Show independence of $X_{\text{bar}}$ and $\frac{\sum_{i=1}^nX_i^2}{X^2_{\text{bar}}}$ for i.i.d exponential variables

Question

Let $X_1, \ldots , X_n$ be i.i.d distributed from Exp($\lambda$). I need to show that

$$X_{bar} = \frac{1}{n}\sum_{i=1}^n X_i$$ and $$\frac{\sum_{i=1}^nX_i^2}{X_{bar}^2}$$

are independent. How do I manage to show this? I know have to find the joint distribution of some form and compute the marginal densities to show that it is independent when multiplying marginal densities is the joint distribution.

Answer 1

Write

$$ T = \sum_{j=1}^{n} X_j \qquad\text{and}\qquad V_k = \frac{\sum_{j=1}^{k} X_j}{T} \quad\text{for}\quad k = 1, \dots, n-1. $$

Then the substitution of the form $(x_1, \dots, x_{n-1}, x_n) = \Phi(t, v_1, \dots, v_{n-1})$ given by

\begin{gather*} x_1 = v_1 t, \quad x_2 = (v_2 - v_1) t, \quad \dots, \quad x_{n-1} = (v_{n-1} - v_{n-2}) t, \quad x_n = (1 - v_{n-1}) t, \end{gather*}

satisfies

\begin{align*} \operatorname{Jac}(\Phi) &= \det \left( \begin{array}{ccccc} v_1 & t & & & \\ v_2-v_1 & -t & t & & \\ \vdots & & \ddots & \ddots & \\ v_{n-1} - v_{n-2} & & & -t & t \\ 1-v_{v-1} & & & & -t \end{array} \right) \\ &= \det \left( \begin{array}{c|ccc} v_1 & t & & \\ \vdots & & \ddots & \\ v_{n-1} & & & t \\ \hline 1 & 0 & \cdots & 0 \end{array} \right) = (-t)^{n-1}. \end{align*}

Also, if we define $\mathcal{D}$

$$ \mathcal{D} = \{ (v_1, \dots, v_{n-1} ) : 0 \leq v_1 \leq v_2 \leq \dots \leq v_{n-1} \leq 1 \}, $$

then the support of the distribution of $(T, V_1, \dots, V_{n-1})$ is $[0, \infty) \times \mathcal{D}$. So, for any $a_1, \dots, a_{n-1}, b > 0$,

\begin{align*} &\mathbb{P} \left( \cap_{k=1}^{n-1} \{ V_k \leq a_k \} \cap \{ T \leq b \} \right) \\ &= \int_{[0,\infty)^n} \mathrm{d}x_1 \dots \mathrm{d}x_n \, \lambda^n e^{-\lambda(x_1+\dots+x_n)} \left( \prod_{k=1}^{n-1} \mathbf{1}_{\{v_k \leq a_k \}} \right) \mathbf{1}_{\{ t \leq b \}} \\ &= \int_{[0,\infty)\times\mathcal{D}} \mathrm{d}t \, \mathrm{d}v_1 \dots \mathrm{d}v_{n-1} \, t^{n-1} \lambda^n e^{-\lambda t} \left( \prod_{k=1}^{n-1} \mathbf{1}_{\{v_k \leq a_k \}} \right) \mathbf{1}_{\{ t \leq b \}} \\ &= \left( \int_{0}^{b} \mathrm{d}t \, \frac{t^{n-1} \lambda^n e^{-\lambda t}}{(n-1)!} \right) \left( \int_{\mathcal{D}} \mathrm{d}v_1 \dots \mathrm{d}v_{n-1} \, (n-1)! \left( \prod_{k=1}^{n-1} \mathbf{1}_{\{v_k \leq a_k \}} \right) \right). \end{align*}

This proves that the following two observations hold:

1. $(V_1, \dots, V_{n-1})$ and $T$ are independent, and

2. $(V_1, V_2, \dots, V_{n-1})$ has the same distribution as the order statics of $(n-1)$ i.i.d. samples from the uniform distribution on $[0, 1]$.

Then the desired claim follows from the first observation by noting that

$$ x_{\text{bar}} = \frac{T}{n} \qquad\text{and}\qquad \frac{\sum_{j=1}^{n}X_j^2}{X_{\text{bar}}^2} = n^2 \sum_{j=1}^{n} (V_j - V_{j-1})^2, $$

where we interpret $V_0 = 0$ and $V_n = 1$.

---

Remark. The above observations can also be proved by using Poisson point processes, and then they admit a nice interpretation in terms of Poisson points.

Answer 2

There is a easy way of showing independence

$$\bar{X}=\frac{1}{n}\sum_{i=1}^n X_i$$ is a sufficient statistics (also minimal) for parameter $\lambda$ by Fisher–Neyman factorization theorem. And $\bar{X}$ also form complete (boundedly) family of statistics. (i.e. carry whole information about parameter $\lambda$).

On the other hand $$\frac{\sum_{i=1}^n X_i^2}{\bar{X}^2}=\frac{\sum_{i=1}^n \frac{X_i^2}{\lambda^2}}{\frac{\bar{X}^2}{\lambda^2}}$$ form a ancillary statistics (i.e. does not carry any information about parameter $\lambda$)

then by Basu's Theorem, two proposed statistics is independent.

Answer 3

You can solve the problem using Basu's theorem.

• First note that $\overline{X}$ is Complete and Sufficient Statistic for $\lambda$

• Second, note that the exponential density belongs to a "scale family"

Thus, as $Z=\frac{\sum_iX_i^2}{(\overline{X})^2}$ is scale invariant, for a well known theorem, $Z$ is ancillary for $\lambda$

• Finally, you can apply Basu's theorem which states that any CSS is independent of any Ancillary statistics