Independent Sums from Disjoint Variable Groups

Why Disjoint Groups Matter

When several random variables are mutually independent, calculations can often be simplified by collecting them into groups. For example, a statistic might add the first few observations and compare that sum with a sum of later observations. If the groups use different independent variables, their sums are independent too. The important condition is that the groups are disjoint: if even one variable is used in both sums, the conclusion need not hold.

The result is not limited to sums. Any measurable function of one group is independent of any measurable function of another disjoint group. Sums are a useful case because they arise in counts, totals, and sample statistics. The proof first establishes independence of the group vectors, then transfers that independence through measurable functions.

From Independent Variables to Independent Groups

Definition: Random variables $X_1,\ldots,X_n$ are mutually independent if, for every choice of Borel sets $B_1,\ldots,B_n$, $$ P(X_1\in B_1,\ldots,X_n\in B_n)=\prod_{i=1}^{n}P(X_i\in B_i). $$ A group of variables indexed by $G\subseteq\{1,\ldots,n\}$ is the random vector $X_G=(X_i:i\in G)$, with its coordinates arranged in a fixed order.

An event specifying that $X_G$ lies in a coordinate rectangle has the form $\{X_i\in B_i\text{ for every }i\in G\}$. Mutual independence gives the product formula when one such rectangle is specified for each group. To obtain the formula for all Borel events involving the group vectors, an extension argument is needed.

Theorem: Suppose $X_1,\ldots,X_n$ are mutually independent, and $G_1,\ldots,G_m$ are nonempty, pairwise disjoint subsets of $\{1,\ldots,n\}$. Then the random vectors $X_{G_1},\ldots,X_{G_m}$ are mutually independent. Consequently, if each $g_j$ is a Borel-measurable function on the space of $X_{G_j}$, then $g_1(X_{G_1}),\ldots,g_m(X_{G_m})$ are mutually independent.

Here, independence of the group vectors means that for all Borel sets $A_j$ in the corresponding vector spaces, $$ P(X_{G_1}\in A_1,\ldots,X_{G_m}\in A_m) =\prod_{j=1}^{m}P(X_{G_j}\in A_j). $$ The functions need only be measurable so that the events defined by their values are measurable.

Proof by Sequential Extension

For each group $G_j$, let $\mathcal C_j$ be the collection of coordinate rectangles in its vector space: sets that restrict each coordinate to a Borel subset of the real line. This is a pi-system, meaning it contains the whole space and is closed under finite intersections. It generates the Borel sets of the finite-dimensional vector space. Mutual independence gives the required product formula whenever $A_j\in\mathcal C_j$ for every $j$.

We extend one group at a time. This order matters: when the lambda-system argument is applied to a group, its whole-space condition must already be known for the other events being held fixed.

Extend the first group.
Fix $A_j\in\mathcal C_j$ for $j=2,\ldots,m$. Let $\mathcal D_1$ be the class of Borel sets $A_1$ for which the product formula holds. It contains $\mathcal C_1$, by mutual independence. For $A_1$ equal to the whole space, the formula also holds: the whole space belongs to $\mathcal C_1$, so this is the same coordinate-rectangle factorization. Thus $\mathcal D_1$ contains the whole space, is closed under complements, and is closed under countable disjoint unions. It is a lambda-system. The pi-lambda theorem gives the formula for every Borel $A_1$.

Extend the second group.
Now fix an arbitrary Borel $A_1$, and fix $A_j\in\mathcal C_j$ for $j=3,\ldots,m$. Let $\mathcal D_2$ be the class of Borel $A_2$ for which the formula holds. It contains $\mathcal C_2$, by the extension from Step 1, which allowed any Borel $A_1$. Its whole-space condition follows from Step 1 applied with $A_2$ equal to the whole space. The same complement and disjoint-union argument makes $\mathcal D_2$ a lambda-system, so the formula extends to all Borel $A_2$.

Continue through the groups.
At stage $k$, fix arbitrary Borel events for groups $1,\ldots,k-1$ and coordinate rectangles for groups $k+1,\ldots,m$. Vary the event for group $k$. The class of events for which the formula holds contains $\mathcal C_k$ by the preceding stage. Its whole-space condition follows from that same preceding stage with group $k$'s event equal to the whole space. It is therefore a lambda-system, and the pi-lambda theorem extends the formula to all Borel events for group $k$. After the final stage, the formula holds for arbitrary Borel $A_1,\ldots,A_m$.

This proves independence of the group vectors. For measurable $g_j$ and Borel sets $D_j$, the event $\{g_j(X_{G_j})\in D_j\}$ is $\{X_{G_j}\in g_j^{-1}(D_j)\}$, where $g_j^{-1}(D_j)$ is Borel. Apply the group-vector product formula to these preimages. The resulting product formula is exactly mutual independence of the transformed variables. $\square$

Sums from Separate Groups

Addition is a Borel-measurable function on every finite-dimensional Euclidean space. Therefore, if $G_1,\ldots,G_m$ are disjoint groups of mutually independent variables, the sums $$ S_j=\sum_{i\in G_j}X_i,\qquad j=1,\ldots,m, $$ are mutually independent. This conclusion concerns independence between the sums; the distribution of each sum is a separate calculation determined by the distributions of the variables in its own group.

