Centering and Scaling Nonidentical Independent Sums

Why Nonidentical Sums Need Their Own Center and Scale

An aggregate can combine risks, measurements, or payments whose distributions are not alike. One component might have a relatively small expected value and another a larger one; their variances can differ as well. In that setting, standardizing as though every component shared one common mean and variance can misstate both the expected aggregate and its spread.

The appropriate center is the sum of the component means. If the components are independent and have finite variances, the aggregate variance is the sum of their variances. A normal approximation, when supported by the structure of the sum, is then applied using this aggregate center and scale.

Definition: Let \(X_1,\ldots,X_n\) be independent random variables with finite means \(\mu_i=E[X_i]\) and finite variances \(\sigma_i^2=\operatorname{Var}(X_i)\). For their sum \(S_n=\sum_{i=1}^n X_i\), define the total mean and variance by \(M_n=\sum_{i=1}^n\mu_i\) and \(V_n=\sum_{i=1}^n\sigma_i^2\). When \(V_n>0\), its standardized form is \(Z_n=(S_n-M_n)/\sqrt{V_n}\).

The word nonidentical means that the variables need not share a common distribution. Their means, variances, and distributional shapes may differ. Independence is still important: without it, covariances may contribute to the variance of the sum.

Centering and Scaling the Aggregate

For any real threshold \(c\), standardization translates the event \(S_n\leq c\) to a threshold on the standardized variable:

P(S_n\leq c) = P\left(\frac{S_n-M_n}{\sqrt{V_n}} \leq \frac{c-M_n}{\sqrt{V_n}}\right).

If the standardized sum is reasonably approximated by a standard normal random variable, this gives \(P(S_n\leq c)\approx\Phi((c-M_n)/\sqrt{V_n})\), where \(\Phi\) is the standard normal cumulative distribution function. For an interval, standardize each endpoint separately. The approximation is not an exact distributional identity: unlike a sum of independent normal variables, a general sum does not become exactly normal just because its mean and variance are known.

List the component models.
Record each distribution’s parameters, mean, and variance. Do not replace different parameters with a pooled value.

Aggregate the centers and variances.
Add the means to obtain \(M_n\), and, under independence, add the variances to obtain \(V_n\).

Standardize the event.
Subtract \(M_n\) from the relevant threshold and divide by \(\sqrt{V_n}\), not by \(V_n\).

Assess the approximation.
Use a normal probability only when the number and behavior of the summands support a central limit approximation; report it as approximate.

Why the Variance Adds

The aggregate center and scale follow directly from expectation and independence. This short derivation also shows where the independence assumption enters.

E[S_n] =E\left[\sum_{i=1}^n X_i\right] =\sum_{i=1}^n E[X_i] =\sum_{i=1}^n\mu_i.

For the variance, expand the variance of a sum into individual variances and covariances. Independence makes each covariance between distinct summands zero:

\begin{aligned} \operatorname{Var}(S_n) &=\sum_{i=1}^n\operatorname{Var}(X_i) +2\sum_{1\leq i<j\leq n}\operatorname{Cov}(X_i,X_j)\\ &=\sum_{i=1}^n\sigma_i^2. \end{aligned}

Consequently, the standard deviation used for scaling is \(\sqrt{\sum_i\sigma_i^2}\). It is generally neither the sum of the individual standard deviations nor a common standard deviation multiplied by \(\sqrt n\). The latter shortcut requires identical variances; the usual independent-identically-distributed central limit theorem also uses a common distribution.

When a Normal Approximation Is Supported

A central limit approximation for a sequence of independent, nonidentically distributed variables requires more than independence. A useful general condition is the Lindeberg condition. Write \(V_n=\sum_i\sigma_i^2\), and suppose \(V_n\) grows without bound. For every \(\varepsilon>0\), the condition is

\frac{1}{V_n} \sum_{i=1}^n E\left[(X_i-\mu_i)^2 \mathbf{1}\{|X_i-\mu_i|>\varepsilon\sqrt{V_n}\}\right] \longrightarrow 0.

