# Showing independence between two functions of a set of random variables

I've been working on the following problem and I'm confused about how to get started:

Let $X_1, X_2,\dots, X_n$ denote i.i.d. real valued random variables, each absolutely continuous with an exponential distribution with parameter $\theta$. Let $a_1,\dots,a_n$ denote real non-zero constants and let $Y = \sum_1^n a_i\log\left(X_i\right)$. Specify the conditions on $\left(a_i\right)_{1\leqslant i \leqslant n}$ under which $Y$ and $\sum_1^n X_i$ are independent.

There is a suggestion to consider the gamma distribution. Based on that I've played around with the Moment Generating functions (mgfs) of $Y$ (which resembles the gamma function) and of $\sum_1^n X_i$, which as the sum of exponentials has a gamma distribution, and my plan was to show that under certain conditions the joint mgf $M(t_1,t_2)$ = $M(t_1,0)*M(0,t_2)$ but I'm not sure how to get a joint distribution of $Y$ and $\sum_1^n X_i$.

Can anyone help me see how this is done? Is this even the right way to go about solving this? I feel like I'm not really making use of the suggestion in more than a trivial way at present, but I'm not sure how to go about using it.

The MGF idea works well.

The scale of the exponential distribution doesn't matter, so we may take $$\theta=1$$ and set the exponential density to

$$f(x) = e^{-x} \mathcal{I}(x \gt 0).$$

Writing

$$Y = \sum_i a_i \log(X_i),\ Z = \sum_i X_i,$$

for $$|s|\lt 1$$ and $$|t|\lt 1$$ compute the joint MGF as

\eqalign{ \phi_{Y,Z}(s,t) &= \mathbb{E}\left[e^{sY + tZ}\right] \\ &= \int\cdots\int \exp(sY+tZ) \prod_i \exp(-x_i)\mathrm{d}x_i \\ &= \prod_i \int_{0}^\infty \exp\left(sa_i \log(x_i) + (t-1) x_i\right) \mathrm{d}x_i\\ &= \prod_i \int_{0}^\infty x_i^{sa_i} e^{(t-1)x_i}\mathrm{d}x_i\\ &= \prod_i (t-1)^{-(sa_i + 1)} \Gamma(sa_i + 1) \\ &= \left((t-1)^{-n}(t-1)^{-s \sum_i a_i}\right)\ \left(\prod_i \Gamma(a_is + 1)\right). }

The term $$(t-1)^{-s\sum_i a_i}$$ shows this factors into separate MGFs $$\phi_Y(s)\phi_Z(t)$$ if and only if $$\sum_i a_i = 0.$$ (When it does factor, we recognize the components $$\phi_Y$$ and $$\phi_Z$$ as the MGFs of a sum of shifted Gumbel distributions and a Gamma distribution, respectively.) Therefore

$$Y$$ and $$Z$$ are independent if and only if the $$a_i$$ sum to zero.

This figure documents a simulation of ten thousand draws of the $$X_i$$ with $$n=3.$$ The four panels are scatterplots of $$Z$$ vs $$Y$$ for four different sets of vectors $$(a_i).$$ The red lines are Lowess smooths: a non-horizontal smooth indicates lack of independence. Can you spot the vector that does not sum to zero?

This R code produced the examples.

#
# Specify the problem, the simulation size, and number of examples to plot.
#
n <- 3
n.sim <- 1e4
n.examples <- 4
#
# Create the examples randomly.
#
a <- matrix(rnorm(n * n.examples), n)
a <- scale(a, scale=FALSE)                              # Will sum to 0
a[, n.examples] <- a[, n.examples] + runif(n, 1/n, 2/n) # Will not sum to 0
#
# Generate the data.
#
x <- matrix(rexp(n*n.sim), n)
#
# Compute Z and various Y.
#
z <- colSums(x)
y <- t(a) %*% log(x)
#
# Show scatterplots and smooths.
#
mfr <- par(mfrow=c(1, n.examples))
invisible(apply(y, 1, function(y) {
plot(y, z, pch=19, col="#00000004")
f <- lowess(z ~ y)
lines(f$$x, f$$y, col="Red", lwd=2)
}))
par(mfrow=mfr)