Candlestick Patterns

Question 1

(a) Proof of $R = B + U + Lw$ (4 marks)

Let $m = \max(O,C)$, $n = \min(O,C)$. Then by definition $U = H - m$, $Lw = n - L$, $B = m - n$ (since $B=|C-O| = m-n$). [2]

$B + U + Lw = (m-n) + (H-m) + (n-L) = H - L = R.$ [1]

This holds regardless of whether $C>O$ (bullish) or $C<O$ (bearish), because the algebra uses only $\max/\min$, which is symmetric in $O,C$. [1]

(b) Classification, $\tau=0.1$ (6 marks) — 2 marks each.

• Candle A: $R=16$, $B=|100.5-100|=0.5$. $\tau R = 1.6$; $0.5 \le 1.6$ → doji. $U = 108-100.5=7.5$, $Lw=100-92=8$. Neither $\le 1.6$ → neutral (long-legged) doji. [2]
• Candle B: $R=8.4$, $B=0.2$. $\tau R=0.84$; $0.2\le0.84$ → doji. $U=50.4-50.2=0.2\le0.84$ ✓; $Lw=50-42=8 > 0.84$ ✗. Only upper wick small → gravestone doji. [2]
• Candle C: $R=8.2$, $B=0.1$. $\tau R=0.82$; $0.1\le0.82$ → doji. $U=38-30.1=7.9$; $Lw=30-29.8=0.2\le0.82$ ✓. Only lower wick small → dragonfly doji. [2]

(c) Classifier (8 marks)

Precise rules (using body $B$, wicks $U,Lw$, range $R$, and $\tau$):

• hammer: $Lw \ge 2B$ and $U \le \tau R$ and $B > \tau R$ (long lower wick ≥ twice body, tiny/no upper wick, non-doji body). [2]
• shooting star: $U \ge 2B$ and $Lw \le \tau R$ and $B > \tau R$. [2]

def classify(o, h, l, c, tau=0.1):
    m, n = max(o, c), min(o, c)
    B, U, Lw, R = m - n, h - m, n - l, h - l
    if R == 0:
        return "other"
    if B <= tau * R:                       # doji family
        if U <= tau * R and Lw > tau * R:  return "dragonfly_doji"
        # (gravestone/neutral omitted for return-set brevity)
        return "spinning_top"
    if U == 0 and Lw == 0:                 return "bullish_marubozu" if c > o else "other"
    if Lw >= 2*B and U <= tau*R:           return "hammer"
    if U  >= 2*B and Lw <= tau*R:          return "shooting_star"
    if U > tau*R and Lw > tau*R:           return "spinning_top"
    return "other"

Function structure & correct branch ordering [2]. Psychology justification: hammer's long lower wick = sellers pushed price down but buyers reclaimed the close near the top → bullish rejection of lows; shooting star's long upper wick = buyers pushed up but sellers slammed close back near the low → bearish rejection of highs. [2]

(d) Marubozu cannot be a hammer (4 marks)

For a marubozu, $U=0$ and $Lw=0$. The hammer rule requires $Lw \ge 2B$. [1] For a real candle $B = R > 0$ (marubozu has a body). [1] So $Lw \ge 2B$ becomes $0 \ge 2R > 0$, a contradiction. [1] Hence the hammer branch is never entered; a marubozu is classified bullish_marubozu (which is checked first). [1]

Question 2

(a) Engulfing vs Harami (6 marks)

Bullish engulfing (prior downtrend): candle 1 bearish, candle 2 bullish, body 2 engulfs body 1: $C_1 < O_1,\quad C_2 > O_2,\quad O_2 \le C_1,\quad C_2 \ge O_1.$ [2]

Bullish harami: candle 1 bearish (large body), candle 2 bullish and body contained inside candle 1's body: $C_1 < O_1,\quad C_2 > O_2,\quad O_2 \ge C_1,\quad C_2 \le O_1.$ [2]

Mutual exclusivity: engulfing needs $O_2 \le C_1$ and $C_2 \ge O_1$; harami needs $O_2 \ge C_1$ and $C_2 \le O_1$. Both sets simultaneously force $O_2 = C_1$ and $C_2 = O_1$, i.e. body 2 = body 1 exactly (degenerate, zero-measure). For any genuine engulfing ($O_2 < C_1$ or $C_2 > O_1$ strictly) the harami condition is violated, so they cannot both hold non-trivially. [2]

(b) Probability of bearish engulfing (9 marks)

Let $X = C_1 - O_1 \sim \mathcal N(0,4)$, $Y = C_2 - O_2 \sim \mathcal N(0,4)$, independent. With $O_2=C_1$.

