CONTENTS

Solution: Log-Returns as an Additive Random Walk

Exercise: Log-Returns as an Additive Random Walk

Part 1 — Distribution of RnR_n

Rn=ln⁡Zn={ln⁡u=ln⁡(1.02)≈0.01980w.p. 0.55ln⁡d=−ln⁡u≈−0.01980w.p. 0.45R_n = \ln Z_n = \begin{cases} \ln u = \ln(1.02) \approx 0.01980 & \text{w.p. } 0.55 \\ \ln d = -\ln u \approx -0.01980 & \text{w.p. } 0.45 \end{cases}

Let ℓ=ln⁡u\ell = \ln u for brevity. Then:

E[Rn]=pℓ+q(−ℓ)=(p−q)ℓ=(0.55−0.45)(0.01980)≈0.00198\mathbb{E}[R_n] = p\ell + q(-\ell) = (p - q)\ell = (0.55 - 0.45)(0.01980) \approx 0.00198 E[Rn2]=pℓ2+qℓ2=ℓ2≈3.921×10−4\mathbb{E}[R_n^2] = p\ell^2 + q\ell^2 = \ell^2 \approx 3.921 \times 10^{-4} Var⁡(Rn)=ℓ2−((p−q)ℓ)2=ℓ2(1−(p−q)2)=ℓ2⋅4pq≈3.921×10−4⋅0.99≈3.882×10−4\operatorname{Var}(R_n) = \ell^2 - ((p - q)\ell)^2 = \ell^2(1 - (p-q)^2) = \ell^2 \cdot 4pq \approx 3.921 \times 10^{-4} \cdot 0.99 \approx 3.882 \times 10^{-4}

Part 2 — Additive structure

Starting from Sn=Sn−1Zn=S0∏i=1nZiS_n = S_{n-1} Z_n = S_0 \prod_{i=1}^{n} Z_i and taking logs:

ln⁡Sn=ln⁡S0+∑i=1nln⁡Zi=ln⁡S0+∑i=1nRi\ln S_n = \ln S_0 + \sum_{i=1}^{n} \ln Z_i = \ln S_0 + \sum_{i=1}^{n} R_i
The RiR_i are i.i.d., so ln⁡Sn−ln⁡S0\ln S_n - \ln S_0 is a simple additive random walk in nn steps — exactly the random-walk structure from the lesson, but on the log scale.

Part 3 — One-year moments

Using linearity and independence with n=250n = 250:

E[ln⁡S250]=ln⁡S0+250E[Ri]=ln⁡100+250(0.00198)≈4.6052+0.4950=5.1002\mathbb{E}[\ln S_{250}] = \ln S_0 + 250\mathbb{E}[R_i] = \ln 100 + 250(0.00198) \approx 4.6052 + 0.4950 = 5.1002 Var⁡(ln⁡S250)=250Var⁡(Ri)≈250(3.882×10−4)≈0.0971\operatorname{Var}(\ln S_{250}) = 250\operatorname{Var}(R_i) \approx 250(3.882 \times 10^{-4}) \approx 0.0971 SD⁡(ln⁡S250)≈0.3116\operatorname{SD}(\ln S_{250}) \approx 0.3116

So the median-ish log-price ends the year around 5.10, equivalently e^{5.10} \approx \164$. The annualised log-volatility is about 31%, which is realistic for a high-volatility single stock.

Part 4 — Expected price vs. exponentiated expected log-return

Expected price (using E[Zn]=pu+qd\mathbb{E}[Z_n] = pu + qd):
E[Zn]=0.55(1.02)+0.45(1/1.02)≈0.5610+0.4412=1.00222\mathbb{E}[Z_n] = 0.55(1.02) + 0.45(1/1.02) \approx 0.5610 + 0.4412 = 1.00222 E[Sn]=S0(E[Zn])n=100(1.00222)n\mathbb{E}[S_n] = S_0(\mathbb{E}[Z_n])^n = 100(1.00222)^n

The expected annual growth rate is ln⁡(1.00222)≈0.00221\ln(1.00222) \approx 0.00221 per step, or 0.5530.553 per year, giving \mathbb{E}[S_{250}] \approx 100 e^{0.553} \approx \174$.

Exponentiated expected log-return:
enE[Ri]=e250⋅0.00198=e0.495≈1.641e^{n\mathbb{E}[R_i]} = e^{250 \cdot 0.00198} = e^{0.495} \approx 1.641

So S_0 e^{n\mathbb{E}[R_i]} \approx \164—smallerthan— smaller than\mathbb{E}[S_{250}] \approx $174$.

The gap is \10 \approx 6%ofthestartingprice.Itisdrivenby∗∗Jensen′sinequality∗∗:thefunctionof the starting price. It is driven by **Jensen's inequality**: the functionx \mapsto e^xisconvex,sois convex, so\mathbb{E}[e^{R_i}] > e^{\mathbb{E}[R_i]}.Equivalently,writing. Equivalently, writing R_i = \mu_R + \varepsilon_iwithwith\mathbb{E}[\varepsilon_i] = 0$:

E[eRi]=eμRE[eεi]≥eμR\mathbb{E}[e^{R_i}] = e^{\mu_R}\mathbb{E}[e^{\varepsilon_i}] \ge e^{\mu_R}
with equality only when Var⁡(Ri)=0\operatorname{Var}(R_i) = 0. Financially: the expected price is pulled up by the fact that good outcomes compound geometrically. The median (and the typical realised path) grows at the slower rate eμRe^{\mu_R} per step — which is μ−σ2/2\mu - \sigma^2/2 in the continuous-time analogue, the infamous "Itô correction" that reappears in the Black-Scholes formula.

Takeaways

  • Multiplicative walk in prices = additive walk in log-prices. This is why log-returns, not raw returns, are the canonical modelling quantity.
  • E[price]≠exp⁡(E[log-price])\mathbb{E}[\text{price}] \neq \exp(\mathbb{E}[\text{log-price}]). The gap is Jensen's inequality, and it reappears in continuous time as the −12σ2-\frac{1}{2}\sigma^2 term in the log-return drift of geometric Brownian motion.
  • The typical path grows more slowly than the expected value. Most simulated paths underperform the expected price; a few lucky paths with high returns pull the average up. This asymmetry is why long-term compounding is dominated by the log-return, not the arithmetic return.
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Solution — Log-Returns as an Additive Random Walk | q4quant.studio