1. ## Help with problem

Hello,

I have a doubt regarding the attached question. I've tried to figure out a way for the question which is to use the law of large numbers but there is a problem. LLN is mainly used for I.I.D but the question says only independent variables so i guess i cannot use the LLN...is there any other way to do this question??

Thanks alot!

2. Hello,
Originally Posted by daknight1987
Hello,

I have a doubt regarding the attached question. I've tried to figure out a way for the question which is to use the law of large numbers but there is a problem. LLN is mainly used for I.I.D but the question says only independent variables so i guess i cannot use the LLN...is there any other way to do this question??

Thanks alot!
Haha, I know this problem ^^

Use Chebyshev's inequality for $\mathbb{P}\left(\left|\frac{\sum_{i=1}^n X_i}{n}-\mu\right|>\varepsilon\right)$ !

You'll need to show that $\mathbb{E}\left(\frac{\sum_{i=1}^n X_i}{n}\right)=\mu$, which is very easy by using the linearity of the expectation.
You'll also need to remember that $\text{var}\left(\sum_{i=1}^n X_i\right)=\sum_{i=1}^n \text{var}(X_i)$, because the X_i's are independent.

3. Don't have a cow, Moo, but the rv's don't have the same mean.

$\mu$ is the limit of the average of n means, i.e., $\mu=\lim_{n\to\infty}{\sum_{k=1}^n \mu_k\over n}$ .

4. Yeah sorry, I didn't see that ><

So then write :
$\frac{\sum_{i=1}^n X_i}{n}-\mu=\frac{\sum_{i=1}^n X_i}{n}-\frac{\sum_{i=1}^n \mu_i}{n}+\frac{\sum_{i=1}^n \mu_i}{n}-\mu$

Hence :
$\left|\frac{\sum_{i=1}^n X_i}{n}-\mu\right|\leq \left|\frac{\sum_{i=1}^n X_i}{n}-\frac{\sum_{i=1}^n \mu_i}{n}\right|+\left|\frac{\sum_{i=1}^n \mu_i}{n}-\mu\right|$

Now, use Chebyshev's inequality and the fact that $\frac{\sum_{i=1}^n \mu_i}{n}$ converges (almost surely) to $\mu$

5. since when do you apologize?
is this a first?
and the $\mu_i$'s and $\mu$ are just constants so they converge anyway you wish.

6. Originally Posted by matheagle
since when do you apologize?
is this a first?
and the $\mu_i$'s and $\mu$ are just constants so they converge anyway you wish.
They converge when n goes to infinity. That's what the text says.

I said "almost surely", because it will help dealing with the epsilon's.

As for apologizing : you don't know me enough to be able to tell !