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So I have a homework problem that my study group and I are stuck on.

The problem goes,"Let $X$ be a random variable with mean $\mu$, and variance $\sigma_2$, and we have a sample $\{X_1, X_2 , \ldots , X_n \}$.

Show that $T= \frac{2}{n(n+1)}\sum_iiX_i$ is a consistent estimator for $\mu$.

Our professor gave us three hints, and the first one we are struggling with is to find the mean and variance of $T$.

We are assuming that $T$ is a discrete random variable since it is a summation.

To find $$E[X] = \sum xp(x)$$ $$= xT$$ $$= \sum x\left[\frac{2}{n(n+1)}\sum_iiX_i\right]$$ $$= \sum x\left[\frac{2}{n(n+1)}\cdot\frac{n(n+1)}{2}\sum_iX_i\right]$$ Since the sum of the $i$'s is $\frac{n(n+1)}{2}$ $$= \sum x\left[\sum_iX_i\right]$$ $$= \sum x(\overline{x}n)$$ If we manipulate the summation $\frac{X_i}{n}=\overline{x}$ formula around to make $n\overline{x}=\sum_iX_i$ $$=\overline{x}\sum xn$$

This is pretty much where we get stuck at and we begin to question whether or not we're going down the right bunny trail with this one.

After speaking with some of my other classmates, there is another thought about on how to set up this problem:

$$= \sum iT$$ $$= \sum i\left[\frac{2}{n(n+1)}\sum_iiX_i\right]$$ $$= \sum i\left[\frac{2}{n(n+1)}\cdot\frac{n(n+1)}{2}\cdot\sum_iX_i\right]$$ Since the sum of the $i$'s is $\frac{n(n+1)}{2}$ $$= \sum[iX_i]$$

This pretty much leaves us with $\frac{n(n+1)}{2}\sum_iX_i$, and I could throw in the mean there again, but I doubt that is going to work for this problem.

Could someone check our logic to see if we are thinking about this problem correctly or if we need to change our approach on how we should calculate this mean?


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I just converted your mathematics into LaTeX. Please do this in the future as it makes math readable. If you aren't sure how I did what I did click the edit button and take a look. – Jim Feb 25 '13 at 22:01
Hi Jim, Thanks! I don't know LaTeX at all, but I will take a look at how you edited this later. – Perdue Feb 25 '13 at 22:09

Consider the random variable $W=\sum iX_i$. By the linearity of expectation, $$E(W)=E\left(\sum_1^n i X_i\right)=\sum_1^n i E(X_i)=\left(\sum_1^n i\right)\mu.$$ But the sum $1+2+3+\cdots +n$ is $\frac{n(n+1)}{2}$. Since $T=\frac{n(n+1)}{2}W$, it follows that $E(T)=\mu$.

For the variance, recall that if $U$ and $V$ are independent random variables, then $\text{Var}(aU+bV)=a^2\text{Var}(U)+b^2\text{Var}(V)$.

There is no reason to think of $T$ as discrete. If the distribution of the $X_i$ is continuous, then so is the distribution of $T$.

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Thanks for the help! So when I get to the Variance, should I end up with frac{4}{n(n+1)}(2n+1)\sigma? – Perdue Feb 26 '13 at 0:04
$\frac{2}{3n(n+1)}$$2n+1$sigma? Sorry, I have not a clue about LaTeX – Perdue Feb 26 '13 at 0:06
The variance is $\frac{2^2}{n^2(n+1)^2}(\sum i^2)\sigma^2$. There is a standard formula for $\sum_1^n i^2$, it is $\frac{n(n+1)(2n+1)}{6}$. Multiply and simplify. We get what you probably intended to write. Don't forget it is $\sigma^2$, not $\sigma$. Note how straightforward it all was. – André Nicolas Feb 26 '13 at 0:50
Yes, and I now feel completely and truly dumb to see how easy it all was. :( Hopefully throwing all of this into Chebychev's will not be as hard...Stay tuned... Thanks again! – Perdue Feb 26 '13 at 1:01
I know this may sound silly, but I just take the square root of the variance to get sigma, right? – Perdue Feb 26 '13 at 1:59

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