, Exit & Trade Management

Question 1 (10)

(a) Derivation (3) Expected value per trade = sum over outcomes of (probability × payoff). Two outcomes: win (prob $W$) pays $R_w$; loss (prob $1-W$) pays $R_l$ (negative). $E = W\cdot R_w + (1-W)\cdot R_l$

• $W$ = win rate (probability of a winner) — 1 mark
• $R_w$ = average win in R, $R_l$ = average loss in R (signed) — 1 mark
• Sum weighted by probabilities gives expectancy per trade — 1 mark

(b) Compute (3) $E = 0.40(2.5) + 0.60(-1.0) = 1.0 - 0.6 = \mathbf{0.40\ R}$ per trade. (setup 1, arithmetic 2)

(c) Profit estimate (3) Expected R per trade = 0.40; total R over 200 trades = $0.40\times200 = 80$R. ₹ per R = ₹2,000 → profit $= 80\times2000 = \mathbf{₹1,60,000}$. Assumption: R (risk) held constant and stats stable/independent across trades. (1 for R total, 1 for ₹, 1 for assumption)

(d) One sentence (1) Because the average winner (2.5R) is much larger than the average loser (1R), the positive payoff asymmetry more than compensates for winning less than half the time.

Question 2 (12)

(a) Stops (6, 2 each)

• (i) Fixed 2%: $500\times0.98 = \mathbf{₹490.00}$
• (ii) 1.5×ATR: $500 - 1.5\times6 = 500-9 = \mathbf{₹491.00}$
• (iii) Structure: $488 - 1 = \mathbf{₹487.00}$

(b) Risk per share (3, 1 each)

• Fixed: $500-490 = ₹10$
• ATR: $500-491 = ₹9$
• Structure: $500-487 = ₹13$

(c) Position size, structure stop (3) Risk budget $= 1\%\times5,00,000 = ₹5,000$. Shares $= 5000/13 = 384.6 \to \mathbf{384}$ shares (round down). (2) Notional $= 384\times500 = \mathbf{₹1,92,000}$. (1)

Question 3 (10)

(a) 1:2 target (4) Risk per share $= 500-487 = ₹13$. Reward required $= 2\times$ risk $= 2\times13 = ₹26$. (2) Target $= 500 + 26 = \mathbf{₹526}$. (2)

(b) Against resistance (3) Available reward to ₹520 $= 520-500 = ₹20$. RR $= 20/13 = 1.54$, which is < 2. (2) Trade does not meet the 1:2 rule against this resistance; reject or find a lower-risk entry. (1)

(c) Explain (3) Expectancy $= W\cdot R_w+(1-W)\cdot R_l$. Below 1:2 the average winner shrinks toward the average loser, so even a decent win rate produces low or negative expectancy; costs/slippage further erode it, and drawdowns deepen because winners no longer cover the accumulated losses. (3 for expectancy reasoning + magnitude point)

Question 4 (12)

(a) Pseudocode (8)

def manage_trade(entry, R, atr, price_series):
    stop = entry - R              # initial stop (1)
    be_done = False               # break-even flag
    high = entry
    for price in price_series:
        high = max(high, price)                    # track high (1)
        if not be_done and price >= entry + R:     # reached +1R
            stop = max(stop, entry)                 # move to break-even (2)
            be_done = True
        if be_done:                                # trail after BE
            trail = high - 1.5 * atr
            stop = max(stop, trail)                 # never lower (2)
        if price <= stop:                          # exit condition (1)
            exit_price = stop
            r_multiple = (exit_price - entry) / R  # (1)
            return exit_price, r_multiple
    # series ended without hit: mark to last price
    exit_price = price_series[-1]
    return exit_price, (exit_price - entry) / R

Marks: init stop 1, BE move 2, trail w/ never-lower 2, high tracking 1, exit 1, R-calc 1.

(b) Trade-off (4) A tight 1×ATR trail locks profits fast and cuts giveback but gets shaken out on normal noise, capping many trades near break-even/small wins. A loose 3×ATR trail lets winners run into large R-multiples but returns more open profit on reversals and produces choppier results. Choice depends on whether the edge is in frequent small wins or occasional large trends. (2 for each side)

Question 5 (10)

(a) Exit prices (3) R = ₹13. +1R = 513, +2R = 526, +3R = 539.

• +1R: ₹513 — +2R: ₹526 — +3R: ₹539 (1 each)

(b) Total profit (4) 100 shares each tranche.

• Tranche 1: $100\times13 = ₹1,300$
• Tranche 2: $100\times26 = ₹2,600$
• Tranche 3: $100\times39 = ₹3,900$ Total $= 1300+2600+3900 = \mathbf{₹7,800}$. (1 each tranche + 1 sum)

(c) Blended R (3) Total R risked initially $= 300\times13 = ₹3,900$ = 1R for full position. Blended R $= 7800/3900 = \mathbf{2.0R}$. (Equivalently $\tfrac{1}{3}(1+2+3)=2.0R$.) (2 method, 1 answer)

Question 6 (6)

(a) Four rule components (3, ¾ each) Any four of: entry trigger/signal, confirmation criteria, stop-loss placement rule, target/exit rule, position-sizing/risk rule, market/time filter (which instruments/sessions).

(b) Cutting losers (3) The initial stop defines 1R; honouring it caps every loss at −1R (or less). Letting losers run turns −1R into −2R/−3R losses, which distorts the loss term in expectancy $(1-W)R_l$ and can flip a positive edge negative. Quick exits preserve the assumed R-distribution the whole system was built on.

[
{"claim":"Q1b expectancy = 0.40R","code":"E=Rational(40,100)*Rational(5,2)+Rational(60,100)*(-1); result=(E==Rational(2,5))"},
{"claim":"Q1c profit = 160000","code":"result=(0.40*200*2000==160000)"},
{"claim":"Q3a 1:2 target = 526","code":"risk=500-487; result=(500+2*risk==526)"},
{"claim":"Q3b RR to 520 < 2","code":"rr=Rational(520-500,500-487); result=(rr<2)"},
{"claim":"Q5b total profit = 7800","code":"result=(100*13+100*26+100*39==7800)"},
{"claim":"Q5c blended R = 2","code":"result=(Rational(7800,300*13)==2)"}
]