Indicators & Oscillators

Question 1 (10)

(a) Multiplier: $k = \dfrac{2}{N+1}$. (1) Recursion: $EMA_t = P_t \cdot k + EMA_{t-1}\cdot(1-k)$. (2)

(b) For $N=3$: $k = 2/4 = 0.5$. (1)

• $EMA_1 = 10$ (seed)
• $EMA_2 = 11(0.5) + 10(0.5) = 10.5$ (1)
• $EMA_3 = 12(0.5) + 10.5(0.5) = 11.25$ (1)
• $EMA_4 = 13(0.5) + 11.25(0.5) = 12.125$ (1)
• $EMA_5 = 14(0.5) + 12.125(0.5) = 13.0625 \approx 13.06$ (1)

(c) EMA weights recent prices more heavily via the multiplier $k$, while SMA weights all $N$ values equally; older data never fully drops out but decays exponentially, so new prices shift the EMA immediately. (2)

Question 2 (12)

(a) $RS = \dfrac{\text{Avg Gain}}{\text{Avg Loss}}$; $RSI = 100 - \dfrac{100}{1+RS}$. (3)

(b) Price changes (6): +0.5, −0.3, +0.8, +1.0, −0.5, +1.0 (2)

• Gains: 0.5, 0.8, 1.0, 1.0 → sum = 3.3; Avg Gain = 3.3/6 = 0.55 (1)
• Losses: 0.3, 0.5 → sum = 0.8; Avg Loss = 0.8/6 = 0.1333 (1)
• $RS = 0.55/0.13333 = 4.125$ (1)
• $RSI = 100 - 100/(1+4.125) = 100 - 19.512 = 80.49$ (1)

(c) Overbought $\geq 70$, oversold $\leq 30$. (1) Bearish divergence: price makes a higher high while RSI makes a lower high, signalling weakening momentum and possible reversal down. (2)

Question 3 (10)

(a)

• MACD line $= EMA_{12}(P) - EMA_{26}(P)$ (2)
• Signal line $= EMA_9(\text{MACD line})$ (1)
• Histogram $= \text{MACD line} - \text{Signal line}$ (1)

(b) (4)

macd_line = ema(prices,12) - ema(prices,26)   # element-wise
signal    = ema(macd_line, 9)
histogram = macd_line - signal

(1 mark macd line, 1 signal, 1 histogram, 1 correct alignment/element-wise.)

(c) Histogram crossing from negative to positive means the MACD line has crossed above the signal line — a bullish momentum shift / buy signal. (2)

Question 4 (10)

(a) Middle $= SMA_{20}$; Upper $= SMA_{20} + 2\sigma$; Lower $= SMA_{20} - 2\sigma$. (3)

(b) Mean $= (20+22+24+22+22)/5 = 110/5 = 22$ (1) Deviations: −2,0,2,0,0 → squares 4,0,4,0,0 → sum 8; variance $=8/5=1.6$; $\sigma=\sqrt{1.6}=1.2649$ (2)

• Upper $= 22 + 2(1.2649) = 24.53$ (1)
• Lower $= 22 - 2(1.2649) = 19.47$ (1)

(c) A squeeze is when the bands contract (low volatility, narrow band width), often preceding a breakout. (1) Pair with ADX (or MACD/volume/OBV) to confirm breakout direction. (1)

Question 5 (10)

(a) $VWAP = \dfrac{\sum (TP_i \cdot V_i)}{\sum V_i}$. (1) Typical prices: Bar1 $(11+9+10)/3=10$; Bar2 $(13+11+12)/3=12$; Bar3 $(15+13+14)/3=14$. (2)

• TP×Vol: $10(100)=1000$; $12(300)=3600$; $14(100)=1400$; sum $=6000$ (1)
• Cumulative Vol $=500$ (1)
• $VWAP = 6000/500 = 12.0$ (1)

(b) $\%K = \dfrac{C - L_{low}}{H_{high} - L_{low}}\times 100 = \dfrac{47-40}{50-40}\times100 = 70$ (2)

(c) %D is the smoothing (typically 3-period SMA) of %K — it acts as the signal line. (2)

Question 6 (8)

(a) ADX measures trend strength (not direction); ADX > 25 conventionally indicates a trending market, below ~20 indicates non-trending/weak. (2)

(b) Up-day (close > prev close): OBV += volume. Down-day (close < prev close): OBV −= volume. Unchanged: OBV stays the same. (3)

(c) Indicator overload = stacking many correlated indicators that give redundant/conflicting signals, causing analysis paralysis and false confidence. (2) Rule: use a small set from different categories (e.g., one trend, one momentum, one volume). (1)

[
  {"claim":"EMA5 for 3-period EMA equals 13.0625","code":"k=Rational(1,2)\ne=10\nfor p in [11,12,13,14]:\n    e=p*k+e*(1-k)\nresult=(e==Rational(209,16)) and (float(e)==13.0625)"},
  {"claim":"RSI over 6 changes equals ~80.49","code":"ag=Rational(33,10)/6\nal=Rational(8,10)/6\nrs=ag/al\nrsi=100-100/(1+rs)\nresult=abs(float(rsi)-80.4878)<0.01"},
  {"claim":"Bollinger upper/lower bands at 2 sigma","code":"import sympy as sp\nvals=[20,22,24,22,22]\nm=sp.Rational(sum(vals),5)\nvar=sp.Rational(sum((v-m)**2 for v in vals),5)\nsd=sp.sqrt(var)\nup=float(m+2*sd)\nlo=float(m-2*sd)\nresult=abs(up-24.5298)<0.01 and abs(lo-19.4702)<0.01"},
  {"claim":"VWAP equals 12.0","code":"num=10*100+12*300+14*100\nden=100+300+100\nresult=Rational(num,den)==12"},
  {"claim":"Stochastic %K equals 70","code":"k=(47-40)/(50-40)*100\nresult=k==70"}
]