Variance At A Fixed Time

What Does Variance at a Fixed Time Describe?

A time series records values in time order, but each value can be viewed as the outcome of a random variable. At a specified time $t$, write that random variable as $X_t$, and write its mean as $\mu_t=E[X_t]$. The variance at that time describes how much $X_t$ varies around its own mean across the possible outcomes at time $t$.

This is different from simply looking at how the observed values change as time passes. The mean can change over time, the variance can change over time, or both can change. Even when the mean is constant, individual observations can still vary randomly around it. Conversely, variation in the observed series over several times does not by itself tell us the variance at any one of those times.

Definition: For a random variable $X_t$ with finite second moment, the variance at time $t$ is $$ \operatorname{Var}(X_t)=E\left[(X_t-\mu_t)^2\right], \qquad \mu_t=E[X_t]. $$ It is the expected squared distance from the time-specific mean. Variance is measured in squared units; the standard deviation $\sqrt{\operatorname{Var}(X_t)}$ is measured in the same units as $X_t$.

The condition of a finite second moment means that $E[X_t^2]$ is finite. Without it, the variance need not be a finite number. The definition is about the distribution of $X_t$, not just one observed value. If $x_t$ is the value recorded at time $t$, then $x_t-\mu_t$ is that realization’s deviation from the mean; the deviation alone is not the variance.

Calculating Variance from a Distribution

For a discrete random variable, multiply each squared deviation from the mean by its probability, then add the results. The probabilities must describe the possible outcomes at the particular time being analyzed and must sum to one. This calculation weights likely outcomes more heavily than unlikely ones.

Worked Example: Variance of a Claim Count at One Quarter

Method: calculate the mean first, then average the squared deviations from it. A hypothetical risk model assigns the number of claims $X_3$ in a particular quarter these probabilities: 0 claims with probability 0.5, 2 claims with probability 0.3, and 4 claims with probability 0.2. Find the mean and variance at that quarter.

Check the distribution. The probabilities add to $0.5+0.3+0.2=1.0$, so they form a probability distribution.

Calculate the mean.

\mu_3=E[X_3]=(0)(0.5)+(2)(0.3)+(4)(0.2)=0+0.6+0.8=1.4.

Calculate each squared deviation and weight it by its probability. The deviations from 1.4 are $-1.4$, $0.6$, and $2.6$; their squares are 1.96, 0.36, and 6.76.

\operatorname{Var}(X_3) =(1.96)(0.5)+(0.36)(0.3)+(6.76)(0.2) =0.98+0.108+1.352=2.44.

The variance is 2.44 claims squared. The standard deviation is $\sqrt{2.44}\approx1.562$ claims. As a check, the weighted deviations sum to zero: $(-1.4)(0.5)+(0.6)(0.3)+(2.6)(0.2)=-0.7+0.18+0.52=0$, consistent with deviations being measured from the mean. This check does not replace the variance calculation, which uses squared deviations.

Common mistake avoided: the mean, 1.4, is not the variance. The mean describes the distribution’s center; the weighted squared deviations describe its spread.

An Equivalent Formula and Why It Works

When the mean and second moment are easier to calculate than every squared deviation, use the equivalent formula below. The result follows directly by expanding the square in the definition; it does not require a special distributional assumption, but it does require a finite second moment.

Result: If $E[X_t^2]$ is finite and $\mu_t=E[X_t]$, then $$ \operatorname{Var}(X_t)=E[X_t^2]-\mu_t^2. $$

Start with the definition.
$\operatorname{Var}(X_t)=E[(X_t-\mu_t)^2]$.

Expand the square.
$(X_t-\mu_t)^2=X_t^2-2\mu_tX_t+\mu_t^2$.

Take expectations.
Since $\mu_t$ is a constant, $E[X_t^2-2\mu_tX_t+\mu_t^2]=E[X_t^2]-2\mu_tE[X_t]+\mu_t^2$.