Here \(\mathbf{1}\{A\}\) equals 1 when event \(A\) occurs and 0 otherwise. Informally, the condition prevents a substantial share of the total variance from being carried by unusually large deviations of individual terms. Under the Lindeberg condition, the standardized sum converges in distribution to a standard normal variable.

A commonly used sufficient condition is Lyapunov’s condition: for some \(\delta>0\),

\frac{\sum_{i=1}^n E[|X_i-\mu_i|^{2+\delta}]} {V_n^{1+\delta/2}} \longrightarrow 0.

For exam problems, a formal limit calculation is not always required. A growing collection of independent contributions, with no single component dominating the total variability and with reasonably controlled tails, is the usual setting that supports the approximation. The phrase “no single component dominates” is a useful warning sign, not a stand-alone guarantee: tail behavior and the overall structure still matter. If there are only a few components, one has an exceptionally heavy-tailed component, or most of the variance comes from one term, a normal approximation may be poor. A normal approximation also does not automatically remove discreteness; adjustments for integer-valued sums are a separate issue.

Worked Example: Adding Uniform Components With Different Ranges

A hypothetical aggregate is the sum of 20 independent amounts, each uniformly distributed on \((0,2)\), and 20 other independent amounts, each uniformly distributed on \((0,1)\). Approximate the probability that the total exceeds 35.

Method. Find the mean and variance for each type, add them across the independent components, then standardize the threshold. This avoids treating all 40 amounts as if they had the same range.

For a uniform random variable on \((a,b)\), the mean is \((a+b)/2\) and the variance is \((b-a)^2/12\). Thus, each \((0,2)\) amount has mean 1 and variance \(4/12=1/3\). Each \((0,1)\) amount has mean \(1/2\) and variance \(1/12\).

M=20(1)+20\left(\frac12\right)=30, \qquad V=20\left(\frac13\right)+20\left(\frac1{12}\right) =\frac{80+20}{12} =\frac{25}{3}.

The total standard deviation is \(\sqrt{25/3}=5/\sqrt3\approx2.8868\). Standardize 35:

z=\frac{35-30}{5/\sqrt3}=\sqrt3\approx1.7321.

The normal approximation gives \(P(S>35)\approx1-\Phi(1.7321)\approx0.0416\), to four decimal places. This is an approximation, not an exact tail probability. The calculation uses the two component types separately and avoids the mistake of assigning the same mean or variance to every amount.

Check. The aggregate mean is 30, below the threshold, and the standardized threshold is positive. The resulting upper-tail probability is therefore less than one-half, consistent with the result.

Worked Example: Combining Exponential Components With Different Rates

A hypothetical total waiting time consists of 30 independent exponential times with rate 1 per hour and 30 independent exponential times with rate \(0.5\) per hour. Approximate the probability that the total exceeds 110 hours.

Method. For an exponential variable with rate \(\lambda\), the mean is \(1/\lambda\) and the variance is \(1/\lambda^2\). Apply these formulas separately to each group before adding. This avoids confusing a rate with a mean.

The first group has mean 1 hour and variance 1 hour squared per component. The second has mean \(1/0.5=2\) hours and variance \(1/(0.5)^2=4\) hours squared per component. Therefore,

M=30(1)+30(2)=90\text{ hours}, \qquad V=30(1)+30(4)=150\text{ hours}^2.

The total standard deviation is \(\sqrt{150}\approx12.2474\) hours. At 110 hours,

z=\frac{110-90}{\sqrt{150}} =\frac{20}{12.2474} \approx1.6330.

Thus \(P(S>110)\approx1-\Phi(1.6330)\approx0.0512\), to four decimal places. The approximation is supported by 60 independent components rather than a single component; it is still not exact, and the aggregate is not asserted to have an exact normal distribution.

Check. The threshold is 20 hours above the aggregate mean, or about 1.63 aggregate standard deviations above it. A tail probability near 0.05 is consistent with that standardized position.

