Actuarial Statistics · Jointly distributed random variables

Transformations and Sums of Random Variables: Jacobian, Convolution and MGF Methods

Q: When should I use convolution rather than MGFs?

Use MGFs when the variables are independent and from standard families, because multiplying is quick. Use convolution when the MGF product does not match a known distribution or when the question asks for it explicitly. Convolution also helps when the densities are simple, such as uniforms.

Q: Why do I need the Jacobian in the change of variables method?

Changing variables stretches or squeezes the region, so probability per unit area changes. The absolute value of the Jacobian of the inverse map corrects for this. Without it, the new pdf will not integrate to 1.

Q: How do I find the distribution of the minimum or maximum of random variables?

For independent variables, the maximum has CDF equal to the product of the individual CDFs. The minimum has survival function equal to the product of the individual survival functions. Differentiate the CDF to get the pdf. For n i.i.d. variables, use [F(m)]ⁿ for the maximum and 1 − [1 − F(m)]ⁿ for the minimum.

Q: Does the MGF method work if the variables are dependent?

The product rule M_{X+Y}(t) = M_X(t) M_Y(t) needs independence. For dependent variables, you must use E[e^{t(X+Y)}] with the joint distribution, or use another method such as change of variables.

Updated 11 October 2026 · Fact-checked

To find the distribution of a function of random variables, use one of three routes: the change of variables method with a Jacobian, the convolution formula for sums of independent variables, or the MGF method, where the MGF of an independent sum is the product of the MGFs. For minima and maxima, work with the CDF.

Understand Transformations and Sums of Random Variables

A transformation asks: if X (or X and Y) has a known distribution, what is the distribution of a new variable such as U = X + Y, or V = X², or the larger of X and Y? You need this whenever you model total claims, combined losses or order statistics.

There are three main tools. The CDF method works for any function. You write P(U ≤ u) as an event about the original variables, then compute it by integrating or summing the joint density. Differentiate to get the pdf. This is the safest method when the function is not one-to-one, such as X² or max(X, Y).

The change of variables method works when the transformation is one-to-one (or can be split into one-to-one pieces). For two variables, you define (U, V) = (g1(X, Y), g2(X, Y)), invert to get X and Y in terms of U and V, and multiply the original joint pdf by the absolute value of the Jacobian of the inverse. If you only want U, you add a dummy second variable such as V = X, then integrate V out to get the marginal of U.

For sums of independent variables, you can use convolution: f_{X+Y}(z) = ∫ f_X(x) f_Y(z − x) dx. For discrete variables, replace the integral with a sum. The MGF method is often faster. If X and Y are independent, M_{X+Y}(t) = M_X(t) M_Y(t). If the product matches the MGF of a known distribution, you have identified the distribution, because an MGF that exists in an interval around 0 determines the distribution uniquely.

Always check the support (the range where the density is non-zero). Most lost marks come from wrong limits of integration, not from wrong algebra.

Key rules to remember

Change of variables (one variable): If Y = g(X), g strictly monotonic with inverse x = h(y): f_Y(y) = f_X(h(y)) × |h′(y)|; Apply on the range of y that corresponds to the range of x.
Change of variables (two variables): f_{U,V}(u, v) = f_{X,Y}(x(u,v), y(u,v)) × |J|, where J = ∂(x,y)/∂(u,v) = (∂x/∂u)(∂y/∂v) − (∂x/∂v)(∂y/∂u); J is the Jacobian of the inverse transformation. Take the absolute value. The transformation must be one-to-one on the support.
Convolution (continuous): f_{X+Y}(z) = ∫ f_X(x) f_Y(z − x) dx; Requires X and Y independent. Limits are set by where both densities are non-zero.
Convolution (discrete): P(X + Y = z) = Σ P(X = x) P(Y = z − x); Requires independence. Sum over all x giving valid values.
MGF of a sum: M_{X+Y}(t) = M_X(t) × M_Y(t); Requires independence. Extends to n variables as a product of n MGFs.
MGF of a linear function: M_{aX+b}(t) = e^{bt} M_X(at); Holds for any random variable whose MGF exists.
Maximum of independent variables: F_max(m) = F_X(m) × F_Y(m); for n i.i.d. variables, F_max(m) = [F(m)]ⁿ; Max ≤ m means every variable ≤ m.
Minimum of independent variables: P(min > m) = [1 − F_X(m)] × [1 − F_Y(m)]; for n i.i.d., F_min(m) = 1 − [1 − F(m)]ⁿ; Min > m means every variable > m.
Mean and variance of a sum: E(X + Y) = E(X) + E(Y); Var(X + Y) = Var(X) + Var(Y) + 2Cov(X, Y); The mean rule always holds. The covariance term is zero when X and Y are independent.

