Show that ${X_n}$ is i.i.d and $E(X_1) = 1$ and $c_n$ is bounded real numbers

then $\frac{c_1 + ... + c_n}{n} \rightarrow 1$ if and only if $\frac{c_1X_1 + ... + c_nX_n}{n} \rightarrow 1$.

In the K.L. Chung A course in probability theory(3rd edition), there is hint s.t. truncate Xn at n and proceed as in Theorem 5.4.2(the Strong law of Large numbers).

But by separating integrals with the area of $j-1<|c_nX_n|<j$, since c_n can be 0 and cn is not same, applying the process of the proof of SLLN is not easy. Is there any another way to prove this?


Thank you for your answer. I solved it using another way.

As in the K.L.chung, define Yn = XnI(|xn| <= n), not Yn = anXn* I(|xn| <= n) Then, the infinite sum of Var(anYn/n) converges similar to the proof of SLLN.


It is a consequence of the (usual) law of large numbers, that is, $S_n/n\to\mathbb E\left[X_1\right]$, where $S_n=\sum_{i=1}^nX_i$. Indeed, \begin{align} \frac 1n\sum_{i=1}^nc_iX_i &=\frac{c_n}nS_n+\frac 1n\sum_{j=1}^{n-1}j\left(c_j-c_{j+1}\right)\frac{S_j}j\\ &=\frac{c_n}nS_n+\frac 1n\sum_{j=1}^{n-1}j\left(c_j-c_{j+1}\right)\left(\frac{S_j}j-1\right)+\frac 1n\sum_{j=1}^{n-1}j\left(c_j-c_{j+1}\right)\\ &=\frac{c_n}nS_n+\frac 1n\sum_{j=1}^{n-1}j\left(c_j-c_{j+1}\right)\left(\frac{S_j}j-1\right)+\frac 1n\sum_{j=1}^{n-1}j c_j -\frac 1n\sum_{j=2}^n\left(j-1\right)c_j \\ &=\color{blue}{\frac{c_n}nS_n-\frac{n-1}nc_n } +\frac 1n\sum_{j=1}^nc_j +\frac 1n\sum_{j=1}^{n-1}j\left(c_j-c_{j+1}\right)\left(\frac{S_j}j-1\right). \end{align}
Now, the conclusion follows if we manage to show that

  • the blue term goes to zero and
  • if $\varepsilon_n\to 0$, then $\frac 1n\sum_{j=1}^{n-1}j\left(c_j-c_{j+1}\right)\varepsilon_n\to 0$.
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  • $\begingroup$ Thank you for the answer! For the last term, I think that e_n should be replaced to e_ j. And I think this makes it hard. If it is a e_n for every j in the summation, the problem is quite easy because $\frac{c_1 + ... + c_n }{n}$ converges. But it is e_j, so, showing the last term converges to 0 is little bit hard( I tried $\frac 1n\sum_{j=1}^{n-1}j\left(c_j-c_{j+1}\right)\varepsilon_j$ <= $\frac 1n\sum_{j=1}^{n-1}|j|\left(|c_j-c_{j+1}|\right)|\varepsilon_j|$ ), but the existence of j makes it hard(changing the sum with $e_j-e_{j+1}$ doesn't help). Could you please explain a little bit more?? $\endgroup$ – artes75 Jun 6 '17 at 10:40

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