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Actuarial Statistics · Central limit theorem

Sums and Means of Independent Random Variables

Updated 11 October 2026 · Fact-checked

For a sum S of random variables, E[S] is the sum of the means. If the variables are independent, Var(S) is the sum of the variances and the MGF of S is the product of the MGFs. For the mean of n independent identical variables, E = μ and Var = σ² ÷ n.

Understand Sums and Means of Independent Random Variables

A sum of random variables is itself a random variable. In actuarial work, total claims in a year is the sum of individual claims. The average claim is the sum divided by the number of claims. You want to know the mean, spread and distribution of these quantities.

The mean is easy. Expectation is linear, so E[X + Y] = E[X] + E[Y] always. This holds whether or not the variables are independent or identically distributed.

The variance needs independence. In general, Var(X + Y) = Var(X) + Var(Y) + 2Cov(X, Y). When X and Y are independent, the covariance is zero, so variances simply add. Note that variances add even for a difference: Var(X − Y) = Var(X) + Var(Y) when independent. Spread never cancels.

The moment generating function gives the whole distribution. If X and Y are independent, M_{X+Y}(t) = M_X(t) × M_Y(t), because E[e^{t(X+Y)}] = E[e^{tX}] × E[e^{tY}]. You then recognise the product as the MGF of a known distribution. This is how you show that a sum of independent normals is normal, or a sum of independent Poissons is Poisson.

For the sample mean X̄ = (X₁ + … + Xₙ) ÷ n of independent, identically distributed variables with mean μ and variance σ², E[X̄] = μ and Var(X̄) = σ² ÷ n. The spread shrinks as n grows. This is the idea the central limit theorem builds on.

Key rules to remember

Expectation of a sum
E[X₁ + … + Xₙ] = E[X₁] + … + E[Xₙ]
Always true. No independence needed.
Linear combination
E[aX + bY] = aE[X] + bE[Y]; Var(aX + bY) = a²Var(X) + b²Var(Y) + 2ab Cov(X, Y)
The covariance term is zero if X and Y are independent (or uncorrelated).
Variance of a sum (independent)
Var(X₁ + … + Xₙ) = Var(X₁) + … + Var(Xₙ)
Needs independence, or at least zero covariances between all pairs.
Sum of n iid variables
E[S] = nμ; Var(S) = nσ²; SD(S) = σ√n
Mean and variance of each variable are μ and σ².
Sample mean
E[X̄] = μ; Var(X̄) = σ² ÷ n; SD(X̄) = σ ÷ √n
SD of X̄ is the standard error. Needs independent, identically distributed variables.
MGF of a sum
M_{X+Y}(t) = M_X(t) × M_Y(t)
For independent X and Y. For n iid variables, M_S(t) = [M_X(t)]ⁿ.
MGF of a linear function
M_{aX+b}(t) = e^{bt} × M_X(at)
Use this for the sample mean: M_X̄(t) = [M_X(t ÷ n)]ⁿ.

How to solve Sums and Means of Independent Random Variables questions

Use this method for any question on the distribution, mean or variance of a sum or average.

  1. 1Write the quantity you need in terms of the individual variables, for example S = X₁ + … + Xₙ or X̄ = S ÷ n.
  2. 2Check what is given: are the variables independent, and are they identically distributed? List each mean and variance.
  3. 3Find the expectation first using linearity. Do not worry about independence here.
  4. 4Find the variance. If independent, add the variances, remembering to square any constants. If not, include the covariance terms.
  5. 5If the full distribution is asked, write the MGF of one variable, then multiply the MGFs (or raise to the power n for iid).
  6. 6Compare the resulting MGF with the standard MGFs in the Tables to name the distribution and its parameters.
  7. 7State the final answer with the parameters, and check that the mean and variance agree with steps 3 and 4.

Quickest way: Mean and variance shortcut for sums and averages

When to use it: Use it when you only need the mean, variance or standard deviation of a total or an average of independent variables, for example in a multiple-choice question.

  1. Total of n iid variables: mean nμ, variance nσ².
  2. Average of n iid variables: mean μ, variance σ² ÷ n.
  3. For aX + bY, multiply each variance by the square of its constant, then add (if independent).
  4. Take the square root only at the end if a standard deviation is asked.
  5. For the distribution, match the MGF product to a known family: normal, Poisson, gamma (same rate) and binomial (same p) are closed under independent sums.

Common mistakes in Sums and Means of Independent Random Variables

  • Adding variances for dependent variables without checking covariance.