Bearish engulfing needs: candle 1 bullish ($X>0$), candle 2 bearish ($Y<0$), and body-2 engulfs body-1: $O_2 \ge C_1 \text{ (auto, } O_2=C_1) \text{ and } C_2 \le O_1.$ Actually the engulf requires $O_2 \ge C_1$ (open above prior close) and $C_2 \le O_1$ (close below prior open). Since $O_2 = C_1$, first holds with equality. Second: $C_2 = O_2 + Y = C_1 + Y \le O_1$, i.e. $Y \le O_1 - C_1 = -X$. [3]

Given $X>0$ we want $P(Y \le -X \mid X>0)$, with $Y<0$ automatically implied since $-X<0$. [1]

$P = \frac{\displaystyle\int_0^\infty \Phi\!\left(\frac{-x}{\sigma}\right) f_X(x)\,dx}{P(X>0)},\qquad \sigma=2.$ [2]

Since $X,Y$ are i.i.d. symmetric, condition on $X>0$: $P(X>0)=\tfrac12$. The numerator $=\int_0^\infty \Phi(-x/2) f_X(x)dx$. By symmetry $P(Y\le -X, X>0) = P(Y<X, X>0$ ... )$; evaluate directly:$Y+X<0$means$Z=X+Y<0$where$Z\sim\mathcal N(0,8)$, but we still need$X>0$. [1]

Numerically the conditional probability $\approx 0.25$. Reasoning: $P(X>0, Y<-X) = P(X>0, X+Y<0)$. By the geometry of two i.i.d. symmetric variables, the event $\{X>0, X+Y<0\}$ has probability $\tfrac18$ (one-eighth of the plane by rotational symmetry of the isotropic Gaussian: the wedge from angle $135°$ to $180°$ spans $45°=\tfrac18$ of $360°$). Dividing by $P(X>0)=\tfrac12$ gives $\boxed{1/4}$. [2]

(c) Morning star predicate + context (7 marks)

Predicate over $(O_1..C_1, O_2..C_2, O_3..C_3)$:

morning_star =
   (C1 < O1)                               # candle1 long bearish
 and (|C1-O1| >= k*R1)                      # long body, e.g. k=0.6
 and (max(O2,C2) < C1)                      # star gaps down
 and (|C2-O2| <= 0.30*|C1-O1|)              # small star body
 and (C3 > O3)                              # candle3 bullish
 and (C3 > (O1+C1)/2)                       # closes above candle1 midpoint

Correct core conditions incl. gap and 30% body rule [4].

Context (3.2.13): the identical three shapes (long down, small star, long up) occurring at the top of an uptrend would not be read as a bullish reversal — a bearish reversal like the evening star is what appears at tops (long up → small star → long down). More importantly, pattern meaning is location-dependent: a reversal pattern only reverses a prior trend; at a top the bullish morning-star shape (if it appeared) would be a weak/failing signal against overhead resistance, whereas its mirror (evening star) at the top is the high-probability reversal. [3]

Question 3

(a) Rising three methods formalisation (6 marks)

Candles $1..5$: $\underbrace{C_1 > O_1}_{\text{long white}},\quad B_1 = C_1-O_1 \text{ large}.$ [1] Small middle candles contained in candle 1's range: $\text{for } i\in\{2,3,4\}:\ H_i \le H_1 \ \wedge\ L_i \ge L_1,\qquad B_i < B_1.$ [2] Drift down (typically bearish/small): $C_2 \le C_1$, and $C_2 > C_3 > C_4$ (mild decline) — accept $\max(H_2,H_3,H_4)\le H_1$ as the containment. [1] Fifth candle long white closing above candle 1: $C_5 > O_5 \ \wedge\ C_5 > C_1.$ [2]

(b) Vectorised scan (6 marks)

import numpy as np
def rising_three(O,H,L,C):
    n = len(C); idx = []
    for i in range(n-4):                 # window of 5
        long1 = C[i] > O[i] and (C[i]-O[i]) > 0.6*(H[i]-L[i])
        mids  = np.all(H[i+1:i+4] <= H[i]) and np.all(L[i+1:i+4] >= L[i])
        small = np.all(np.abs(C[i+1:i+4]-O[i+1:i+4]) < (C[i]-O[i]))
        last  = C[i+4] > O[i+4] and C[i+4] > C[i]
        if long1 and mids and small and last:
            idx.append(i)
    return np.array(idx)

Fully vectorised alternative uses sliding windows; either accepted. [4] Complexity: each of $N-4$ windows does $O(1)$ (constant 3-element) work → $O(N)$ time, $O(N)$ output space. [2]

(c) Proof + counterexample (4 marks)

The conditions force $C_5 > C_1$ directly (it is an explicit clause), so that inequality is guaranteed. [1]

$C_1 > \max(C_2,C_3,C_4)$ is not guaranteed: containment only requires $H_i \le H_1$ and $L_i \ge L_1$ and $B_i<B_1$; a middle candle can close above $C_1$ while still sitting below $H_1$. [1]

Counterexample: Candle 1: $O_1=100, H_1=112, L_1=99, C_1=110$ (body 10). Candle 2: $O_2=109, H_2=111, L_2=105, C_2=110.5$ — contained ($H_2=111\le112$, $L_2=105\ge99$, body $1.5<10$) yet $C_2=110.5 > C_1=110$. So $C_1 \not> \max(C_2,C_3,C_4)$.