# How is $\text{Cov}(\bar{Y}, Y_i - \bar{Y}) = \dfrac{1}{n^2} \text{Cov} \left( \sum_{j = 1}^n Y_j, nY_i - \sum_{j = 1}^n Y_j \right)$?

I have this example of sufficiency:

Let $$Y_1, \dots, Y_n$$ be i.i.d. $$N(\mu, \sigma^2)$$. Note that $$\sum_{i = 1}^n (y_i - \mu)^2 = \sum_{i = 1}^n (y_i - \bar{y})^2 + n(\bar{y} - \mu)^2$$. Hence

\begin{align} L(\mu, \sigma; \mathbf{y}) &= \prod_{i = 1}^n \dfrac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{1}{2\sigma^2}(y_i - \mu)^2} \\ &= \dfrac{1}{(2\pi \sigma^2)^{n/2}}e^{-\frac{1}{2\sigma^2}\sum_{i = 1}^n (y_i - \bar{y})^2}e^{-\frac{1}{2\sigma^2}n(\bar{y} - \mu)^2} \end{align}

From Theorem 1, it follows that where $$T(\mathbf{Y}) = (\bar{Y}, \sum_{i = 1}^n (Y_i - \bar{Y})^2)$$ is a sufficient statistic for $$(\mu, \sigma)$$.

It then says the following:

We now show that $$\bar{Y} \sim N(\mu, \frac{\sigma^2}n)$$.

It is clear that

$$Y_1 + \dots + Y_n \sim N(n\mu, n\sigma^2)$$

and so

$$\bar{Y} \sim N\left( \mu, \frac{\sigma^2}n \right)$$

And then the following:

We show that $$Y$$ and $$\sum_{i = 1}^n (Y_i - \bar{Y})^2$$ are independent.

One can show that

\begin{align} \text{Cov}(\bar{Y}, Y_i - \bar{Y}) &= \dfrac{1}{n^2} \text{Cov} \left( \sum_{j = 1}^n Y_j, nY_i - \sum_{j = 1}^n Y_j \right) \\ &= \dfrac{1}{n^2} \left( (n - 1)\text{Var}(Y_i) - \sum_{j = 1, j \not= i}^n \text{Var}(Y_j) \right) \\ &= \dfrac{1}{n^2} ((n - 1) \sigma^2 - (n - 1)\sigma^2) \\ &= 0 \end{align}

Since $$(\bar{Y}, Y_i - \bar{Y})$$ is normally distributed and this implies $$\bar{Y}$$ and $$Y_i - \bar{Y}$$ are independent for all $$i$$. So $$\bar{Y}$$ and $$(Y_1 - \bar{Y}, \dots, Y_n - \bar{Y})$$ are also independent. This implies $$\bar{Y}$$ and $$\sum_{i = 1}^n (Y_i - \bar{Y})^2$$ are independent.

How is $$\text{Cov}(\bar{Y}, Y_i - \bar{Y}) = \dfrac{1}{n^2} \text{Cov} \left( \sum_{j = 1}^n Y_j, nY_i - \sum_{j = 1}^n Y_j \right)$$? It seems like it should be $$\text{Cov}(\bar{Y}, Y_i - \bar{Y}) = \dfrac{1}{n} \text{Cov} \left( \sum_{j = 1}^n Y_j, nY_i - \sum_{j = 1}^n Y_j \right)$$, no?

• Covariance is bilinear. In your "How is" formula, the right hand side has been multiplied once by $n$ (the first term in the covariance), multiplied again by $n$ (the second) term, and then divided by $n^2.$ Consequently, you are asking why $(n)(n)/(n^2)=1.$ Check it out for yourself with a simple calculation: use a dataset of two observations, for instance.
– whuber
Apr 7, 2022 at 22:53
• @whuber Is this the same property as $\text{Var}(aX) = a^2 \text{Var}(X)$? I suspected that it was the same property, but I couldn't find anything in my research that explicitly confirmed it. Apr 7, 2022 at 22:55
• Essentially the same: $\text{Cov}(aX,bY)=ab\,\text{Cov}(X,Y)$ so $\text{Cov}(X/n,Y/n)=\frac1{n^2}\,\text{Cov}(X,Y)$ Apr 7, 2022 at 23:43
• See stats.stackexchange.com/a/142472/919 for explicit confirmation and this site search for more explanation of how variances and covariances are two aspects of the same thing.
– whuber
Apr 8, 2022 at 14:32

\begin{align} & \operatorname{Cov}(\bar{Y}, Y_i - \bar{Y}) \\[12pt] = {} & \operatorname{Cov} \left( \frac 1 n \sum_{j = 1}^n Y_j,\,\, \frac 1 n \left( nY_i - \sum_{j = 1}^n Y_j \right) \right) \\[12pt] = {} & \frac 1 n \operatorname{Cov} \left( \sum_{j = 1}^n Y_j,\,\, \frac 1 n \left( nY_i - \sum_{j = 1}^n Y_j \right) \right) \\[12pt] = {} & \frac 1 n\cdot\frac 1 n \operatorname{Cov} \left( \sum_{j = 1}^n Y_j,\,\, nY_i - \sum_{j = 1}^n Y_j \right) \end{align}