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I have the following example:

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

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 did the authors get from $\dfrac{1}{n^2} \text{Cov} \left( \sum_{j = 1}^n Y_j, nY_i - \sum_{j = 1}^n Y_j \right)$ to $\dfrac{1}{n^2} \left( (n - 1)\text{Var}(Y_i) - \sum_{j = 1, j \not= i}^n \text{Var}(Y_j) \right)$?

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It comes from the distributive property of covariance: $$\begin{align}C&=\text{Cov} \left( \sum_{j = 1}^n Y_j, nY_i - \sum_{j = 1}^n Y_j \right)=\operatorname{cov}\left(\sum_{j=1}^n Y_j, nY_i\right)-\operatorname{cov}\left(\sum_{j=1}^n Y_j,\sum_{k=1}^n Y_k\right)\\&= n\sum_{j=1}^n\operatorname{cov}\left(Y_j,Y_i\right)-\sum_{j=1}^n\sum_{k=1}^n \operatorname{cov}(Y_j,Y_k)\\&=n\operatorname{cov}(Y_i,Y_i)-\sum_{j=1}^n\operatorname{cov}(Y_j,Y_j)=n\operatorname{var}(Y_i)-\sum_{j=1}^n\operatorname{var}(Y_j)\\&=n\operatorname{var}(Y_i)-\operatorname{var}(Y_i)-\sum_{j=1,j\neq i}^n\operatorname{var}(Y_j)\\&=(n-1)\operatorname{var}(Y_i)-\sum_{j=1,j\neq i}^n\operatorname{var}(Y_j)\end{align}$$

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  • $\begingroup$ Thanks again! There seems to be an interesting property of covariance here that I was not familiar with: $\sum_{j = 1}^n \text{cov}(Y_j, Y_i) = \text{cov}(Y_i, Y_i)$. $\endgroup$ Mar 14, 2020 at 21:08
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    $\begingroup$ it doesn't directly come from properties of covariance. the covariance between $Y_i$ and $Y_j$ is $0$ if $i\neq j$ (because in the question it's said iid), so in the summation only $j=i$ remains. $\endgroup$
    – gunes
    Mar 14, 2020 at 21:27
  • $\begingroup$ Ok, thanks for the clarification. $\endgroup$ Mar 14, 2020 at 21:28
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    $\begingroup$ Both of them are due to distributive property, i.e. $\operatorname{cov}(A,B+C)=\operatorname{cov}(A,B)+\operatorname{cov}(A,C)$. $\endgroup$
    – gunes
    Apr 14, 2021 at 12:33
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    $\begingroup$ We have $\operatorname{cov}(A,nB)=n\operatorname{cov}(A,B)$ $\endgroup$
    – gunes
    Apr 14, 2021 at 15:20

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