Mean And Variance Of A Random Walk

Why Track the Mean and Variance?

A random walk builds a sequence of levels by adding a new random increment at each time step. Even if the increments have the same distribution at every step, the level can become more variable as it accumulates those increments. Calculating the mean and variance at a specified time makes this accumulation precise.

The mean describes the expected level, including any systematic drift in the increments. The variance describes the spread of the level around that mean. These are different features: a walk can have no drift and still have increasing variance.

Definition: A random walk with starting value \(X_0=x_0\) is defined by \(X_t=X_{t-1}+Z_t\) for \(t=1,2,\ldots\), where \(Z_t\) is the increment at time \(t\). In this tutorial, \(x_0\) is fixed, and the increments \(Z_1,Z_2,\ldots\) are independent and identically distributed with finite mean \(\mu=E[Z_t]\) and finite variance \(\sigma^2=\operatorname{Var}(Z_t)\).

The assumption that the increments are identically distributed gives them a common mean and variance. Independence ensures that variation in one increment does not systematically move with variation in another. Both conditions matter: if increment means or variances differ over time, the formulas below must be adjusted.

The Mean and Variance at Time \(t\)

Repeatedly applying the defining equation shows that the level at time \(t\) is the starting value plus the first \(t\) increments:

X_t=x_0+\sum_{j=1}^{t}Z_j.

Linearity of expectation gives the mean. For the variance, adding a fixed constant does not change variance, and the variances of independent random variables add. Thus:

E[X_t]=x_0+t\mu \qquad\text{and}\qquad \operatorname{Var}(X_t)=t\sigma^2.

At time \(t=0\), these expressions give \(E[X_0]=x_0\) and \(\operatorname{Var}(X_0)=0\), as expected for a fixed starting value. If \(\mu=0\), the mean remains at \(x_0\), but the variance still grows whenever \(\sigma^2>0\). If \(\mu\ne0\), the mean changes by \(\mu\) per time step.

Why the Variance Adds

The variance formula follows from expanding the variance of the sum. For any finite collection of random variables, the variance of a sum is the sum of the individual variances plus twice the sum of their pairwise covariances. Here, independence makes every covariance between distinct increments zero.

Write the level as a sum.
Since \(x_0\) is fixed, \(\operatorname{Var}(X_t)=\operatorname{Var}\left(\sum_{j=1}^{t}Z_j\right)\).

Expand the variance.
\(\operatorname{Var}(X_t)=\sum_{j=1}^{t}\operatorname{Var}(Z_j)+2\sum_{1\le j<k\le t}\operatorname{Cov}(Z_j,Z_k)\).

Use the increment assumptions.
Each \(\operatorname{Var}(Z_j)=\sigma^2\). Independence gives \(\operatorname{Cov}(Z_j,Z_k)=0\) for \(j\ne k\), leaving \(t\) terms, each equal to \(\sigma^2\).

Consequently, \(\operatorname{Var}(X_t)=t\sigma^2\). The variance grows linearly with the number of increments. The standard deviation, which is the square root of the variance, is \(\sigma\sqrt{t}\); it does not grow linearly with \(t\).

Worked Example: A Walk With Discrete Increments and Drift

A hypothetical monitoring index starts at \(X_0=100\). At each month, its increment \(Z_t\) is independently drawn from the same distribution: \(-2\) with probability \(0.20\), \(0\) with probability \(0.50\), and \(4\) with probability \(0.30\). Find the mean and variance of the index after four months.

Method. First calculate the increment mean and variance from the stated distribution, then use the random-walk formulas. The fixed starting value is 100, and there are four increments.

Step 1: Calculate the increment mean.

\mu=E[Z_t]=(-2)(0.20)+(0)(0.50)+(4)(0.30) =-0.40+0+1.20=0.80.

Step 2: Calculate the increment variance. The second moment is \(E[Z_t^2]\):

$$ E[Z_t^2]=(-2)^2(0.20)+0^2(0.50)+4^2(0.30) =0.80+0+4.80=5.60. $$

Therefore, \(\sigma^2=E[Z_t^2]-\mu^2=5.60-(0.80)^2=5.60-0.64=4.96\).

Step 3: Calculate the level’s mean and variance.

E[X_4]=100+4(0.80)=103.20, \qquad \operatorname{Var}(X_4)=4(4.96)=19.84.

Check. The mean is 3.20 above the starting value, equal to four times the mean monthly increment. The variance is four times the variance of one increment. The common mistake is to use \(5.60\), the increment’s second moment, as its variance; subtracting the squared mean gives the correct increment variance of \(4.96\).

Worked Example: Normal Increments and the Distribution of the Level

A hypothetical monthly index starts at \(X_0=50\). Its increments are independent with the same normal distribution, \(Z_t\sim N(0.4,1.44)\), where the second parameter is the variance. Find the mean and variance after five months, and state the distribution of \(X_5\).

