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First, let

\begin{equation} SSE = \overrightarrow{y}'(I - H) \overrightarrow{y} \end{equation}

where

\begin{equation} \overrightarrow{y} \sim MN(\textbf{X} \overrightarrow{\beta}, \sigma^{2}I) \\ H = X(X'X)^{-1}X' \end{equation}

then

\begin{equation} E(SSE) = E(\overrightarrow{y}'(I - H) \overrightarrow{y}) = k_{1} \end{equation}

\begin{equation} => k_{1} = tr[(I-H) \sigma^{2}I] + (\textbf{X}\overrightarrow{\beta})'(I-H)\textbf{X}\overrightarrow{\beta} \\ = \sigma^{2}(tr(I)-tr(H)) + 0 \\ = \sigma^{2}[n-(k+1)] \end{equation}

\begin{equation} Var(SSE) = Var(\overrightarrow{y}'(I - H) \overrightarrow{y}) = k_{2} \end{equation}

\begin{equation} => k_{2} = 2tr[((I-H) \sigma^{2}I)^{2}] + 2(\textbf{X}\overrightarrow{\beta})'(I-H)(\sigma^{2}I(I-H))\textbf{X}\overrightarrow{\beta} \end{equation} and using a similar argument as the expectation: \begin{equation} => k_{2} = \sigma^{4}[n-(k+1)] \end{equation}

I know my mean is correct but I'm thinking my variance is wrong. Any thoughts?

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1 Answer 1

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Looks good to me except you forgot to multiply by $2$ for the first term in the expression for $\mathrm{Var}(SSE)$. So the final answer for the variance should be double what you wrote. By the way, the formula for $\mathrm{Var}(\mathbf{y}'(I-H)\mathbf{y})$ should have a coefficient of $4$ rather than $2$ for the second term (see this), but it doesn't matter since that part becomes $0$ anyway.

Also, another way to check your answer would be to recall that $$\frac{(n-(k+1))\widehat{\sigma}^2}{\sigma^2}\sim \chi^2_{n-(k+1)}$$ in this model. You can then take the mean and variance of both sides (using the formula for mean and variance of a chi-squared random variable).

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