When to Trade — Timing & Sessions

Question 1

(a) Expand and integrate: $V=\int_0^1\left[a\left(t-\tfrac12\right)^2+b\right]dt = a\left[\frac{(t-\frac12)^3}{3}\right]_0^1 + b.$ At $t=1$: $(\tfrac12)^3=\tfrac18$; at $t=0$: $(-\tfrac12)^3=-\tfrac18$. Difference $=\tfrac{1}{4}$. $\boxed{V=\frac{a}{12}+b}$ Marks: integral setup (2), correct evaluation $a/12+b$ (2).

(b) $0.0009 = \dfrac{a}{12}+0.0004 \Rightarrow \dfrac{a}{12}=0.0005 \Rightarrow a=0.006.$ Marks: substitution (2), $a=0.006$ (2).

(c) Opening-range variance: $\int_0^{0.08}\!\!\left[a\left(t-\tfrac12\right)^2+b\right]dt = a\left[\frac{(t-\frac12)^3}{3}\right]_0^{0.08}+b(0.08).$ At $t=0.08$: $(0.08-0.5)=-0.42$, $(-0.42)^3=-0.074088$; at $t=0$: $(-0.5)^3=-0.125$. Difference $=(-0.074088-(-0.125))/3 = 0.050912/3 = 0.0169707$. Times $a=0.006$: $0.006\times0.0169707 = 1.01824\times10^{-4}$. Plus $b\cdot0.08 = 0.0004\times0.08 = 3.2\times10^{-5}$. Numerator $=1.01824\times10^{-4}+3.2\times10^{-5} = 1.33824\times10^{-4}$. $f = \frac{1.33824\times10^{-4}}{0.0009} = 0.14869 \approx 14.9\%.$ Marks: integral setup (2), cube evaluations (2), $f\approx0.149$ (2).

(d) The first 30 minutes is $8\%$ of clock time but carries $\approx14.9\%$ of the day's variance — nearly 1.9× its time-share. So it is a disproportionately high-move (and high-risk) window. A risk-averse or slippage-sensitive trader who "avoids the first 30 minutes" thus discards a large slice of realizable move; conversely a scalper seeking movement should trade it. This quantifies the "times to avoid" principle: avoidance decisions should be weighted by variance density, not clock minutes. Full marks require: (i) compare 8% vs 14.9% (2), (ii) compute the ~1.9× ratio (2), (iii) link to trader style / times-to-avoid (2).

Question 2

(a) VaR (fraction) $= z_{0.99}\,\sigma_G$ (since $\mu=0$).

• Normal: $2.326\times0.008 = 0.018608 \approx 1.861\%$.
• Earnings: $\sigma_G'=3.5\times0.008=0.028$; VaR $=2.326\times0.028=0.065128\approx6.513\%$. Marks: formula (2), normal value (2), earnings value (2).

(b) Constraint: $\text{VaR}\cdot P \le 0.02E = ₹20{,}000$.

• Normal: $P_{max}=\dfrac{20000}{0.018608}=₹1{,}074{,}806$.
• Earnings: $P_{max}=\dfrac{20000}{0.065128}=₹307{,}087$.
• Ratio $=\dfrac{P_{norm}}{P_{earn}} = 3.5$ (equals $k$, since VaR scales linearly with $\sigma_G$). Marks: constraint (2), both $P_{max}$ (2), ratio = 3.5 with reasoning (2).

(c) Mixture variance (mean return $0$ in each regime): $\sigma^2_{eff}=p(9\sigma_0^2)+(1-p)\sigma_0^2 = \sigma_0^2[9p+(1-p)] = \sigma_0^2(1+8p).$ With $p=0.2$: $1+8(0.2)=2.6$, so $\sigma^2_{eff}=2.6\sigma_0^2 = 2.6\times0.0001=2.6\times10^{-4}$. $\sigma_{eff}=\sqrt{2.6\times10^{-4}}=0.016125.$ The mixture of two zero-mean normals with different variances is leptokurtic (excess kurtosis $>0$): the high-variance regime fattens the tails and the low-variance regime peaks the center — hence "fat tails" on expiry, justifying wider stops / smaller size. Marks: mixture formula (2), $\sigma^2_{eff}=2.6\sigma_0^2$ & $\sigma_{eff}\approx0.0161$ (2), leptokurtic argument (2).

(d)

import math
def expiry_var(P, sigma0, p, z):
    var_eff = sigma0**2 * (1 + 8*p)   # mixture variance
    sigma_eff = math.sqrt(var_eff)
    return z * sigma_eff * P

For $P=100000,\ \sigma_0=0.01,\ p=0.2,\ z=2.326$: $\sigma_{eff}=0.01\sqrt{2.6}=0.0161245$; VaR $=2.326\times0.0161245\times100000 = ₹3750.55$. Marks: correct mixture var in code (2), returned value $\approx₹3750.6$ (2).

