Algebraic manipulation question coming from Method of Moments application

Question

So today in my statistical inference class the professor wrote on the board:

Using the Method of Moments:

$$\sigma^2 = E[Y_1^2|\theta] - E[Y_1|\theta]^2$$ $$= \frac{1}{n}\sum_{j = 1}^{n}{Y_j^2} - \left( \frac{1}{n}\sum_{j = 1}^{n}{Y_j}\right)^2$$ $$= \frac{1}{n}\sum_{j = 1}^{n}{\left( Y_j - \bar{Y}\right)^2 }$$

where $\bar{Y} = \frac{1}{n}\sum_{j = 1}^{n}Y_j$(average values of Y's)

I am confused at how he arrived at the last equation from the one above that and I've been trying to figure it out.

Here's what I have so far:

$$= \frac{1}{n}\sum_{j = 1}^{n}{Y_j^2} - \left( \frac{1}{n}\sum_{j = 1}^{n}{Y_j}\right)^2$$ $$ = \frac{1}{n}\sum_{j = 1}^{n}{Y_j^2} - \frac{1}{n^2}\left(n\bar Y\right)^2$$ $$ = \frac{1}{n}\sum_{j = 1}^{n}{Y_j^2} - \bar Y^2$$

Which is fundamentally different that the equation that my professor had on the board. My equation takes the difference of squares while his takes the square of a difference and then sums those up...

Where did I go so wrong? Is there just some statistical insight that I don't know and haven't used?

Answer 1

Perhaps it's easier to go the other way around. Let $S^2$ be the sample variance and note

\begin{align*} S^2 &= \frac{1}{n}\sum_{j=1}^n(Y_j-\bar Y)^2 \\ &= \frac{1}{n}\sum_{j=1}^n(Y_j^2-2\bar Y Y_j + \bar Y^2)\\ &= \frac{1}{n}\left(\sum_{j=1}^nY_j^2-2\bar Y \left(\sum_{j=1}^n Y_j\right) + n\bar Y^2\right)\\ &= \frac{1}{n}\left(\sum_{j=1}^nY_j^2-2\bar Y (n \bar Y) + n\bar Y^2\right)\\ & = \frac{1}{n}\sum_{j=1}^nY_j^2 - \bar Y^2. \end{align*}

Essentially the Method of Moments here is suggesting to estimate $\sigma^2$ by $S^2$, where the two population moments $E(Y_1^2|\theta),E(Y_1|\theta)$ are replaced by the corresponding sample moments.

Answer 2

Let's define $c:=\bar{Y}=\frac{1}{n}\sum_{j=1}^nY_j$. As already mentioned in the comments, it holds $\sum_{j=1}c=nc$ for a constant $c\in\mathbb{R}$.

It is easy to see, that

$$ \begin{aligned} \left( Y_j - \bar{Y}\right)^2&=Y_j^2-2\bar{Y}Y_j+\bar{Y}^2=Y_j^2-2cY_j+c^2\\ \end{aligned}$$

From this, one finds

$$ \begin{aligned} \sum_{j=1}^n\left( Y_j - \bar{Y}\right)^2&=\sum_{j=1}^nY_j^2-2c\sum_{j=1}^nY_j+\sum_{j=1}^n c^2\\ &=\sum_{j=1}^nY_j^2-2c(nc)+nc^2\\ &=\sum_{j=1}^nY_j^2 -nc^2.\\ \end{aligned}$$

This implies

$$ \begin{aligned} \frac{1}{n}\sum_{j=1}^n\left( Y_j - \bar{Y}\right)^2 &=\left(\frac{1}{n}\sum_{j=1}^nY_j^2 \right) -c^2\\ &=\left(\frac{1}{n}\sum_{j=1}^nY_j^2 \right) -\left(\frac{1}{n}\sum_{j=1}^n Y_j \right)^2.\\ \end{aligned}$$

However, it is also true, that

$$ \begin{aligned} \sum_{j=1}^n\left( Y_j - \bar{Y}\right)^2&=\sum_{j=1}^nY_j^2 -nc^2\\ &=\sum_{j=1}^nY_j^2 -\sum_{j=1}^nc^2\\ &=\sum_{j=1}^nY_j^2 -\sum_{j=1}^n\bar{Y}^2\\ &= \sum_{j=1}^n(Y_j^2-\bar{Y}^2), \end{aligned}$$

such that

$$ \begin{aligned} \frac{1}{n}\sum_{j=1}^n\left( Y_j - \bar{Y}\right)^2&= \frac{1}{n}\sum_{j=1}^n(Y_j^2-\bar{Y}^2). \end{aligned}$$