# Estimator of mu for gaussian variable with penalty

I have a question about the following problem: estimate $\mu=\mu_1,...\mu_n$ when $Y_i \sim N(\mu_i,\sigma^2)$ using a ridge like penalty

$$\min_\mu \sum_i(Y_i-\mu_i)^2 + \lambda\sum_i\mu_i^2$$

I have calculated the closed form decomposition of the generalized cross validation error and I know the optimal $\lambda$ depends on $\mu$ and $\sigma$.

My question is how calculate the optimal $\lambda$ and the cross validation error with a cross validation procedure ? I don't see how to do that.

Thanks for your help

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I'm not sure I quite understand your question yet. Is $\mu$ a vector, as in $\mu = (\mu_1,\ldots,\mu_n)$? And, is the notation $Y_i \to N(\mu_i,\sigma^2)$ meant to indicate that $Y_i$ has a normal distribution with mean $\mu_i$ and $\sigma^2$. (I ask because this is usually denoted by $\sim$, i.e., `\sim` and so I want to make sure your question is being interpreted properly.) The description after the display equation could also be improved with a little more elaboration, I think. – cardinal Oct 22 '11 at 13:18
hi, thanks for the answer. Your right , i had to use \sim... sorry. In fact i ask the question for $\mu$ constant and more specifically if $\mu$ is not constant like this kind of model: $$Y_t = \mu_t + \epsilon_1$$ $$\mu_t = \mu_{ t-1} + \epsilon_2$$ where $\epsilon_1$ and $\epsilon_2$ are two independents gaussian noise. like times series as example. I know Kalman filter can estimate the vector $\mu$, i ask me if there is an other way thanks to cross validation – Trevor Oct 24 '11 at 22:22
What is the relation between $\mu$ and $\mu_i$? Without it the expression $\min_{\mu}f(\mu_1,...,\mu_n,Y_1,...,Y_n)$ does not make sense. – mpiktas Oct 25 '11 at 11:07
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