AIC equivalent to Mallows' Cp and Mallows' Cp unbiased for test MSE Part 1:
The goal is to show that with Gaussian errors and a linear model, Mallows' $C_p$ and $AIC$ are equivalent.
Using our definition of Mallows' $C_p$: $$C_p=1/n(RSS+2d\hat\sigma^2)$$
and $AIC$: $$AIC=-2\log L+2d$$ where $L$ is the maximised value of the likelihood function.
My approach was to find the loglikelihood function for Gaussian errors:
$$-\frac{n}2 \log(2\pi)-\frac{n}2\log(\sigma^2)-\frac{\sum (y_i-\mu)^2}{2\sigma^2}$$
and plugging in the expression for the MLE estimators for $\hat\mu=\frac{1}n\sum y_i$, but not for $\hat\sigma^2$. 
This leads to:
$$-\frac{n}2\log(2\pi)-\frac{n}2\log(\hat\sigma^2)-\frac{RSS}{2\hat\sigma^2}$$
However, if I now plug this in, I get:
$$AIC=n\log(2\pi)+n\log(\hat\sigma^2)+\frac{ RSS}{\hat\sigma^2}+2d$$
This doesn't look similar to the $C_p$ formula though, and I am especially concerned about the second term. Where do I go wrong?
Part 2: A further task is to show the unbiasedness of Mallows' $C_p$ if $\hat\sigma^2$ is an unbiased estimator of $\sigma^2$. I am stuck here completely: could someone provide an initial starting point?
 A: For the Gaussian model (with variance $\sigma_{\epsilon}^2 = \hat{\sigma}_{\epsilon}^2$ assumed known), the AIC statistic is equivalent to $C_p$, and so we refer to them collectively as AIC.
Suppose a Gaussian model with variance $\sigma_{\epsilon}^2 = \hat{\sigma}_{\epsilon}^2 = \sigma^2$
Mallow's $C_p$
[
C_p = \frac{1}{n} (RSS + 2 d \sigma^2)
]
AIC is
\begin{equation*}
\begin{aligned}
\text{AIC} & = -2 \log L + 2 d \\
& = -2 (- \frac{n}{2} \log(2 \pi \sigma^2) - \frac{\sum_{i=1}^{n} (y_i - \hat{y}_i   ) ^ 2 }{2\sigma^2} ) + 2 d \\
& = n \log(2 \pi \sigma^2) + \frac{\sum_{i=1}^{n} (y_i - \hat{y}_i  ) ^ 2 }{\sigma^2}  + 2 d \\
& =  n \log(2 \pi \sigma^2) + \frac{RSS}{\sigma^2}  + 2 d \\
& = n \log(2 \pi \sigma^2) + \frac{n}{\sigma^2} C_p
\end{aligned}
\end{equation*}
Since they differ by a constant which is only a function of costants $\sigma^2, n$, it doesn't affect model selection by using AIC or $C_p$ in such a situation (equivalance).
A: Regarding showing Mallows' Cp is unbiased, use


*

*$\text{tr}(\mathbf{A}) = \text{rank}(\mathbf{A})$ for idempotent matrices $\mathbf{A}$, and

*the following technique to get expectations of quadratic forms
\begin{align*}
E\left[\mathbf{y}^\intercal \mathbf{A} \mathbf{y}\right] &= E\left[\text{tr}(\mathbf{y}^\intercal \mathbf{A} \mathbf{y})\right] \\
&= E\left[\text{tr}( \mathbf{A} \mathbf{y}\mathbf{y}^\intercal)\right] \\
&= \text{tr}(E\left[ \mathbf{A} \mathbf{y}\mathbf{y}^\intercal\right] ) \\
&= \text{tr}( \mathbf{A} E\left[  \mathbf{y}\mathbf{y}^\intercal\right] ) \\
&= \text{tr}( \mathbf{A} V[\mathbf{y}] + \mathbf{A}(E\left[  \mathbf{y}\right])(E\left[  \mathbf{y}\right])^\intercal )
\end{align*}
Regarding the showing that these two criteria are equivalent--they're not.
If you plug in the ML estimators into the normal density, you should get
$$
\log L = -\frac{n}{2}\log(2\pi) - \frac{n}{2}\log(\hat{\sigma}^2 ) - \frac{1}{2 \hat{\sigma}^2}\sum_{i=1}^n \left( y_i - \hat{y}_i \right)^2 = \underbrace{-\frac{n}{2}(\log(2\pi) +1)}_{\text{often ignored}}- \frac{n}{2}\log\left(\frac{RSS}{n} \right) .
$$
Take note of 1.) $\hat{\sigma}^2 = SS_{Res}/n$, the MLE, being different from mean squared error, and 2.) there's a lot of cancellation in the sum term.
Let $d_1$ and $d_2$ be the number of degrees of freedom in two models. Using the following definitions of AIC and Mallows' CP, respectively (there are several)


*

*$n  \log\left(\frac{SS_{Res}}{n} \right) + 2p$,

*$ 1/n(RSS+2d\hat\sigma^2)$
equivalence, to me, would mean
$$
n  \log\left(\frac{RSS(M_1)}{n} \right) + 2d_1 \le n  \log\left(\frac{RSS(M_2)}{n} \right) + 2d_2
$$
if and only if
$$
1/n(RSS(M_1)+2d_1 \hat\sigma^2) \le 1/n(RSS(M_2)+2d_2\hat\sigma^2).
$$
You could rearrange the Mallows' inequality to 
$$
\left(RSS(M_1) - RSS(M_2)\right) \le 2c (d_2 - d_1)
$$
and the AIC one too
$$
\left(\log RSS(M_1) - \log RSS(M_2)\right) \le 2c'(d_2 - d_1).
$$
They're close, but not the same.
A: They are equivalent if you use  the normalized Mallows' C*p which is 
$$RSS/\sigma^2+2d-n$$ since they differ by a constant which is only a function of $$\sigma^2,n$$ and so doesn't affect model selection (variance is estimated with the full population). unnormalized mallows is just an affine transorm of normalized mallows:
$$C_p = 1
/n \sigma^2(C^*_
p + n)$$ 
