Let $X_1,X_2,\dots$ be independent random variables with $\lim\limits_{n\to\infty}\mathbb E[X_n]=c\in\mathbb R$ and $\sum\limits_{n=1}^\infty \frac1{\sigma_n^2}=\infty$.

I need to show that as $n\to\infty$ $${\sum\limits_{k=1}^n\frac{X_k}{\sigma_k^2}}\big/{\sum\limits_{k=1}^n\frac{1}{\sigma_k^2}}\overset{a.s.}\longrightarrow c.$$

I know that $\mathbb V\left[\frac{X_k}{\sigma_k^2}\right] = \frac{\mathbb V[X_k]}{\sigma_k^4} = \frac{1}{\sigma_k^2}$ from which follows with independence and the condition above that $\sum\limits_{k=1}^\infty\mathbb V\left[\frac{X_k}{\sigma_k^2}\right] = \infty$. How can I continue?

  • $\begingroup$ $n$ versus $k$. $\endgroup$
    – RobPratt
    Feb 1 at 4:14
  • $\begingroup$ You may want to write instead $${\sum\limits_{k=1}^n \frac{X_k}{\sigma_k^2}}\big/{\sum\limits_{k=1}^\infty \frac{1}{\sigma_k^2}}\overset{a.s.}\longrightarrow c.$$if this is the right expression then it is $0$ everywhere. If the lower sum is from $1$ to $n$ then it might give a different result. Which one of the two are you looking at ? $\endgroup$
    – P. Quinton
    Feb 1 at 10:05
  • $\begingroup$ @P.Quinton I hope I have made it more clear. $\endgroup$
    – HelloWorld
    Feb 1 at 12:13
  • $\begingroup$ Much better now. $\endgroup$
    – P. Quinton
    Feb 1 at 12:21
  • 1
    $\begingroup$ Title still needs correction $\endgroup$
    – RobPratt
    Feb 1 at 12:36

2 Answers 2


Observe that the variance goes to $0$ : \begin{align*} \mathbb V\left[ \frac{\sum_{k=1}^n \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^n \frac{1}{\sigma_k^2}} \right] &= \frac{\sum_{k=1}^n \frac{\sigma_k^2}{\sigma_k^4}}{\left(\sum_{k=1}^n \frac{1}{\sigma_k^2}\right)^2}\\ &=\frac{1}{\sum_{k=1}^n \frac{1}{\sigma_k^2}}\\ &\to 0 \end{align*}

and therefore \begin{align*} \frac{\sum_{k=1}^n \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^n \frac{1}{\sigma_k^2}}\overset{a.s.}\longrightarrow& \mathbb E\left[ \frac{\sum_{k=1}^n \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^n \frac{1}{\sigma_k^2}} \right]\\ =&\frac{\sum_{k=1}^n \frac{c}{\sigma_k^2}}{\sum_{k=1}^n \frac{1}{\sigma_k^2}}\\ =&c \end{align*}


By Chebyshev Inequality,

$$P \left(\left|\frac{\sum_{k=1}^n \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^n\frac{1}{\sigma_k^2}} -c\right|^2 > \epsilon^2 \right) \leq \frac{Var\left( \frac{\sum_{k=1}^n \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^n\frac{1}{\sigma_k^2}} \right) }{\epsilon^2}\leq \frac{\left( \frac{\sum_{k=1}^n \frac{1}{\sigma_k^2}}{\left(\sum_{k=1}^n\frac{1}{\sigma_k^2}\right)^2} \right) }{\epsilon^2} \leq \frac{1}{\epsilon^2 \left(\sum_{k=1}^n\frac{1}{\sigma_k^2}\right)}$$

$$\lim_{n \rightarrow \infty} P \left(\left|\frac{\sum_{k=1}^n \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^n\frac{1}{\sigma_k^2}} -c\right|^2 > \epsilon^2 \right) \leq \lim_{n \rightarrow \infty} \frac{1}{\epsilon^2 \left(\sum_{k=1}^n\frac{1}{\sigma_k^2}\right)} = 0$$

If $$\frac{\sum_{k=1}^{\infty} \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^{\infty}\frac{1}{\sigma_k^2}}$$ exists then,

For all, $ n \geq N_{\delta,\epsilon,m}$ with $\lim_{m \rightarrow \infty} N_{\delta,\epsilon,m} = \infty$ such that $$ P \left(\left|\frac{\sum_{k=1}^n \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^n\frac{1}{\sigma_k^2}} -c\right|^2 > \epsilon^2 \right) \leq \frac{\delta}{2^{m}} \implies$$

$$ P \left( \cup_{m=1}^{\infty} \left \{ \left|\frac{\sum_{k=1}^{N_{\delta,\epsilon,m}} \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^{N_{\delta,\epsilon,m}}\frac{1}{\sigma_k^2}} -c\right|^2 > \epsilon^2 \right \} \right) \leq \delta \implies$$

$$ P \left( \left \{ \left|\frac{\sum_{k=1}^{\infty} \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^{\infty}\frac{1}{\sigma_k^2}} -c\right|^2 > \epsilon^2 \right \} \right) \leq \delta \implies$$

Since $\delta$ is arbitrary, $$ P \left( \left \{ \left|\frac{\sum_{k=1}^{\infty} \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^{\infty}\frac{1}{\sigma_k^2}} -c\right|^2 > \epsilon^2 \right \} \right) = 0$$


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