How to solve Transformations and Sums of Random Variables questions

Use this method for any question asking for the distribution of a function of one or more random variables.

1Write down the joint (or single) pdf and its support. Sketch the region if there are two variables.
2Choose the method. For a sum of independent variables, try the MGF first, then convolution. For max or min, use the CDF. For a one-to-one transformation, use change of variables. For a non-monotonic function, use the CDF method.
3For change of variables, define a second variable (for example V = X) so the map is one-to-one. Solve for the original variables in terms of the new ones.
4Compute the Jacobian of the inverse map and take its absolute value. Write the new joint pdf.
5Transform the support carefully. Rewrite the inequalities in terms of u and v and draw the new region.
6If you only need the distribution of U, integrate out the dummy variable over its range, which may depend on u. Split into cases if the limits change.
7For MGF or CDF methods, multiply MGFs or compute P(U ≤ u), then identify the distribution or differentiate to get the pdf.
8Check the answer: the pdf must be non-negative and integrate to 1, and the mean should agree with simple rules such as E(X + Y) = E(X) + E(Y).

Quickest way: MGF shortcut for independent sums

When to use it: Use when the variables are independent and come from standard families (normal, Poisson, gamma, binomial, exponential) and the question asks for the distribution of a sum or linear combination.

Write each MGF from the Formulae and Tables book or from memory.
Multiply the MGFs. Combine powers and exponents.
Match the product to a known MGF. Read off the parameters.
Quote the result with its conditions, for example: same p for binomials, same rate for gammas.

Common mistakes in Transformations and Sums of Random Variables

Using convolution or the product of MGFs when the variables are not independent.
The formulas look general and the independence condition is easy to skip.
Fix: State independence explicitly before using the formula. If it is not given, use the joint pdf and the CDF or change of variables method.
Forgetting the absolute value of the Jacobian, or using the Jacobian of the forward map instead of the inverse.
Students remember the word Jacobian but not which way the map goes.
Fix: Write x and y as functions of u and v first, then differentiate. Always write |J|. If you used the forward map, divide by it instead.
Wrong limits of integration in convolution, such as integrating from 0 to ∞ regardless of z.
The condition that z − x must also lie in the support of the second variable is overlooked.
Fix: List the inequalities 0 < x and 0 < z − x. This gives 0 < x < z. For uniform variables, split z into ranges such as 0 < z < 1 and 1 < z < 2.
Writing the pdf of the maximum as f(x) f(y), or the CDF of the minimum as F(x) F(y).
The max and min cases are mixed up.
Fix: Max ≤ m needs all variables ≤ m, so multiply CDFs. Min > m needs all variables > m, so multiply survival functions. Then differentiate.
Assuming the sum of two independent variables from the same family stays in that family without checking the parameters.
Normal and Poisson sums work, so students over-generalise.
Fix: Check the MGF product. Binomials need the same p. Gammas need the same rate. Exponentials with the same rate give a gamma, not an exponential. Lognormals do not add to a lognormal.
Adding standard deviations instead of variances for independent variables.
Variance rules are confused with mean rules.
Fix: Add variances, then take the square root. For aX + bY, use a²Var(X) + b²Var(Y).

