suppose that for a sequence of reals $(x_t)_{t\in\mathbb{N}}$ it holds that $\frac{1}{T}\sum_{t=1}^T x_t \rightarrow 0$, for $T\rightarrow \infty$. How do I show (sorry, this might be an embarrassing question) that $\frac{1}{T^2}\sum_{t=1}^T t x_t \rightarrow 0$, for $T\rightarrow \infty$? Many thanks for any help, it is much appreciated!
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I think just use $t \leqslant T$ – uncookedfalcon Sep 07 '13 at 22:28
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@uncookedfalcon The $x_i$ are not assumed to be nonnegative. – Andrés E. Caicedo Sep 07 '13 at 22:29
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ah great point! – uncookedfalcon Sep 07 '13 at 22:29
2 Answers
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Does not seem like an embarrassing question to me.
Hint $$\frac 1 {T^2} \sum_{t=1}^T t x_t = \operatorname{Average}\left\{\frac 1 T \sum_{t=1}^T x_t, \frac 1 T \sum_{t=2}^T x_t, \ldots, \frac 1 T \sum_{t=T}^T x_t \right\}$$
Yoni Rozenshein
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On top of Yoni's hint, seems still a bit tricky, I was toying around with estimates of the form
$\left| \frac{1}{T} \sum_{t=k}^T x_t \right| - \left| \frac{1}{T} \sum_{t=1}^{k-1} x_t \right| \leq \left| \frac{1}{T} \sum_{t=1}^T x_t \right| < \epsilon $
and using two estimates:
$\left| \frac{1}{T} \sum_{t=1}^{k} x_t \right| \leq \begin{cases} \epsilon k/T & k>k_0 \\ kM/T & k \leq k_0 \end{cases}$
where $M$ is a bound on the $|x_t|$. Here $k_0$ is the threshold after which for $T > k_0$, $\left| \frac{1}{T} \sum_{t=1}^T x_t \right| \leq \epsilon$
Then if $T$ is sufficiently large, I think everything in Yoni's average can be brought down below a target $\epsilon$ (how large should $T$ be? $T$ for which $k_0 M/T < \epsilon$)
Evan
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