# Whether or not $X_1$,…,$X_n$ are independent and exchangeable

For some n = 1,2,..., let $Y_1$,...,$Y_{n+1}$ denote iid real-valued random variables.

Define $X_j$ = $Y_j$$Y_{j+1}, \hspace{10mm}j=1,...,n a) Are X_1,X_2,...,X_n independent? b) Are X_1,X_2,...,X_n exchangeable? Attempts: a) No. Counterexample: Let Y be a Bernoulli (0,1) distribution, with p = 1/2, where Y_1 = 0 and Y_2 = 1. Then X_1 = Y_1$$Y_{2}$

Now $E(Y_1*Y_2)%$ = $E(0*1)$ = $E(0)$ = 1/2

However, $E(Y_1)%$$E(Y_2)% = 1/2*1/2 = 1/4 Since E(Y_1*Y_2)% \neq E(Y_1)%$$E(Y_2)%$, the $X_i$'s are not independent.

b) Not sure.

• Your reasoning on a) is not valid; expressing $X_2$ in terms of $X_1$, $Y_1$, and $Y_3$ does not establish a dependence among $X_1$ and $X_2$ alone. – Dustan Levenstein Jun 20 '14 at 8:08
• Ok. Is the resulting product of two IID variables also IID? – statsguyz Jun 20 '14 at 8:09
• I don't actually know the solution; on gut intuition I suspect they're not exchangable in general, since the way the $Y_i$'s are linked together by the $X_i$'s is not symmetric up to permutation. – Dustan Levenstein Jun 20 '14 at 8:13
• Answers: a) Not in general (but this requires an explicit counterexample). b) Not in general (but this requires an explicit counterexample). Hint: Try Bernoulli {0,1}-valued random variables. – Did Jun 20 '14 at 8:15

Your reasoning on a) is not valid yet the good idea is there. Let's keep your example with the $X_{i}$'s being independant Bernouilli variables with $p=0.5$.

Then, $\mathbb{E} X_{1}X_{2} = \mathbb{E} Y_{1}Y_{2}^{2} Y_{3} = \mathbb{E}Y_{1} \mathbb{E}Y_{2}^{2} \mathbb{E}Y_{3}$ because of the independance of the $Y_{i}$s. Since $\mathbb{E}Y_{2}^{2} = p$, we have $\mathbb{E} X_{1}X_{2} = p^{3}$. On the other side, it is easy to see that $\mathbb{E}X_{i} = p^{2}$.

If $X_{1}$ and $X_{2}$ were independant, we should have $p^{3} = \mathbb{E} X_{1}X_{2} = \mathbb{E} X_{1} \mathbb{E}X_{2} = p^{4}$, which is not true.

What do you mean by "exchangeable variables ?" Here is my definition : $X_{1}, ... , X_{n}$ are exchangeable if, for every measurable function $f : \mathbb{R}^{n} \to \mathbb{R}_{+}$, and for every permutation $\sigma \in \mathfrak{S}_{n}$, we have : $$\mathbb{E}f(X_{\sigma(1)}, ..., X_{\sigma(n)}) = \mathbb{E}f(X_{1}, ..., X_{n}).$$

Let's try with $n=2$. We have $f(X_{1}, X_{2}) = f(Y_{1}Y_{2}, Y_{2}Y_{3}) = f \circ g (Y_{1}, Y_{2}, Y_{3})$, where we defined $g : \mathbb{R}^{3} \to \mathbb{R}^{2}$ by : $$g : (x,y,z) \to (xy,yz)$$

Then, $f \circ g$ is measurable and since $(Y_{1}, Y_{2}, Y_{3})$ is i.i.d., we have : $$\mathbb{E}f f(X_{1}, X_{2}) = \mathbb{E}f \circ g (Y_{1}, Y_{2}, Y_{3}) = \mathbb{E}f \circ g (Y_{3}, Y_{2}, Y_{1}) = \mathbb{E}f (Y_{3}Y_{2}, Y_{2}Y_{1}) = \mathbb{E}f(X_{2}, X_{1})$$

Therefore, $X_{1}$ and $X_{2}$ are exchangeable. I'm not sure the same result holds with $n > 2$, but you should try to adapt the proof (which might reveal tricky).

edit. Okay, here is a counterexample with $n=3$. Let's have $f(x,y,z) = xz$, nd $\sigma = (2,3,1)$. We have : $$\mathbb{E}f(X_{\sigma(1)}, X_{\sigma(2)}, X_{\sigma(3)}) = \mathbb{E}f(X_{2}, X_{3}, X_{1}) = \mathbb{E}Y_{2}Y_{3}Y_{1}Y_{2} = p^{3}$$

On the other side, we have : $$\mathbb{E}f(X_{1}, X_{2}, X_{3}) = \mathbb{E}Y_{1}Y_{2}Y_{3}Y_{4} = p^{4}$$

In conclusion, $X_{1}, X_{2}$ and $X_{3}$ are not exchangeable. The $n=2$ cas was deceptive !