Question 3

(a) Event window $[\tau-w,\tau+w]$. Asymmetry rationale: pre-release volatility is suppressed (traders flatten/hedge, spreads widen but moves are small — "calm before"), while post-release volatility spikes as the surprise is priced in and can persist for many minutes. So the effective volatility window is longer after $\tau$; a symmetric $w$ either wastes the quiet pre-window or truncates the post-move. Practically use $w_{pre}<w_{post}$. Marks: window definition (2), asymmetry pre/post reasoning (2).

(b)

def event_move(bars, tau, w):
    lut = {b["t"]: b for b in bars}          # index by minute
    def nearest(target):                      # graceful missing-bar handling
        if target in lut: return lut[target]
        cands = [b for b in bars if abs(b["t"]-target) <= 2]
        if not cands: return None
        return min(cands, key=lambda b: abs(b["t"]-target))
    lo, hi = nearest(tau - w), nearest(tau + w)
    if lo is None or hi is None:
        return None
    abs_move = abs(hi["price"] - lo["price"])
    window = [b for b in bars if tau - w <= b["t"] <= tau + w]
    vols = [b["vol"] for b in window]
    avg_vol = sum(vols)/len(vols) if vols else 0.0
    return {"abs_move": abs_move, "avg_vol": avg_vol}

Marks: correct endpoint prices & abs move (3), average volume over window (2), missing-bar handling / empty guard (3).

(c) Assumptions: (i) slippage cost per trade $\approx c\cdot s$ where $s$ = half-spread, monotone increasing in local volatility; (ii) in the pre-release minute spreads widen and depth thins, so $s_{pre}>s_{normal}$; (iii) expected fill quality worsens as move magnitude grows. Let day standard deviation be $\sigma_D$. If the event window absolute move satisfies $|P_{\tau+w}-P_{\tau-w}| > N\,\sigma_D,$ then expected adverse slippage of entering just before $\tau$ is $\ge \alpha N\sigma_D$ (partial fill at post-jump prices), while deferring entry to $\tau+w$ caps slippage at the (lower) post-event spread cost $c\,s_{post}$. Since $s_{pre}$ is elevated and jump exposure is avoided, the deferral rule gives $\mathbb{E}[\text{cost}_{avoid}] = c\,s_{post} < \alpha N\sigma_D + c\,s_{pre} = \mathbb{E}[\text{cost}_{enter}],$ which holds whenever $\alpha N\sigma_D > c(s_{post}-s_{pre})$; with $s_{post}\le s_{pre}$ (spreads normalize after the print) the RHS is $\le 0$, so the inequality holds for any $N>0$. Hence avoidance strictly reduces expected slippage. Marks: assumptions (2), inequality construction (2), conclusion under stated condition (2).

[
  {"claim":"Q1a total variance V = a/12 + b via integration of a(t-1/2)^2+b on [0,1]","code":"t,a,b=symbols('t a b'); V=integrate(a*(t-Rational(1,2))**2+b,(t,0,1)); result= simplify(V-(a/12+b))==0"},
  {"claim":"Q1b solving 0.0009=a/12+0.0004 gives a=0.006","code":"a=symbols('a'); sol=solve(Eq(Rational(9,10000),a/12+Rational(4,10000)),a)[0]; result= sol==Rational(6,1000)"},
  {"claim":"Q1c opening range variance fraction f ~ 0.1487","code":"t=symbols('t'); a=Rational(6,1000); b=Rational(4,10000); num=integrate(a*(t-Rational(1,2))**2+b,(t,0,Rational(8,100))); f=num/Rational(9,10000); result= abs(float(f)-0.14869)<0.001"},
  {"claim":"Q2a earnings overnight VaR = 2.326*0.028 = 0.065128","code":"result= abs(2.326*0.028-0.065128)<1e-9"},
  {"claim":"Q2b position ratio normal/earnings equals 3.5","code":"pn=20000/(2.326*0.008); pe=20000/(2.326*0.028); result= abs(pn/pe-3.5)<1e-9"},
  {"claim":"Q2c effective mixture variance = 2.6*sigma0^2 for p=0.2","code":"s0,p=symbols('s0 p'); veff=p*9*s0**2+(1-p)*s0**2; result= simplify(veff.subs(p,Rational(2,10))-Rational(26,10)*s0**2)==0"},
  {"claim":"Q2d expiry VaR ~ 3750.55 for P=1e5,sigma0=0.01,p=0.2,z=2.326","code":"import math; var=(0.01**2)*(1+8*0.2); v=2.326*math.sqrt(var)*100000; result= abs(v-3750.55)<0.5"}
]