Indicators & Oscillators

Question 1

(a) (6 marks) Unroll: $E_t=\alpha P_t+(1-\alpha)[\alpha P_{t-1}+(1-\alpha)E_{t-2}]=\dots$ $E_t=\alpha\sum_{k=0}^{\infty}(1-\alpha)^k P_{t-k}.$

• Correct unrolling to two/three levels (2)
• General form with weight $w_k=\alpha(1-\alpha)^k$ on $P_{t-k}$ (2)
• Sum of weights: $\sum_{k=0}^\infty \alpha(1-\alpha)^k=\alpha\cdot\frac{1}{1-(1-\alpha)}=\alpha\cdot\frac1\alpha=1$ ✓ (geometric series, ratio $1-\alpha<1$) (2)

(b) (7 marks) $L=\sum_{k\ge0}k\,\alpha(1-\alpha)^k=\alpha\cdot\frac{(1-\alpha)}{(1-(1-\alpha))^2}=\alpha\cdot\frac{1-\alpha}{\alpha^2}=\frac{1-\alpha}{\alpha}.$ Using $\sum k r^k=\frac{r}{(1-r)^2}$ with $r=1-\alpha$. (3) Substitute $\alpha=\frac2{N+1}$: $L=\frac{1-\frac2{N+1}}{\frac2{N+1}}=\frac{\frac{N-1}{N+1}}{\frac{2}{N+1}}=\frac{N-1}{2}.$ (2) For $N=19$: $L_{EMA}=\frac{18}{2}=9$; $L_{SMA}=\frac{N-1}{2}=9$. They are equal in center-of-mass. (2) (Note/insight: EMA has same average lag but a fatter recent-weight tail, so it responds faster to new data — leads into (c).)

(c) (7 marks)

def ema(prices, N):
    alpha = 2/(N+1)
    e = [prices[0]]              # seed with first price
    for p in prices[1:]:
        e.append(alpha*p + (1-alpha)*e[-1])
    return e

• Correct recursion & alpha (3); valid seeding stated (1)
• Explanation (3): Although average lag matches SMA, the EMA weight $w_0=\alpha$ on the most recent price exceeds the SMA's flat weight $\frac1N$ (since $\alpha=\frac2{N+1}>\frac1N$ for $N>1$). A price shock at $t$ therefore moves the EMA more on the same bar, whereas the SMA weights it only $1/N$ and must also wait for the shock to enter/exit the fixed window. Hence faster reaction to shocks.

Question 2

(a) (5 marks) As $RS=\overline U/\overline D\in[0,\infty)$, $1+RS\in[1,\infty)$, so $\frac{100}{1+RS}\in(0,100]$, giving $RSI=100-\frac{100}{1+RS}\in[0,100)$. Limits: $RS=0\Rightarrow RSI=0$; $RS\to\infty\Rightarrow RSI\to100$. Bounds proven. (3) For $RSI=70$: $70=100-\frac{100}{1+RS}\Rightarrow\frac{100}{1+RS}=30\Rightarrow 1+RS=\frac{100}{30}\Rightarrow RS=\frac{7}{3}\approx2.333$. (2)

(b) (8 marks) Changes: $+1,+1,-1,+2,-1,+2,+2$. Gains: $1,1,2,2,2$ → sum $=8$; Losses: $1,1$ → sum $=2$. Over 7 periods: $\overline U=8/7$, $\overline D=2/7$. (3) $RS=\dfrac{8/7}{2/7}=4$. (2) $RSI=100-\frac{100}{1+4}=100-20=80.0$. (3)

(c) (7 marks) Bearish divergence: price makes a higher high while RSI makes a lower high over the same interval — momentum weakening despite rising price. (2)

def detect_bearish_divergence(price, rsi, lookback=20):
    highs = local_maxima_indices(price, lookback)   # swing highs
    for a, b in consecutive_pairs(highs):
        if price[b] > price[a] and rsi[b] < rsi[a]:
            return True, (a, b)
    return False, None

• Correct definition (2); pseudocode identifying two swing highs & comparing price↑ vs RSI↓ (3)
• False-signal reason (2): in a strong uptrend RSI can stay pinned high (repeated overbought) and print lower peaks purely from smoothing/scaling while price keeps advancing legitimately — divergence fires early and the trend continues.

Question 3

(a) (6 marks) Histogram $H=$ MACD $-$ signal. A bullish MACD/signal crossover is defined as MACD rising through signal, i.e. transition from MACD $<$ signal to MACD $>$ signal $\iff$ $H$ transitions from $<0$ to $>0$. Thus crossover $\equiv$ histogram zero-crossing (negative→positive). (3) Since signal $=EMA_9(\text{MACD})$ lags MACD, $H\approx$ MACD $-$ (its own smoothed value) $\propto$ the recent rate of change of MACD; $H>0$ and rising indicates MACD accelerating upward. Sign change of $H$ marks where MACD's momentum turns positive relative to its average. (3)

(b) (8 marks) Mean $=\frac{20+22+21+23+24}{5}=\frac{110}{5}=22$. (2) Deviations: $-2,0,-1,1,2$; squares $4,0,1,1,4$, sum $=10$. Population variance $=10/5=2$; $\sigma=\sqrt2\approx1.4142$. (3) Upper $=22+2\sqrt2\approx24.828$; lower $=22-2\sqrt2\approx19.172$. $BW=\frac{4\sqrt2}{22}=\frac{2\sqrt2}{11}\approx0.2571$. (2) Falling bandwidth = squeeze: contracting volatility, bands tighten; often precedes a volatility expansion / breakout. (1)

(c) (6 marks) RSI, Stochastic %K and Williams %R are all bounded momentum oscillators built from recent price range/change; %K and Williams %R are algebraically near-mirror images ($\%R=\%K-100$ over the same $N$), and both correlate strongly with RSI. Stacking them adds near-zero independent information but multiplies visual noise and gives a false sense of "confirmation" (correlated signals agreeing is not independent evidence) — classic indicator overload. (3) Low-redundancy dashboard: Trend = ADX (or MA crossover) — measures directional strength; Momentum = RSI or MACD — one oscillator only; Volatility = Bollinger Bandwidth or ATR — measures dispersion. Each captures an orthogonal dimension (direction / speed / spread), minimizing correlation and maximizing information per indicator. (3)

[
  {"claim":"EMA center of mass = (N-1)/2 for N=19 equals 9","code":"N=19; alpha=Rational(2,N+1); L=(1-alpha)/alpha; result=(L==9)"},
  {"claim":"RS for RSI=70 is 7/3","code":"RS=symbols('RS'); sol=solve(Eq(100-100/(1+RS),70),RS)[0]; result=(sol==Rational(7,3))"},
  {"claim":"Q2b RSI equals 80","code":"U=Rational(8,7); D=Rational(2,7); RS=U/D; RSI=100-100/(1+RS); result=(RSI==80)"},
  {"claim":"Bollinger population sigma sqrt2 and bandwidth 2*sqrt2/11","code":"vals=[20,22,21,23,24]; m=Rational(sum(vals),5); var=sum((v-m)**2 for v in vals)/5; sd=sqrt(var); bw=(4*sd)/m; result=(simplify(sd-sqrt(2))==0 and simplify(bw-2*sqrt(2)/11)==0)"}
]