Worked Example: A Small Upper Tail and a Standardization Trap

A hypothetical aggregate \(S\) is the sum of 65 independent uniform variables. Twenty have distribution \(U(1-\sqrt3,1+\sqrt3)\), and 45 have distribution \(U(4/9-\sqrt3,4/9+\sqrt3)\). Approximate \(P(S>70)\).

Method. Each variable has interval width \(2\sqrt3\), so its variance is \((2\sqrt3)^2/12=1\). Add the means and variances, then standardize. Keeping the center and scale separate prevents an arithmetic or aggregation error.

The total mean is \(20(1)+45(4/9)=20+20=40\). Since the 65 variables are independent and each has variance 1, the total variance is \(65\), and its standard deviation is \(\sqrt{65}\approx8.0623\). The threshold’s standardized value is

z=\frac{70-40}{\sqrt{65}} =\frac{30}{8.0623} \approx3.7210.

The normal approximation gives \(P(S>70)\approx1-\Phi(3.7210)\approx0.0001\), to four decimal places. This is a very small approximate tail probability; rounding to four decimals hides its additional precision.

Common mistake and correction. A calculation such as \(30/8.0623\approx3.7204\) is an arithmetic error: \(8.0623(3.7204)\approx29.9950\), not 30. The corrected value is \(30/8.0623\approx3.7210\). Although both values lead to the same four-decimal tail probability here, the standardized value itself must still be calculated correctly.

Check. The threshold is 30 above the mean, which is more than three aggregate standard deviations. A very small upper-tail probability is therefore plausible.

Common Errors and Practical Checks

Before using a normal table or calculator, verify that the sum’s center and scale actually correspond to the stated component models. In particular:

Do not assume identical parameters. In the exponential example, the mean is 90, not \(60(1)\), because half the components have mean 2.
Add variances, not standard deviations. For independent terms, the standard deviation of the sum is the square root of the sum of the variances.
Do not scale by \(\sqrt n\) without checking the model. The shortcut \(\sigma\sqrt n\) applies when all independent components have a common variance \(\sigma^2\), not to arbitrary nonidentical terms.
Do not treat independence as a guarantee of normality. A small number of terms or a dominant, extreme component can make the approximation unreliable.
Keep approximation separate from exact distribution. Unless a distributional result establishes exact normality, write an approximate probability.

Key takeaway: For an independent, nonidentical sum, use \(M_n=\sum_i\mu_i\) and \(\sqrt{V_n}=\sqrt{\sum_i\sigma_i^2}\) to center and scale. A normal approximation is justified by the behavior of the collection of summands, not by the standardization calculation alone.

Check Your Understanding

For probability questions, give answers to four decimal places. Treat any requested normal probability as an approximation.

A sum contains 12 independent \(U(0,2)\) variables and 12 independent \(U(0,1)\) variables. Find its mean and variance, then approximate the probability that the sum exceeds 21. (Answer: mean \(18\), variance \(5\), probability \(0.0899\))
A total is the sum of 20 independent exponential variables with rate 1 and 10 independent exponential variables with rate \(0.5\), with rates measured per hour. Approximate the probability that the total exceeds 50 hours. (Answer: mean \(40\) hours, variance \(60\) hours squared, probability \(0.0984\))
Ten independent variables are each uniform on \((-\sqrt3,\sqrt3)\), and ten more are each uniform on \((-2\sqrt3,2\sqrt3)\). Approximate the probability that their sum is between \(-5\) and 10. (Answer: mean \(0\), variance \(50\), probability \(0.6816\))
A sum consists of 10 independent \(U(0,1)\) variables and 10 independent \(U(0,3)\) variables. Approximate the probability that the total exceeds 25. (Answer: mean \(20\), variance \(25/3\), probability \(0.0416\))
Suppose a sequence of independent sums has total variance \(V_n\), and one centered component contributes a fixed positive fraction of \(V_n\) as \(n\) increases. Is the “no dominant component” rationale for a normal approximation satisfied? (Answer: No; that component remains dominant, so independence alone does not support the approximation.)

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