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?

  • 1
    $\begingroup$ 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. $\endgroup$
    – whuber
    Apr 7, 2022 at 22:53
  • $\begingroup$ @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. $\endgroup$ Apr 7, 2022 at 22:55
  • 3
    $\begingroup$ Essentially the same: $\text{Cov}(aX,bY)=ab\,\text{Cov}(X,Y)$ so $\text{Cov}(X/n,Y/n)=\frac1{n^2}\,\text{Cov}(X,Y)$ $\endgroup$
    – Henry
    Apr 7, 2022 at 23:43
  • $\begingroup$ 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. $\endgroup$
    – whuber
    Apr 8, 2022 at 14:32

1 Answer 1


\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}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.