11
$\begingroup$

Prove or provide a counterexample:

If $X_n$ $\,{\buildrel a.s. \over \rightarrow}\,$ $X$, then $(\prod_{i=1}^{n}X_i)^{1/n}$ $\,{\buildrel a.s. \over \rightarrow}\,$ $X$

My attempt:

FALSE: Suppose $X$ can take on only negative values, and suppose $X_n \equiv X$ $\forall$ $n$

THEN $X_n$ $\,{\buildrel a.s. \over \rightarrow}\,$ $X$, however for even $n$, $(\prod_{i=1}^{n}X_i)^{1/n}$ is not strictly negative. Instead, it alternates negative to posotive and negative. Therefore, $(\prod_{i=1}^{n}X_i)^{1/n}$ does not converge almost surely to $X$.

Is this a reasonable answer?? If not, how can I improve my answer?

$\endgroup$
4
  • 4
    $\begingroup$ $X_i$ has to be strictly positive for this to be meaningful. $\endgroup$
    – user765195
    Apr 3 '15 at 1:51
  • 2
    $\begingroup$ Of course, you need $X_i>0$ a.s. to define $G_n=(\prod_{i=1}^n X_i)^{1/n}$ properly. First prove that $A_n=\sum_{i=1}^n X_n/n$ converges to $X$ a.s. (google "Cesaro mean" in Real Analysis and adapt the argument). Then, consider $L_n=\log G_n$. $\endgroup$
    – Zen
    Apr 3 '15 at 3:08
  • 1
    $\begingroup$ The needed Real Analysis result is this: If $x_n\to L$, then $\sum_{i=1}^n x_i/n\to L$. Proof: for any $\epsilon>0$, there is an $n_0\geq 1$ such that $|x_n-L|<\epsilon/2$, for every $n\geq n_0$. Therefore, $|\sum_{i=1}^n x_i/n - L|\leq \sum_{i=1}^{n_0}|x_i-L|/n+\sum_{i=n_0+1}^{n}|x_i-L|/n<n_0\max_{1\leq i\leq n_0}|x_i-L|/n + \epsilon/2$. Hence, if we pick $n_1>2\,n_0\max_{1\leq i\leq n_0}|x_i-L|/\epsilon$, then $|\sum_{i=1}^n x_i/n - L|<\epsilon$, for every $n\geq n_1$. $\endgroup$
    – Zen
    Apr 3 '15 at 21:31
  • $\begingroup$ The intuition is that you're computing the average with more and more $x_i$'s that are closer and closer to $L$, and they end up dominating the result. $\endgroup$
    – Zen
    Apr 3 '15 at 21:31
3
$\begingroup$

Before proving something of interest, notice that $X_i >0$ almost surely for all $i$ is not a necessary condition for both statements to make sense, which the deterministic sequence $(-1, -1, 1, 1, 1, \dots)$ illustrates.

Moreover, the statement is indeed false in general, as the following deterministic sequence proves: $(0, 1, 1, \dots)$.

Now, suppose $X_i >0$ almost surely for all $i$, then the statement is true by the following argument:

Define $$S_n = \frac{1}{n}\sum_{i=1}^n\log(X_i).$$ By contuity of $x\mapsto \log(x)$, $\log(X_n)\to\log(X)$ almost surely. Thus, $S_n \to\log(X)$ almost surely by a result for Cesaro means also proven in the comments above. Thus, by continuity of $x\mapsto \exp(x)$, $$\left(\prod_{i=1}^nX_i\right)^{1/n}\to X,$$ almost surely.

$\endgroup$
0
$\begingroup$

This claim is false. I give proof by providing a counterexample.

Suppose the random sequence $X_i$ is defined as follows:

\begin{align} Z_i &\sim N(0,1/i), iid, \; \forall i \in \mathbb{N} \\ X_i &= 1_{\{i \neq 1\}} + 1_{\{i \neq 1\}}Z_i, \; i \in \mathbb{N} \end{align}

Clearly, $X_i$ is (1) degenerate and (2) converges almost surely to $X=1$ as $i \longrightarrow \infty$ by Chebyshev's strong law of large numbers. (To see this, rewrite $Z_i = i^{-0.5}Z$ for $Z \sim N(0,1)$.)

However, since $X_1 = 0$, $\Pi_{i=1}^nX_i = 0, \; \forall n \in \mathbb{N}$. Consequently, $(\Pi_{i=1}^nX_i)^{1/n} = 0, \forall n \in \mathbb{N}$, so it will in the limit trivially converge to $0$, that is $lim_{n\longrightarrow \infty}(\Pi_{i=1}^nX_i)^{1/n} = 0$. $\square$

$\endgroup$
5
  • 2
    $\begingroup$ You appear to have forgotten the $1/n$ exponent. $\endgroup$
    – whuber
    Feb 4 '16 at 14:13
  • $\begingroup$ Thanks whuber, I fixed it :) I should really work on reading things more carefully... I also first proved that the statement also doesn't hold for $\Pi_{i=1}^nX_i^{1/i}$ because I didn't read properly. $\endgroup$
    – Jeremias K
    Feb 4 '16 at 18:51
  • $\begingroup$ Thanks. All these calculations seem to obscure a simple idea: if $X$ is nonzero you will not change the limit by changing any finite number of the $X_i$ to zero, but that will make the product zero and you get a contradiction. Fair enough. However, unless we are told otherwise, statements about infinite products should be understood as statements about infinite sums of the logarithms. In particular, the interest in this question focuses on the case where every $X_i$ is almost surely strictly positive. $\endgroup$
    – whuber
    Feb 4 '16 at 19:20
  • $\begingroup$ @whuber that last comment is interesting. Is it indeed the case that limits of products are by convention, or perhaps by definition (?), understood in terms of the logarithms? If so, I would change the wording of my answer above, too. In particular, the last appeal to continuity would be superfluous. $\endgroup$
    – ekvall
    Feb 4 '16 at 19:27
  • $\begingroup$ @Student The reasoning in your answer is fine. In statistical applications it's rare that anyone would be looking at such a limit of geometric means unless they were already thinking in terms of the logarithms. $\endgroup$
    – whuber
    Feb 4 '16 at 19:34

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.