Worked Example: Two Bernoulli Group Sums

Let $X_1,X_2,X_3,X_4$ be mutually independent Bernoulli variables with success probability $1/2$. Define $S=X_1+X_2$ and $T=X_3+X_4$. The groups $\{1,2\}$ and $\{3,4\}$ are disjoint, so the theorem implies that $S$ and $T$ are independent.

Each sum has a binomial distribution with two trials. In particular, $$ P(S=0)=\frac14,\quad P(S=1)=\frac12,\quad P(S=2)=\frac14, $$ and $T$ has the same probabilities. For example, the joint probability that $S=1$ and $T=2$ is $$ P(S=1,T=2)=P(S=1)P(T=2)=\frac12\cdot\frac14=\frac18. $$ The same result can be checked from the original variables: $S=1$ has two possible outcomes for $(X_1,X_2)$, each of probability $1/4$, while $T=2$ requires $(X_3,X_4)=(1,1)$, of probability $1/4$. The combined event has probability $2(1/4)(1/4)=1/8$.

Worked Example: Sums of Independent Poisson Variables

Let $X_1,X_2,X_3$ be mutually independent, with $X_1\sim\operatorname{Poisson}(2)$, $X_2\sim\operatorname{Poisson}(3)$, and $X_3\sim\operatorname{Poisson}(1)$. Set $S=X_1+X_2$ and $T=X_3$. The two groups are disjoint, so $S$ and $T$ are independent.

For $k=0,1,2,\ldots$, the distribution of $S$ follows by summing over the possible values of $X_1$: $$ \begin{aligned} P(S=k) &=\sum_{r=0}^{k}P(X_1=r)P(X_2=k-r)\\ &=\sum_{r=0}^{k}e^{-2}\frac{2^r}{r!}\,e^{-3}\frac{3^{k-r}}{(k-r)!}\\ &=e^{-5}\frac{1}{k!}\sum_{r=0}^{k}\binom{k}{r}2^r3^{k-r} =e^{-5}\frac{5^k}{k!}. \end{aligned} $$ Thus $S\sim\operatorname{Poisson}(5)$, while $T\sim\operatorname{Poisson}(1)$. For a specific joint probability, $$ P(S=2,T=1) =\left(e^{-5}\frac{5^2}{2!}\right)(e^{-1}) =\frac{25}{2}e^{-6}. $$ The first factor is $P(S=2)$; the second is $P(T=1)$. The factorization is justified by the disjoint groups, not by the Poisson calculations themselves.

Worked Example: Exponential Variables and a Joint Threshold

Let $X_1,X_2,X_3,X_4$ be independent exponential variables, each with rate $1$, and set $S=X_1+X_2$, $T=X_3+X_4$. Their groups are disjoint, hence $S$ and $T$ are independent.

For $t\geq 0$, the density of $S$ is obtained by convolution: $$ f_S(t)=\int_0^t e^{-x}e^{-(t-x)}\,dx =\int_0^t e^{-t}\,dx =te^{-t}. $$ Therefore, $$ P(S\leq 1)=\int_0^1 te^{-t}\,dt =\left[-(t+1)e^{-t}\right]_{0}^{1} =1-\frac{2}{e}. $$ The same calculation gives $P(T\leq 1)=1-2/e$. Independence now yields $$ P(S\leq 1,\ T\leq 1) =P(S\leq 1)P(T\leq 1) =\left(1-\frac{2}{e}\right)^2. $$ This is a probability of approximately $0.0698$, since $1-2/e\approx0.2642$.

Common Mistake: Reusing a Variable

Disjointness is essential. If $X_1$ and $X_2$ are independent Bernoulli variables with success probability $1/2$, compare $S=X_1+X_2$ with $T=X_2$. The two expressions share $X_2$, so the theorem does not apply. Indeed, $$ P(S=0,T=1)=0, $$ because $S=0$ requires $X_2=0$, whereas $T=1$ requires $X_2=1$. But $P(S=0)=1/4$ and $P(T=1)=1/2$, so their product is $1/8$, not $0$. Thus $S$ and $T$ are not independent.

A related error is to infer independence merely because two sums have zero covariance. Independence does imply zero covariance when the relevant second moments are finite, but the converse is false in general. Here the valid argument is stronger and simpler: identify the underlying groups, check that they are disjoint, and use mutual independence of the original variables.

Key takeaway: Mutually independent variables can be partitioned into disjoint groups; the resulting group vectors are independent, as are measurable functions of those vectors. In particular, sums formed from separate groups are independent. Always check that no variable appears in more than one group.

Check Your Understanding

Answer each question using the group-independence result and the conditions in its proof.

Why is disjointness of the index groups required for the theorem about independent sums?
Three mutually independent variables are split into groups $\{1\}$ and $\{2,3\}$. Name one sum from each group and state why the sums are independent.
If $X_1,\ldots,X_6$ are mutually independent Bernoulli variables with success probability $p$, what are the distribution and mean of $X_1+X_2+X_3$? Why is this sum independent of $X_4+X_5+X_6$?
In the sequential pi-lambda proof, why must the whole-space case be established before using complement closure?
Suppose two sums share one variable. Does the theorem guarantee that they are dependent? Explain what the theorem does and does not conclude.