Use $E[X_t]=\mu_t$.
$E[X_t^2]-2\mu_t^2+\mu_t^2=E[X_t^2]-\mu_t^2$, as claimed.

The second moment $E[X_t^2]$ is not itself the variance. The mean-squared term must be subtracted. For the claim-count example, $E[X_3^2]=(0^2)(0.5)+(2^2)(0.3)+(4^2)(0.2)=0+1.2+3.2=4.4$. Therefore $4.4-(1.4)^2=4.4-1.96=2.44$, matching the direct calculation.

Worked Example: Same Mean, Different Variance

Method: calculate the mean and variance separately for each time. A hypothetical monthly index has the following distributions at two specified months. At time 1, $X_1$ is 0 or 2, each with probability 0.5. At time 2, $X_2$ is 0 with probability 0.5, 1 with probability 0.25, and 3 with probability 0.25. Compare their means and variances.

Time 1. The mean is $E[X_1]=(0)(0.5)+(2)(0.5)=1$. Its second moment is $E[X_1^2]=(0^2)(0.5)+(2^2)(0.5)=2$. Thus $\operatorname{Var}(X_1)=2-1^2=1$.

Time 2. The mean is $E[X_2]=(0)(0.5)+(1)(0.25)+(3)(0.25)=0+0.25+0.75=1$. Its second moment is $E[X_2^2]=(0^2)(0.5)+(1^2)(0.25)+(3^2)(0.25)=0+0.25+2.25=2.5$. Thus $\operatorname{Var}(X_2)=2.5-1^2=1.5$.

Both time-specific means equal 1, but the variances differ: 1 at time 1 and 1.5 at time 2. The distributions are not equally concentrated around their common mean. The variance at time 2 is greater even though the center has not moved.

Common mistake avoided: equal means do not imply equal variances. A mean describes location, not spread; each time’s distribution must be examined to determine its variance.

Location, Scale, and Units

Adding a constant to a random variable shifts every possible value and its mean by the same amount, so it does not change the squared deviations from the mean. Multiplying by a constant changes each deviation by that constant and each squared deviation by its square. In particular, for constants $a$ and $b$,

\operatorname{Var}(a+bX_t) =E\left[\left(a+bX_t-(a+b\mu_t)\right)^2\right] =b^2E[(X_t-\mu_t)^2] =b^2\operatorname{Var}(X_t).

This identity applies whenever $X_t$ has finite variance. The additive constant $a$ has no effect on variance, while the scale factor $b$ multiplies variance by $b^2$. The square also explains why variance has squared units: if observations are measured in dollars, variance is in dollars squared. Standard deviation returns to dollars.

Worked Example: Rescaling a Time-Specific Measurement

Method: find the transformed distribution, then calculate its mean and variance. A hypothetical sensor reports an error score $Z_5$ that equals $-1$, 0, or 2 with probabilities 0.2, 0.5, and 0.3, respectively. A reported quantity is $W_5=5-2Z_5$. Find the mean and variance of $W_5$.

Check the probabilities. They sum to $0.2+0.5+0.3=1.0$.

Transform the possible values. If $Z_5=-1$, then $W_5=5-2(-1)=7$. If $Z_5=0$, then $W_5=5$. If $Z_5=2$, then $W_5=5-4=1$. Thus $W_5$ has values 7, 5, and 1 with probabilities 0.2, 0.5, and 0.3.

Calculate the mean. $E[W_5]=(7)(0.2)+(5)(0.5)+(1)(0.3)=1.4+2.5+0.3=4.2$.

Calculate the second moment. $E[W_5^2]=(7^2)(0.2)+(5^2)(0.5)+(1^2)(0.3)=9.8+12.5+0.3=22.6$. Hence $\operatorname{Var}(W_5)=22.6-(4.2)^2=22.6-17.64=4.96$.

