3
$\begingroup$

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 \begin{equation} 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], \end{equation} 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?

Many thanks in advance for your discussion and comments.

$\endgroup$
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I think OP is looking for examples where the differences are independent, rather than a counterexample to the general claim. $\endgroup$ – angryavian Jun 22 '17 at 4:37
  • $\begingroup$ I'm seeking both examples, but finding an example with independent differences is, I think, harder than finding counter examples. $\endgroup$ – Jae Young Lee Jun 22 '17 at 4:54
  • $\begingroup$ @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 $Y_2$? $\endgroup$ – Jae Young Lee Jun 22 '17 at 4:59
  • $\begingroup$ @JaeYoungLee Consider $Y_1=-1+\delta$, where $\delta$ is small; then $X_2<\delta$, $Y_2\geq -\delta$ $\endgroup$ – kludg Jun 22 '17 at 5:11
  • $\begingroup$ @kludg Thanks for your comments, and you mean $Y_1 \leq -1 + \delta$ $\Longrightarrow$ $X_2 < \delta$ and $Y_2 \geq - \delta$? $\endgroup$ – Jae Young Lee Jun 22 '17 at 9:59
1
$\begingroup$

$\newcommand{\cov}{\operatorname{cov}}$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.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ A quick way of completing the last part is to put s=-t which gives abs(\phi_Y)=1 in a neighborhood of 0. $\endgroup$ – Kavi Rama Murthy Jun 22 '17 at 11:22
  • $\begingroup$ 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. $\endgroup$ – Jae Young Lee Jun 22 '17 at 18:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.