    The rule 'variances add' is memorised without its condition.

    Fix: Always state independence (or zero covariance) before adding. Otherwise include 2Cov terms.

  • Writing Var(X − Y) = Var(X) − Var(Y).

    Students copy the sign from the expectation.

    Fix: Var(X − Y) = Var(X) + Var(Y) for independent variables. Variances never subtract.

  • Forgetting to square the constant: Var(3X) = 3Var(X).

    Treating variance like expectation.

    Fix: Var(aX) = a²Var(X). So Var(3X) = 9Var(X).

  • Confusing nX with the sum of n independent copies of X.

    Both have mean nμ, so they look the same.

    Fix: Var(nX) = n²σ², but Var(X₁ + … + Xₙ) = nσ². The first uses one variable scaled up. The second uses n independent variables.

  • Using σ² instead of σ² ÷ n for the variance of the sample mean, or using σ ÷ n for the standard error.

    Mixing up variance and standard deviation.

    Fix: Variance of X̄ is σ² ÷ n. Standard error is σ ÷ √n.

  • Adding MGFs instead of multiplying them.

    Mixing up the sum of variables with the sum of functions.

    Fix: For independent variables, the MGF of the sum is the product of MGFs.

Worked examples

Example 1

Claims X₁, X₂, …, X₂₅ are independent, each with mean ₹8,000 and standard deviation ₹3,000. Find the mean and standard deviation of (a) total claims S and (b) average claim X̄.

Show the solution
  1. (a) E[S] = 25 × 8,000 = 2,00,000.
  2. Var(S) = 25 × 3,000² = 25 × 90,00,000 = 22,50,00,000.
  3. SD(S) = 3,000 × √25 = 3,000 × 5 = 15,000.
  4. (b) E[X̄] = 8,000.
  5. Var(X̄) = 90,00,000 ÷ 25 = 3,60,000.
  6. SD(X̄) = 3,000 ÷ 5 = 600.

Answer: Total: mean ₹2,00,000, standard deviation ₹15,000. Average: mean ₹8,000, standard deviation ₹600.

Example 2

X ~ Poisson(2) and Y ~ Poisson(3) are independent. The MGF of a Poisson(λ) variable is exp(λ(eᵗ − 1)). Find the distribution of X + Y and the value of Var(X + Y).

Show the solution
  1. M_X(t) = exp(2(eᵗ − 1)) and M_Y(t) = exp(3(eᵗ − 1)).
  2. By independence, M_{X+Y}(t) = M_X(t) × M_Y(t).
  3. Multiply: exp(2(eᵗ − 1) + 3(eᵗ − 1)) = exp(5(eᵗ − 1)).
  4. This is the MGF of a Poisson(5) variable. The MGF determines the distribution uniquely, so X + Y ~ Poisson(5).
  5. Check: Var(X + Y) = Var(X) + Var(Y) = 2 + 3 = 5, which equals the Poisson(5) variance.

Answer: X + Y ~ Poisson(5), and Var(X + Y) = 5.

Exam tips

  • Always write down the independence assumption when you add variances or multiply MGFs. Examiners give marks for stating it.
  • In MCQs, check whether the question asks for variance or standard deviation, and for the sum or the mean. Wrong options are often the right answer to a nearby question.
  • For MGF questions, write the product first, then simplify the exponents. Compare with the Tables to name the distribution.
  • Be careful with differences and scaled variables, such as 2X − 3Y. Square the constants and add the variances.
  • Use the check E[S] = nμ and Var(S) = nσ² on your final answer. It catches slips quickly.

Practice questions from Central limit theorem

Sums and Means of Independent Random Variables in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

Sums and Means of Independent Random Variables: frequently asked questions

What is the variance of the sample mean?

For n independent, identically distributed observations with variance σ², the variance of the sample mean is σ² ÷ n. The standard error is σ ÷ √n. It falls as the sample size grows.

Do I need independence to find the expected value of a sum?

No. The expectation of a sum is always the sum of the expectations. Independence is needed for variances to add, and for MGFs to multiply.

How do I find the MGF of a sum of independent random variables?

Multiply the individual MGFs. If all n variables have the same MGF M(t), the sum has MGF [M(t)]ⁿ. Then match the result to a standard MGF to identify the distribution.

How does this topic link to the central limit theorem?

The central limit theorem says the standardised sum or mean of many independent identical variables is approximately normal. This topic gives you the mean and variance needed to standardise: nμ and nσ² for the sum, μ and σ² ÷ n for the mean.