As a check using the scale rule, $E[Z_5]=(-1)(0.2)+(0)(0.5)+(2)(0.3)=-0.2+0+0.6=0.4$. Also, $E[Z_5^2]=(1)(0.2)+(0)(0.5)+(4)(0.3)=1.4$, so $\operatorname{Var}(Z_5)=1.4-(0.4)^2=1.24$. Therefore $\operatorname{Var}(5-2Z_5)=(-2)^2(1.24)=4.96$, in agreement. The constant 5 shifts the mean but does not add variance.

Common mistake avoided: multiplying the variance by $-2$ would give the wrong scale. Variance is multiplied by the square of the scale factor, so the multiplier is 4.

Fixed-Time Variability Versus Variation Across Time

A time-indexed model can have a different mean or variance at different times. In notation, $\mu_t=E[X_t]$ and $\operatorname{Var}(X_t)$ are both indexed by $t$. A fixed mean does not force a fixed variance, as the preceding comparison showed. Likewise, observing a different value at each time does not prove that the variance changes: observations can differ simply because a random variable produces different outcomes.

It is useful to keep three quantities distinct:

The mean at time $t$: the expected value $\mu_t$, describing the center of the distribution at that time.
The variance at time $t$: the expected squared deviation of $X_t$ from $\mu_t$, describing the distribution’s spread at that time.
The realized value: the single outcome $x_t$ observed at time $t$, which may lie above or below $\mu_t$.

A single observed value cannot reveal the full distribution of $X_t$, and therefore cannot by itself determine its variance. If one computes a spread from observations collected across different times, that spread can reflect both random deviations from the time-specific means and changes in those means. It is not automatically the variance at a fixed time. Estimating a time-specific variance requires an appropriate source of information about the distribution at that time, such as repeated observations under comparable conditions or a justified model.

The practical distinction matters whenever the degree of uncertainty may change over time. In a hypothetical claim-count process, a stable expected count could coexist with greater variability in some periods. Reporting only the mean would miss that difference. A careful analysis asks separately where the distribution is centered and how widely its outcomes vary around that center.

Key takeaway: At a fixed time, variance is the expected squared distance of that time’s random variable from its own mean. Calculate it from the distribution, or use $E[X_t^2]-\{E[X_t]\}^2$. Do not confuse it with the mean, a single observation’s deviation, or spread across observations from different times.

Check Your Understanding

For each problem, identify the time-specific distribution or transformation before calculating. Answers are given at the end of each question.

At time $t=6$, a hypothetical count $X_6$ is 0 with probability 0.6, 1 with probability 0.2, and 3 with probability 0.2. Find $E[X_6]$ and $\operatorname{Var}(X_6)$. (Answer: $E[X_6]=0.8$, $\operatorname{Var}(X_6)=1.36$.)
At time 1, $X_1$ is equally likely to be 4 or 8. At time 2, $X_2$ is equally likely to be 2 or 10. Calculate each mean and variance, then state whether the means or variances differ. (Answer: both means are 6; $\operatorname{Var}(X_1)=4$ and $\operatorname{Var}(X_2)=16$, so the means match but the variances differ.)
At a fixed time, $Z$ equals $-1$, 0, or 2 with probabilities 0.2, 0.5, and 0.3. Let $Y=5-2Z$. Find $E[Y]$ and $\operatorname{Var}(Y)$. (Answer: $E[Y]=4.2$, $\operatorname{Var}(Y)=4.96$.)
At time 4, a hypothetical measurement $X_4$ equals 1, 3, or 5 with probabilities 0.25, 0.5, and 0.25. If the realized value is 5, calculate the distribution’s mean and variance, and distinguish the realized value’s deviation from the variance. (Answer: $E[X_4]=3$, $\operatorname{Var}(X_4)=2$; the realized deviation is $5-3=2$, not the variance.)
A hypothetical series has mean 10 at each of two times. Its first-time distribution is 8 or 12 with equal probabilities; its second-time distribution is 6 or 14 with equal probabilities. Calculate both variances and explain what the comparison establishes. (Answer: the first variance is 4 and the second is 16; equal means do not imply equal variance.)

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