2
$\begingroup$

Consider $X_1, ..., X_n,...$ independent random variables that satisfy: $E[X_i]=1+\frac{1}{1+i^2}$ and $Var[X_i]=\sqrt{i}$. Show that the chance of $\frac{1}{n}\sum\limits_{i=1}^n X_i$ converges to $1$ as $n \rightarrow \infty$.

My question is whether I can apply the strong law of large numbers, as far as I understand that law it requires $E[X_i]=\mu \neq \mu_i$ Idem for $Var[X_i]$. Could anyone give a hint here?

$\endgroup$
  • 1
    $\begingroup$ You might find this useful: math.stackexchange.com/q/418537/35983. Also start with $P[|\frac{\sum_{k=1}^nX_k}{n} -1| > \epsilon]$ and use Chebyshev's Inequality. Let us know what happens. As for LLN, it cannot be directly applied as the variables are not identically distributed but that doesn't mean the result can't be proved... $\endgroup$ – Gautam Shenoy Jun 15 '15 at 10:24
  • 1
    $\begingroup$ Hint: Show this in the more general setting where $(X_i)$ are independent, $E(X_i)=1+u_i$ and Var$(X_i)=v_i$ with $\sum\limits_{i=1}^nu_i\ll n$ and $\sum\limits_{i=1}^nv_i\ll n^2$. $\endgroup$ – Did Jun 15 '15 at 10:33
  • $\begingroup$ Thanks for your answers. After applying the chebyshev's inequality, I get: $P[|\frac{\sum\limits_{k=1}^n X_k}{n}-1|]|>\epsilon]\leq \frac{\sigma^2}{\epsilon^2}$. Let $\epsilon=k\sigma$, then $P[|\frac{\sum\limits_{k=1}^n X_k}{n}-1|]|>k\sigma]\leq \frac{1}{k^2}$. However, I am stuck here. Could you give me another hint? $\endgroup$ – Holograph Jun 15 '15 at 11:02
  • $\begingroup$ Since only for $k=1$ the statement would hold. $\endgroup$ – Holograph Jun 15 '15 at 11:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.