# Choosing constant to minimize mean square error

Can you please help me show the statement below? I am not exactly sure where to start. As a function of $c$, the quadratic $\mathrm E(S^4)c^2-2\sigma^2\mathrm E(S^2)c+\sigma^4$ is minimum at $c=\sigma^2\mathrm E(S^2)/\mathrm E(S^4)$. To compute this ratio, the simplest approach could be to compute $\mathrm E(S^2)$ and $\mathrm E(S^4)$.
For starters, would you know how to compute $\mathrm E(S^2)$? Hint: the value of $\mathrm E(S^2)$ does not depend on the sample being drawn from a normal distribution, but only on the parameters $\mu$ and $\sigma^2$.
• (1) What is the definition of $S^2$ you want to use? (2) Answering that $E(S^2)=\sigma^2$ because $S^2$ is [an] unbiased [estimator of the variance], is akin to playing with words and this is not what I suggested. My suggestion was rather to reprove this comparatively simple fact (or to remember how its proof goes), in order to be ready to tackle $E(S^4)$. In the end, this is up to you: either you wait for a complete answer to appear on this page (something which may very well happen, unfortunately), or you start doing maths yourself... – Did Oct 15 '11 at 6:56
• (1) (bis) What is the definition of $S^2$ you want to use? There are several options and as long as you have not said which one you use, the discussion is pointless. (3) For homogeneity reasons, $E(S^4)$ should be of degree $4$ in the variables $\mu$ and $\sigma$ hence the formula for $E(S^4)$ in your last comment cannot hold, whatever option you choose to define $S^2$. (4) NEVERTHELESS, I suggest you write here (either in your post or in an answer) how you got $\mu^2+\sigma^2$ for $E(S^4)$ and $\sigma^2$ for $E(S^2)$, so that we have something concrete to discuss. – Did Oct 16 '11 at 7:50