Trading Strategies

Question 1

(a) VWAP (4 marks) $VWAP = \frac{\sum (\text{TypPrice}\times \text{Vol})}{\sum \text{Vol}}$ Numerator $= 100.00(4000)+100.50(6000)+101.00(10000)$ $= 400{,}000 + 603{,}000 + 1{,}010{,}000 = 2{,}013{,}000$ (2 marks) Denominator $= 4000+6000+10000 = 20{,}000$ $VWAP = \frac{2{,}013{,}000}{20{,}000} = 100.65$ (2 marks)

(b) VWAP filter (3 marks) Entry price $101.35 > VWAP\ 100.65$ ⇒ price is above VWAP, so the long filter confirms the trade. (3 marks: correct comparison + conclusion)

(c) Stop, risk, target, reward (5 marks) OR midpoint $= \frac{H_{OR}+L_{OR}}{2} = \frac{101.20+99.80}{2} = 100.50$ (stop). (1) Risk per share $= 101.35 - 100.50 = 0.85$. (1) Reward per share $= R\times\text{risk} = 2.5\times 0.85 = 2.125$. (1) Target $= 101.35 + 2.125 = 103.475 \approx 103.48$. (2)

(d) Expectancy (4 marks) $E = p\cdot(2.5) + (1-p)(-1)$ $= 0.42(2.5) + 0.58(-1) = 1.05 - 0.58 = 0.47$ units of risk. (3) $E>0$ ⇒ profitable system (expected $+0.47R$ per trade). (1)

(e) Break-even win rate (5 marks) Set expectancy to zero: $p^\*(R) + (1-p^\*)(-1) = 0 \Rightarrow p^\*R - 1 + p^\* = 0$ $p^\*(R+1) = 1 \Rightarrow \boxed{p^\* = \frac{1}{R+1}}$ (3) For $R=2.5$: $p^\* = \frac{1}{3.5} = 0.2857$. (1) Actual $p=0.42 > 0.2857$, giving a comfortable margin of safety (~13 percentage points above break-even). (1)

(f) Position sizing (3 marks) Risk budget = 1\%\times 50{,}000 = \500$. *(1)* Shares$= \dfrac{500}{0.85} = 588.24 \Rightarrow 588$ shares (round down). (2)

Question 2

(a) Constant SMA difference in linear trend (8 marks) For $P_t = a+bt$, the SMA of period $k$ ending at $t$ (using bars $t-k+1,\dots,t$): $\text{SMA}_k(t) = \frac{1}{k}\sum_{i=0}^{k-1} P_{t-i} = \frac{1}{k}\sum_{i=0}^{k-1}\big(a + b(t-i)\big)$ $= a + bt - \frac{b}{k}\sum_{i=0}^{k-1} i = a + bt - \frac{b}{k}\cdot\frac{(k-1)k}{2} = a + bt - \frac{b(k-1)}{2}.$ (4 marks) Therefore: $\text{SMA}_n(t) - \text{SMA}_m(t) = -\frac{b(n-1)}{2} + \frac{b(m-1)}{2} = \frac{b(m-n)}{2}.$ (2 marks) This is constant (no $t$). Since $b>0$ and $m>n$, the difference is positive: the fast SMA sits a fixed distance above the slow SMA. Interpretation: in a perfectly linear trend the SMAs never cross — crossover signals arise only from curvature / changes in slope, not from steady trends. (2 marks)

(b) SMA computation (6 marks) Last 3 prices (indices 7,8,9): $17,19,18$ ⇒ $\text{SMA}_3 = \frac{17+19+18}{3} = \frac{54}{3} = 18.00$. (3) Last 5 prices (indices 5–9): $16,15,17,19,18$ ⇒ $\text{SMA}_5 = \frac{85}{5} = 17.00$. (2) $18.00 > 17.00$ ⇒ fast SMA is above slow SMA (bullish). (1)

