Hello given dataset $Y$ with $a$ factors and $n$ observations. How can I show $E(MSE) = \sigma^2$ from the one-way ANOVA of Y?
Where I'm at, $$E(MSE) = \sigma^2$$ $$E\left(\frac{SSE}{a(n-1)}\right) = \frac{\sum_{i=0}^n(y_i - \bar{y})^2}{n-1}$$ $$\frac{1}{a(n-1)}E(SSE) = \frac{\sum_{i=0}^n(y_i - \bar{y})^2}{n-1}$$ $$\frac{1}{a}E(SSE) = \sum_{i=0}^n(y_i - \bar{y})^2$$ $$\frac{1}{a} \sum_{i=1}^a \sum_{j=1}^n E((y_{ij}-\bar{y}_i)^2) = \sum_{i=0}^n(y_i - \bar{y})^2$$
Now, do I simply apply the null assumptions of the one-way ANOVA meaning $\bar{y_i} = \bar{y}$? Anything is helpful.
I assume your model is
$$y_{ij}=\mu_i+\varepsilon_{ij} \quad\small,\,i=1,\ldots,a\,;j=1,\ldots,n$$
where $\mu_i$ is a fixed effect and $\varepsilon_{ij}$'s are i.i.d random errors with mean $0$ and variance $\sigma^2$.
Equivalently, the model is
$$y_{ij}=\mu+\alpha_i+\varepsilon_{ij} \quad\small,\,i=1,\ldots,a\,;j=1,\ldots,n $$
Here $\mu$ is a general effect and $\alpha_i$ is an additional fixed effect subject to $\sum_{i=1}^a \alpha_i=0$.
Keeping in mind $\overline y_{i0}=\frac1n\sum_{j=1}^n y_{ij}$, verify that
$$\text{SSE}=\sum_{i,j}(y_{ij}-\overline y_{i0})^2=\sum_{i,j}(\varepsilon_{ij}-\overline \varepsilon_{i0})^2=\sum_{i,j}\varepsilon_{ij}^2-n\sum_i \overline \varepsilon_{i0}^2$$
Noting that $\overline\varepsilon_{i0}$ has mean $0$ and variance $\frac{\sigma^2}{n}$, you have
$$\operatorname E\left[\text{SSE}\right]=na\sigma^2-a\sigma^2=(n-1)a\sigma^2$$
That is,
$$\operatorname E\left[\text{MSE}\right]=\operatorname E\left[\frac{\text{SSE}}{a(n-1)}\right]=\sigma^2$$