Problem with verifying variance of residual

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I am supposed to show the following:

$$Var(e_{ij}) = \sigma^{2}\left(1-\frac{1}{n_i}\right)$$

Where the follwing is known:

$$y_{ij} = \mu + \alpha_{i} + \varepsilon_{ij}$$

$$e_{ij} = y_{ij} - \hat{y}_{ij}$$

$$E(e_{ij}) = 0$$

$$\varepsilon_{ij} \in N(0,\sigma^{2})$$

$$ j = 1,...,n_{i}$$

$$ i = 1,...,a$$

I've started but now I'm stuck, would be glad for some guidance.

$Var(e_{ij}) = Var(y_{ij}-\hat{y}_{ij}) = Var(y_{ij}) + Var(\hat{y}_{ij}) - 2Cov(y_{ij},\hat{y}_{ij}) = $

$= Var(\mu + \alpha_{i} + \varepsilon_{ij}) + Var(\hat{y}_{ij}) - 2Cov(y_{ij},\hat{y}_{ij}) = $

$Var(\varepsilon_{ij}) + Var(\bar{y}_{i.}) - 2Cov(y_{ij},\bar{y}_{i.}) = $

$\sigma^{2} + Var(\frac{1}{n_{i}}y_{i.}) - 2Cov(y_{ij},\frac{1}{n_{i}}y_{i.}) = \dots $

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I managed to solve it.

$Var(e_{ij}) = Var(y_{ij}-\hat{y}_{ij}) = Var(y_{ij} - \bar{y}_{i.}) = $

$ = Var(y_{ij}) + Var(\bar{y}_{ij}) - 2Cov(y_{ij},\bar{y}_{i.}) = $

$ = Var(y_{ij}) + Var(\frac{1}{n_i}\sum_{i=1}^{n_i}{y_{ij}}) - 2Cov(y_{ij},\frac{1}{n_i}\sum_{i=1}^{n_i}{y_{ij}})$

$Var(y_{ij}) = \sigma^{2}$

$Var(\frac{1}{n_i}\sum_{i=1}^{n_i{y_{ij}}}) = \frac{1}{n_{i}^{2}}Var(\sum_{i=1}^{n_i}{y_{ij}})$

Assume all $y_{ij}$ are independent:

$\frac{1}{n_{i}^{2}}\sum_{j=1}^{n_i}Var(y_{ij}) = \frac{1}{n_{i}^{2}}n_{i}\sigma^{2} = \frac{1}{n_i}\sigma^2$

$Cov(y_{ij},\frac{1}{n_i}\sum_{i=1}^{n_i}{y_{ij}}) = \frac{1}{n_{i}}Cov(y_{ij},{y_{ij}}) = \frac{1}{n_i}Var(y_{ij}) = \frac{1}{n_i}\sigma^2$

$\Rightarrow Var(e_{ij})=\sigma^2 + \frac{1}{n_i}\sigma^2 - \frac{2}{n_i}\sigma^{2} = \sigma^{2}\left(1-\frac{1}{n_i}\right)$