# Proving/disproving that differences $X_2 - X_1, \ldots, X_n - X_{n-1}$ of independent RVs $X_1, \ldots, X_n$ are also independent.

Let $X_1,\ldots,X_n$ be pairwise independent RVs in a Euclidean space. My question is:

Is the set of differences $\{X_{j+1} - X_j \}_{j=1}^{n-1}$ also pairwise independent?

(Edited: by the discussion, it turns out that the proof below is wrong--see the answers)

If they are all Gaussian, then $$X_j, \, X_k \textrm{ independent } \; \Longleftrightarrow \; X_j, X_k \textrm{ uncorrelated } \; \Longleftrightarrow \; \mathbb{E}[X_jX_k] = \mathbb{E} [X_j] \mathbb{E} [X_k],$$ so the above statement can be proven as follows.

Let $\mu_i = \mathbb{E}[X_i]$ for $i = 1,2,\ldots,n$. Then, since $X_{j-1}$, $X_j$, and $X_{j+1}$ are all uncorrelated, \begin{align} \mathbb{E}\big [ \, (X_j - X_{j-1}) \cdot (X_{j+1} - X_j) \, \big ] &= \mu_j \mu_{j+1} - \mu_j^2 - \mu_{j-1}\mu_{j+1} + \mu_{j-1} \mu_j \\ &= (\mu_j - \mu_{j-1}) \cdot (\mu_{j+1} - \mu_j ) \\ &= \mathbb{E}[X_j - X_{j-1}]\cdot \mathbb{E}[X_{j+1} - X_j]. \end{align} Hence, $X_j - X_{j-1}$ and $X_{j+1} - X_j$ are also uncorrelated. Since $X_j$'s are all Gaussian, so are $X_{j+1} - X_{j}$ for all $j=1,2,\ldots,n-1$. Therefore, $\{X_{j+1} - X_j \}_{j=1}^{n-1}$ is independent.

(Edited: The statement below regarding non-Gaussian RVs is also not true--see the answer)

In general non-Gaussian situations, the same proof-line shows that $\{X_{j+1} - X_j \}_{j=1}^{n-1}$ is uncorrelated, but not necessarily independent since independent RVs are always uncorrelated, but not vice versa in general.

I wonder if there is any other specific non-Gaussian distributions that make the statement true? A counter example? Or, is there any general proof of the statement w/o assuming RVs are Gaussian?

Let $X_1, X_2, X_3$ be independent identical random variables uniformly distributed on $[0,1]$. We construct random variables $Y_1=X_2-X_1$, $Y_2=X_3-X_2$.

If I tell you that $Y_1=-1$, then $X_2=0$, so $Y_2\geq 0$, consequently $Y_1$ and $Y_2$ are dependent.

PS: Your derivation is wrong:

\begin{align} \mathbb{E}\big [(X_j - X_{j-1}) \cdot (X_{j+1} - X_j)] &= \mathbb{E}[X_j X_{j+1}-X_j^2-X_{j-1}X_{j+1}+X_{j-1}X_j] \\ &= \mu_j \mu_{j+1}-\mathbb{E}[X_j^2]-\mu_{j-1}\mu_{j+1} + \mu_{j-1} \mu_j \\ &= \mu_j \mu_{j+1}-\mu_j^2-\mathbb{var}[X_j^2]-\mu_{j-1}\mu_{j+1} + \mu_{j-1} \mu_j \\ &\neq \mu_j \mu_{j+1} - \mu_j^2 - \mu_{j-1}\mu_{j+1} + \mu_{j-1} \mu_j \\ &= (\mu_j - \mu_{j-1}) \cdot (\mu_{j+1} - \mu_j ) \\ &= \mathbb{E}[X_j - X_{j-1}]\cdot \mathbb{E}[X_{j+1} - X_j]. \end{align}

so the assumption is wrong for Gaussians too. Actually differences of independent random variables are never independent, except for the case when some middle r.v's are constants.

• I think OP is looking for examples where the differences are independent, rather than a counterexample to the general claim. – angryavian Jun 22 '17 at 4:37
• I'm seeking both examples, but finding an example with independent differences is, I think, harder than finding counter examples. – Jae Young Lee Jun 22 '17 at 4:54
• @kludg In your example, $\mathbb{P}(Y_1 = -1) = 0$, so I think $\mathbb{P}(Y_1 = -1) \cdot \mathbb{P}(Y_2 \in S) = 0$ and also, $\mathbb{P}(Y_1 = -1, Y_2 \in S) = 0$ since it is a set of measure zero. Can you give a little more detail about your dependency of $Y_1$ and $Y_2$? – Jae Young Lee Jun 22 '17 at 4:59
• @JaeYoungLee Consider $Y_1=-1+\delta$, where $\delta$ is small; then $X_2<\delta$, $Y_2\geq -\delta$ – kludg Jun 22 '17 at 5:11
• @kludg Thanks for your comments, and you mean $Y_1 \leq -1 + \delta$ $\Longrightarrow$ $X_2 < \delta$ and $Y_2 \geq - \delta$? – Jae Young Lee Jun 22 '17 at 9:59

$\newcommand{\cov}{\operatorname{cov}}$Suppose $X_1,X_2,X_3$ are independent and have respective variances $1,2,3.$

Then \begin{align} \cov(X_1-X_2,X_2-X_3) & = \cov(X_1,X_2) - \cov(X_2,X_2) - \cov(X_1,X_3) + \cov(X_2,X_3) \\[10pt] & = 0 - 2 - 0 + 0 = -2. \end{align} So they're certainly not independent.

Assume that $X, Y, Z$ are $\mathbb{R}$-valued random variables such that

• $X, Y, Z$ are mutually independent,
• $X-Y$ and $Y-Z$ are independent.

Claim. Under the assumptions above, $Y$ is constant.

Indeed, the above assumption tells that the characteristic functions of $X, Y, Z$ satisfy

\begin{align*} \varphi_X(-s)\varphi_Y(s+t)\varphi_Z(-t) &= \mathbb{E}[e^{-isX}e^{i(s+t)Y}e^{-itZ}] \\ &= \mathbb{E}[e^{is(Y-X)}e^{it(Y-Z)}] \\ &= \varphi_X(-s)\varphi_Y(s)\varphi_Y(t)\varphi(-t). \end{align*}

Since $\varphi_X(-s)$ and $\varphi_Z(-t)$ are non-zero if $s, t$ are sufficiently close to $0$, there exists $\delta > 0$ such that

$$\varphi_Y(s+t) = \varphi_Y(s)\varphi_Y(t) \qquad \forall s, t \in (-\delta, \delta).$$

Although it requires a bit of justification (which I skip here), this tells that $\varphi_Y(s)$ is an exponential function of the form $\varphi_Y(s) = e^{is\alpha}$ for some $\alpha \in \mathbb{R}$. This forces $Y = \alpha$ with probability 1.

• A quick way of completing the last part is to put s=-t which gives abs(\phi_Y)=1 in a neighborhood of 0. – Kavi Rama Murthy Jun 22 '17 at 11:22
• The standard Brownian motion $X_j = B_{t_j}$ ($t_j < t_{j+1}$) gives an example of RVs that have independent differences, but in this case, $X_1$, $\ldots$, $X_n$ themselves are not independent, so your claim also works in this case. – Jae Young Lee Jun 22 '17 at 18:01