Worked examples

Example 1

X and Y are independent Poisson random variables with means 2 and 3. Using MGFs, find the distribution of S = X + Y, and calculate P(S = 2).

Show the solution

The Poisson(λ) MGF is M(t) = exp[λ(eᵗ − 1)].
By independence, M_S(t) = exp[2(eᵗ − 1)] × exp[3(eᵗ − 1)] = exp[5(eᵗ − 1)].
This is the MGF of a Poisson distribution with mean 5, so S ~ Poisson(5) by uniqueness of the MGF.
P(S = 2) = e⁻⁵ × 5² ÷ 2! = 12.5 e⁻⁵.
e⁻⁵ ≈ 0.0067379, so P(S = 2) ≈ 12.5 × 0.0067379 ≈ 0.08422.

Answer: S ~ Poisson(5) and P(S = 2) ≈ 0.0842.

Example 2

X and Y are independent, each with pdf f(x) = 1 for 0 < x < 1. Find the pdf of Z = X + Y, and find P(Z ≤ 1.5).

Show the solution

Convolution: f_Z(z) = ∫ f_X(x) f_Y(z − x) dx, where both densities equal 1 only if 0 < x < 1 and 0 < z − x < 1.
The second condition gives z − 1 < x < z. Combined with 0 < x < 1, the range depends on z.
For 0 < z < 1: x runs from 0 to z, so f_Z(z) = z.
For 1 ≤ z < 2: x runs from z − 1 to 1, so f_Z(z) = 1 − (z − 1) = 2 − z.
Otherwise f_Z(z) = 0. Check: the triangle has base 2 and height 1, so area = 1.
P(Z ≤ 1.5) = 1 − P(Z > 1.5) = 1 − ∫ from 1.5 to 2 of (2 − z) dz.
∫ from 1.5 to 2 of (2 − z) dz = [2z − z²/2] from 1.5 to 2 = (4 − 2) − (3 − 1.125) = 2 − 1.875 = 0.125.
So P(Z ≤ 1.5) = 1 − 0.125 = 0.875.

Answer: f_Z(z) = z for 0 < z < 1, f_Z(z) = 2 − z for 1 ≤ z < 2, and 0 otherwise. P(Z ≤ 1.5) = 0.875.

Exam tips

In written answers, state independence and the support at the start. Examiners award marks for these conditions.
For sums of standard distributions, the MGF method is usually the shortest. Quote the MGFs from the Tables and name the uniqueness property when you identify the result.
For change of variables with two variables, draw the region in the (x, y) plane and the (u, v) plane. Marks are often for the correct new support.
For max and min questions, go through the CDF first and differentiate last. Do not go straight to densities.
In the computer-based paper, you can check an analytic answer by simulation in R. Generate many values, apply the transformation and compare the sample mean and variance with your formulas.

Practice questions from Jointly distributed random variables

Transformations and Sums of 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.

Transformations and Sums of Random Variables: frequently asked questions

When should I use convolution rather than MGFs?

Use MGFs when the variables are independent and from standard families, because multiplying is quick. Use convolution when the MGF product does not match a known distribution or when the question asks for it explicitly. Convolution also helps when the densities are simple, such as uniforms.

Why do I need the Jacobian in the change of variables method?

Changing variables stretches or squeezes the region, so probability per unit area changes. The absolute value of the Jacobian of the inverse map corrects for this. Without it, the new pdf will not integrate to 1.

How do I find the distribution of the minimum or maximum of random variables?

For independent variables, the maximum has CDF equal to the product of the individual CDFs. The minimum has survival function equal to the product of the individual survival functions. Differentiate the CDF to get the pdf. For n i.i.d. variables, use [F(m)]ⁿ for the maximum and 1 − [1 − F(m)]ⁿ for the minimum.

Does the MGF method work if the variables are dependent?

The product rule M_{X+Y}(t) = M_X(t) M_Y(t) needs independence. For dependent variables, you must use E[e^{t(X+Y)}] with the joint distribution, or use another method such as change of variables.