I have a random variable denoted $N$. Then I have random variables denoted $X_1$, $X_2$, ..., $X_N$ distributed according to a uniform distribution. I have also random variables denoted $Y_1$, $Y_2$, ..., $Y_N$ distributed according to this same probability distribution.

If we consider the variable $S_X$ such that $S_X=\sum_{i=1}^{N}X_i$ and $S_Y$ such that $S_Y=\sum_{i=1}^{N}Y_i$.

Are $S_X$ and $S_Y$ mutually independent ?

If we extend this problem to the case of any number $A$ of sums, for instance $S_X$, $S_Y$ and $S_Z$ when $A=3$. Are these $A$ variables mutually independent ?

Thank you very much.

  • 2
    $\begingroup$ You might like to take a look at the law of total covariance, which in your case is: $\text{cov}(S_X,S_Y)=\text{E}(\text{cov}(S_X,S_Y \mid N))+\text{cov}(\text{E}(S_X\mid N),\text{E}(S_Y\mid N))$. Pay particular attention to the second term, which is a covariance of two functions of $N$. $\endgroup$
    – Glen_b
    Jul 11, 2013 at 5:51

1 Answer 1


Even assuming the $X_i$ and $Y_j$ are independent of $N$ and of one another, $S_X$ and $S_Y$ will be positively correlated and (therefore) not independent. We can calculate that correlation given the distribution of $N$.

To illustrate, let $N$ take on either the value $1$ or $12$ with equal probability. Then half the time $(S_X, S_Y)$ is a point in the unit square (when $N=1$) and half the time it is a point near $(6,6)$ and--according to the Central Limit Theorem--is dispersed randomly around that location in an approximately bivariate Normal manner.


This scatterplot depicts a simulation of 10,000 independent $(S_X, S_Y)$ pairs. The colors distinguish the values of $N$.

The correlation should be obvious in this plot: $S_X$ and $S_Y$ both tend to be small when $N$ is small and large when $N$ is large, whence they tend to be small together or large together: that's positive correlation.

The correlation can be computed using formulas for nested (iterated) expectations. For example,

$$\mathbb{E}_{N;X_i}[S_X] = \mathbb{E}_N[\mathbb{E}_{X_i|N}[S_X | N]] = (1/2 + 12(1/2))/2 = 13/4.$$

In a similar manner all relevant multivariate moments of $(S_X, S_Y)$ can be computed based on knowing that $\mathbb{E}[X_i] = \mathbb{E}[Y_j] = 1/2$ and $\mathbb{E}[X_i^2] = \mathbb{E}[Y_j^2] = 1/3$ (if we assume the $X_i$ and $Y_j$ are all independent). The variances are $56/3 - (13/4)^2 \approx 8.104$ and the covariance is $145/8 - (13/4)^2 = 7.5625$, whence the correlation is $363/389 \approx 0.9332$. Indeed, in this simulation the observed correlation was $0.9316$, apparently differing from this theoretical value only by chance variation.

This answer obviously extends to more than two such sums. It provides a nice example of variables that can be conditionally independent (which will be the case when the $X_i$ are independent of the $Y_j$) but not themselves independent.

Simulation Code

The simulation was carried out in R:

N <- 10^4
m <- 12
n <- ifelse(runif(N) < 1/2, 1, m)
x <- matrix(runif(m*N), ncol=m)
y <- matrix(runif(m*N), ncol=m)

s <- t(sapply(1:N, function(i) c(sum(x[i, 1:n[i]]), sum(y[i, 1:n[i]]))))
col = ifelse(n==1, "Blue", "Red")
plot(s, col=col, pch=19, cex=.5, xlab="X", ylab="Y")
  • $\begingroup$ Thank you very much for this simulation. It's more than I could expect. $\endgroup$
    – Dingo13
    Jul 10, 2013 at 18:24

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