(c) Code (8 marks)

def crossover_signals(prices, n, m):
    def sma(seq, k, i):
        return sum(seq[i-k+1:i+1]) / k
    signals = [0] * len(prices)
    warmup = m - 1  # need m bars for slow SMA
    for i in range(len(prices)):
        if i < warmup:
            signals[i] = 0
            continue
        fast_now  = sma(prices, n, i)
        slow_now  = sma(prices, m, i)
        fast_prev = sma(prices, n, i-1)
        slow_prev = sma(prices, m, i-1)
        if fast_prev <= slow_prev and fast_now > slow_now:
            signals[i] = 1    # golden cross
        elif fast_prev >= slow_prev and fast_now < slow_now:
            signals[i] = -1   # death cross
        else:
            signals[i] = 0
    return signals

Marks: warm-up handling (2), correct prev-vs-now cross detection (4), golden/death labelling (2).

Question 3

(a) RSI computation (5 marks) $RS = \frac{1.20}{0.80} = 1.5$. (2) $RSI = 100 - \frac{100}{1+1.5} = 100 - \frac{100}{2.5} = 100 - 40 = 60.$ (2) $60$ is between 30 and 70 ⇒ neutral. (1)

(b) Threshold g for RSI=70 (4 marks) $70 = 100 - \frac{100}{1+g} \Rightarrow \frac{100}{1+g} = 30 \Rightarrow 1+g = \frac{100}{30} = 3.333\ldots$ $g = \frac{100}{30} - 1 = \frac{70}{30} = 2.333\ldots \approx 2.3333.$ (3) So $g = 7/3$ marks the overbought line (avg gain = $2.33\times$ avg loss). (1)

(c) Expectancy of RSI-oversold long (5 marks) $E = 0.60(1.5) + 0.40(-1) = 0.90 - 0.40 = 0.50$ units of risk. (3) $E = +0.50R > 0$ ⇒ positive-edge system; the RSI<30 filter produces a profitable expectancy. (2)

[
  {"claim":"VWAP equals 100.65","code":"num=100*4000+100.5*6000+101*10000; den=4000+6000+10000; result = (Rational(num,den)==Rational(20130,200)) and (num/den==100.65)"},
  {"claim":"ORB risk 0.85, reward 2.125, target 103.475","code":"stop=Rational('100.50'); entry=Rational('101.35'); risk=entry-stop; reward=Rational(5,2)*risk; target=entry+reward; result = (risk==Rational('0.85')) and (reward==Rational('2.125')) and (target==Rational('103.475'))"},
  {"claim":"Expectancy R=2.5 p=0.42 is 0.47","code":"p=Rational(42,100); E=p*Rational(5,2)+(1-p)*(-1); result = (E==Rational(47,100))"},
  {"claim":"Break-even win rate for R=2.5 is 2/7","code":"R=Rational(5,2); pstar=1/(R+1); result = (pstar==Rational(2,7))"},
  {"claim":"Shares = floor(500/0.85) = 588","code":"import math; result = (math.floor(500/Rational('0.85'))==588)"},
  {"claim":"SMA3=18 SMA5=17 at last bar","code":"p=[10,11,13,12,14,16,15,17,19,18]; s3=Rational(sum(p[7:10]),3); s5=Rational(sum(p[5:10]),5); result = (s3==18) and (s5==17)"},
  {"claim":"RSI with RS=1.5 equals 60","code":"RS=Rational(120,80); RSI=100-100/(1+RS); result = (RSI==60)"},
  {"claim":"g for RSI=70 equals 7/3","code":"g=symbols('g'); sol=solve(Eq(100-100/(1+g),70),g); result = (sol[0]==Rational(7,3))"},
  {"claim":"RSI-oversold expectancy = 0.5","code":"E=Rational(60,100)*Rational(3,2)+Rational(40,100)*(-1); result = (E==Rational(1,2))"}
]