Method. The increment mean is \(\mu=0.4\), and its variance is \(\sigma^2=1.44\). Apply the mean and variance formulas. Because a sum of independent normal random variables is normal, the resulting level is also normally distributed.

Step 1: Find the mean.

$$ E[X_5]=50+5(0.4)=50+2=52. $$

Step 2: Find the variance.

\operatorname{Var}(X_5)=5(1.44)=7.20.

Step 3: State the distribution. The fixed starting value shifts the sum’s mean but does not change its variance. Thus \(X_5\sim N(52,7.20)\). Its standard deviation is \(\sqrt{7.20}\approx2.683\), rounded to three decimal places.

Check. Five increments contribute \(5(0.4)=2\) to the expected level and \(5(1.44)=7.20\) to its variance. The common mistake is to treat the second normal parameter as a standard deviation; by the stated convention it is the variance, so it is multiplied directly by five.

Worked Example: Finding the Largest Horizon Under a Variance Limit

A hypothetical risk index starts at \(X_0=12\). Its independent, identically distributed increments are \(+3\) and \(-3\), each with probability \(0.50\). What is the largest whole-number horizon for which the variance of the level is no more than 90? Also give the mean and variance at that horizon.

Method. Calculate the increment mean and variance, then solve \(t\sigma^2\le90\) for the largest whole-number \(t\).

Step 1: Calculate the increment moments.

\mu=(3)(0.50)+(-3)(0.50)=0, \qquad \sigma^2=E[Z_t^2]-\mu^2 =9(0.50)+9(0.50)-0^2=9.

Step 2: Apply the variance constraint.

\operatorname{Var}(X_t)=9t\le90 \quad\Longrightarrow\quad t\le10.

Therefore, the largest eligible whole-number horizon is \(t=10\). At that horizon, \(E[X_{10}]=12+10(0)=12\), and \(\operatorname{Var}(X_{10})=10(9)=90\). For confirmation, at \(t=11\) the variance would be \(11(9)=99\), which exceeds the limit.

Check. The zero increment mean leaves the expected level at 12, but does not eliminate its variance. The common mistake is to conclude that a zero-mean walk has zero variance; each increment still has variance 9.

Interpretation and Common Traps

The formulas give a direct way to separate the expected movement from accumulated uncertainty. The mean moves by \(t\mu\) from the starting value, while each independent increment adds \(\sigma^2\) to the level’s variance. A nonzero drift is not required for variance to grow.

Do not square the number of steps. Under independent increments, the variance is \(t\sigma^2\), not \(t^2\sigma^2\). The latter incorrectly treats the same random increment as if it were added \(t\) times.
Do not confuse variance and standard deviation. Variance grows as \(t\sigma^2\); standard deviation grows as \(\sigma\sqrt{t}\).
Use the variance, not the standard deviation, in the formula. If a problem gives the increment standard deviation \(\sigma\), first square it to obtain \(\sigma^2\).
Check the starting-value assumption. These formulas take \(X_0\) to be fixed. A random starting value would require its variability to be included.
Check independence and identical distribution. If increments are dependent, the cross-covariance terms need not vanish. If their means or variances differ over time, multiplying one common mean or variance by \(t\) is not justified.

Key takeaway: For a random walk with fixed starting value \(x_0\) and iid increments of mean \(\mu\) and variance \(\sigma^2\), \(E[X_t]=x_0+t\mu\) and \(\operatorname{Var}(X_t)=t\sigma^2\). The mean reflects drift; the variance reflects the uncertainty accumulated across steps.

Check Your Understanding

For each problem, identify the increment mean and variance before applying the random-walk formulas.

A walk starts at 4. Its iid increments are 0 and 2, each with probability \(0.50\). Find the mean and variance of the level after seven steps. (Answer: mean \(11\); variance \(7\))
A walk starts at 3. Each increment is \(-1\) with probability \(0.25\) and \(1\) with probability \(0.75\), independently over time. Find the mean and variance after four steps. (Answer: mean \(5\); variance \(3\))
A hypothetical measure starts at 20 and has independent normal increments \(Z_t\sim N(-0.2,0.25)\), with the second parameter denoting variance. Find the distribution of the measure after six steps. (Answer: \(N(18.8,1.5)\))
A walk starts at 5. Each iid increment is 0, 2, or 6 with probabilities \(0.50\), \(0.25\), and \(0.25\), respectively. Find the mean and variance after nine steps. (Answer: mean \(23\); variance \(54\))
A walk starts at 0 and has iid increments \(+2\) and \(-2\), each with probability \(0.50\). What is the largest whole-number horizon for which the variance is at most 50? Give the variance at that horizon. (Answer: \(t=12\); variance \(48\))

What You'll Learn