Expected Stock Price Under Multiplicative Random Walk

Expectation · Easy · Free problem

A stock starts at price $\

$. Each day, independently:

  • With probability $p$, the stock rises by proportion $\alpha$: the new price is $(1 + \alpha)$ times the previous price.
  • With probability
    - p$, the stock falls by proportion $\beta$: the new price is $(1 - \beta)$ times the previous price.

After $n$ days, what is the expected stock price?

Compute the numerical answer for $n = 10$, $\alpha = 0.05$, $\beta = 0.1$, $p = 1/2$, rounded to the nearest thousandth.

Hints

  1. The stock price after $n$ days is $(1+\alpha)^R(1-\beta)^{n-R}$ where $R \sim \text{Bin}(n, p)$ counts the number of up days.
  2. Factor out $(1-\beta)^n$ and recognize the remaining expectation as $E[f^R]$ where $f = (1+\alpha)/(1-\beta)$. Apply LOTUS and the binomial theorem.
  3. Alternatively, write a one-step recurrence: $E[S_k] = [p(1+\alpha) + (1-p)(1-\beta)] \cdot E[S_{k-1}]$, which gives the answer directly.

Worked Solution

How to Think About It: This is a multiplicative random walk -- the stock price is a product of independent random factors. Each day, the price gets multiplied by either $(1 + \alpha)$ or $(1 - \beta)$. The expected price after $n$ days is NOT the same as the price you would "typically" see (that would be the geometric mean, which is the median in log-space). The expected value is driven by the arithmetic mean of the daily multiplier, raised to the $n$-th power.

Quick Estimate: The expected daily multiplier is $p(1 + \alpha) + (1-p)(1 - \beta) = 0.5 \times 1.05 + 0.5 \times 0.90 = 0.975$. So the expected price after

0$ days is roughly $0.975^{10} \approx 0.776$. The stock drifts down on average because the loss proportion (
0\%$) exceeds the gain proportion ($5\%$).

Approach: Use the law of total expectation, conditioning on the number of up days $R \sim \text{Bin}(n, p)$, and apply the binomial theorem.

Formal Solution:

Let $R$ be the number of days the stock rises. Then $R \sim \text{Bin}(n, p)$, and the stock price after $n$ days is:

$S_n = (1 + \alpha)^R \cdot (1 - \beta)^{n - R}$

The expected price is:

$E[S_n] = E\left[(1 + \alpha)^R (1 - \beta)^{n-R}\right]$

Factor out the non-random part:

$= (1 - \beta)^n \cdot E\left[\left(\frac{1 + \alpha}{1 - \beta}\right)^{\!R}\right]$

Let $f = \frac{1 + \alpha}{1 - \beta}$. By LOTUS:

$E[f^R] = \sum_{k=0}^{n} f^k \binom{n}{k} p^k (1-p)^{n-k} = \sum_{k=0}^{n} \binom{n}{k} (pf)^k (1-p)^{n-k}$

By the binomial theorem:

$E[f^R] = (pf + 1 - p)^n$

Multiplying by $(1-\beta)^n$:

$E[S_n] = (1-\beta)^n \cdot \left(\frac{p(1+\alpha)}{1-\beta} + 1 - p\right)^{\!n} = \left[p(1+\alpha) + (1-p)(1-\beta)\right]^n$

Alternative derivation (recurrence): Let $E[S_k]$ denote the expected price on day $k$. By the law of total expectation over day $k

s move:

$E[S_k] = p(1+\alpha) E[S_{k-1}] + (1-p)(1-\beta) E[S_{k-1}] = \left[p(1+\alpha) + (1-p)(1-\beta)\right] E[S_{k-1}]$

With $E[S_0] = 1$, this gives $E[S_n] = \left[p(1+\alpha) + (1-p)(1-\beta)\right]^n$ immediately.

Alternative derivation (MGF): Recognize $E[f^R] = E[e^{R \ln f}] = M_R(\ln f)$, which is the MGF of $\text{Bin}(n, p)$ evaluated at $\theta = \ln f$. The binomial MGF is $M_R(\theta) = (pe^\theta + 1 - p)^n$, giving the same result.

Numerical answer: With $n = 10$, $\alpha = 0.05$, $\beta = 0.1$, $p = 1/2$:

$E[S_{10}] = \left[\frac{1}{2}(1.05) + \frac{1}{2}(0.90)\right]^{10} = (0.975)^{10}$

$0.975^{10} = e^{10 \ln 0.975} = e^{10 \times (-0.02532)} = e^{-0.2532} \approx 0.776$

Answer: $E[S_n] = \left[p(1+\alpha) + (1-p)(1-\beta)\right]^n$. For the given parameters, $E[S_{10}] = 0.975^{10} \approx 0.776$.

Intuition

This problem highlights a subtle but crucial distinction in finance: the expected value of a multiplicative process is not what a "typical" path looks like. The expected price is governed by the arithmetic mean of the daily multiplier, $p(1+\alpha) + (1-p)(1-\beta)$, raised to the $n$-th power. But the typical (median) path is governed by the geometric mean, $(1+\alpha)^p (1-\beta)^{1-p}$, which is always smaller when the multiplier is random (by AM-GM). In our example, the expected price after 10 days is about $0.776$, but the median path does even worse. This is the essence of volatility drag: even if the arithmetic expected return is slightly negative, the geometric return (what you actually experience on a single path) is even more negative. Every quant who works with leveraged products, option pricing, or portfolio simulation must internalize this distinction -- confusing arithmetic and geometric means is one of the most common and costly mistakes in quantitative finance.

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