Yes, indeed there is. Please see the work by Mond and Pec̆arić here. They established the AM-GM inequality for positive semi-definite matrices. Here is a link to the paper that contains the proof:

https://www.sciencedirect.com/science/article/pii/0024379595002693

After downloading the paper, the proof is on pages 450-452, in the Main Result section.

Here is a citation in case you need it:

Mond, B., and Pec̆arić, J. E. (1996), “A mixed arithmetic-mean-harmonic-mean matrix inequality,” Linear Algebra and its Applications, Linear Algebra and Statistics: In Celebration of C. R. Rao’s 75th Birthday (September 10, 1995), 237–238, 449–454. https://doi.org/10.1016/0024-3795(95)00269-3.

I hope this helps you.

Best,
=K=

It helps to think carefully about exactly what type of objects $\hat \theta$ and $\hat g$ are.

In the top case, $\hat \theta$ would be what I would call an estimator of a parameter. Let's break it down. There is some true value we would like to gain knowledge about $\theta$, *it is a number*. To estimate the value of this parameter we use $\hat \theta$, which consumes a sample of data, and produces a number which we take to be an estimate of $\theta$. Said differently, $\hat \theta$ is a function which consumes a set of training data, and produces a number

$$ \hat \theta: \mathcal{T} \rightarrow \mathbb{R} $$

Often, when only one set of training data is around, people use the symbol $\hat \theta$ to mean the numeric estimate instead of the estimator, but in the grand scheme of things, this is a relatively benign abuse of notation.

OK, on to the second thing, what is $\hat g$? In this case, we are doing much the same, but this time we are estimating a *function* instead of a number. Now we consume a training dataset, and are returned a *function from datapoints to real numbers*

$$ \hat g: \mathcal{T} \rightarrow (\mathcal{X} \rightarrow \mathbb{R}) $$

This is a little mind bending the first time you think about it, but it's worth digesting.

Now, if we think of our samples as being distributed in some way, then $\hat \theta$ becomes a random variable, and we can take its expectation and variance and whatever we want, with no problem. But what is the variance of a *function* valued random variable? It's not really obvious.

The way out is to think like a computer programmer, *what can functions do*? They can be evaluated. This is where your $x_i$ comes in.

In this setup, $x_i$ is just a solitary fixed datapoint. The second equation is saying *as long as you have a datapoint $x_i$ fixed, you can think of $\hat g$ as an estimator that returns a function, which you immediately evaluate to get a number*. Now we're back in the situation where we consume datasets and get a number in return, so all our statistics of number values random variables comes to bear.

I've discussed this in a slightly different way in this answer.

Is it correct to think of this as each observation/fitted value having its own variance and bias?

Yup.

You can see this in confidence intervals around scatterplot smoothers, they tend to be wider near the boundaries of the data, as there the predicted value is more influenced by the neighborly training points. There are some examples in this tutorial on smoothing splines.

## Best Answer

The latter equation is closest to correct; $MSE$ is a scalar. (It is a little weird having $\hat{Y}$ inside the expectation since $\hat{Y}=X\hat\beta$.) That said, many people might ignore the expectation.

Therefore, it might be easier to remember that $MSE = \frac{RSS}{df}$ where RSS is the residual sum of squares $RSS=(Y-X\hat\beta)^T(Y-X\hat\beta)$

and the degrees of freedom is $df=n-p$.

Finally, one point of convention: We typically say there are $p$ covariates, not including the intercept. Thus the $X$ used in the model is typically $n\times (p+1)$. Hence we

usuallywrite that $df=n-p-1$.