Demonstrate maximum likelihood estimates do not depend on linear transformations prove that the likelihood functions $f(\tilde{y}/x)$ and $f(y/x)$ coincide.
Given:
$\tilde{y}$ = Ty = T Hx+ T v ,
T is a square non-singular $m x m$,
v is a Gaussian random vector.
My first idea was:
The estimate $x^{m}$ is calculated by selecting the value of maximizing with the necessary maximum condition for equation \begin{align}\label{1} f(y/x) = ln\Big( p(y/x) \Big)=- \frac{1}{2} (y-Hx)^{T} R^{-1} (y-Hx): \end{align}
\begin{align*} \frac{d }{dx} \Big(ln \quad \mathit{p}(y/x) \Big) \Big\vert_{ {\hat{x}^{m}(y)} } = 0 \\ \frac{d \Big( - \frac{1}{2} (y-Hx)^{T} R^{-1} (y-Hx) \Big)}{dx} \Big\vert_{ {\hat{x}^{m}(y)} } = 0 \\ -\frac{1}{2} \Big( -H^{T} R^{-1} y + H^{T} R^{-1} H \hat{x}^{m} \Big) = 0\\ H^{T} R^{-1} H \hat{x}^{m} = H^{T} R^{-1} y \\ H \hat{x}^{m} = y \\ \rightarrow \hat{x}^{m} = H^{-1}y \\ \text{or} \qquad \hat{x}^{m} = (H^{T} R^{-1} H)^{-1} H^{T} R^{-1} y \end{align*} This same expression follows directly for equation with $\tilde{y}$ = Ty : \begin{align} \label{2} f( \tilde{y}/x) = ln\Big( p(\tilde{y}/x) \Big)= - \frac{1}{2} (Ty-Hx)^{T} R^{-1} (Ty-Hx): \end{align}
\begin{align*} \frac{d }{dx} \Big(ln \quad \mathit{p}(\tilde{y}/x) \Big) \Big\vert_{ {\hat{x}^{m}(y)} } = 0 \\ \frac{d \Big( - \frac{1}{2} (Ty-Hx)^{T} R^{-1} (Ty-Hx) \Big)}{dx} \Big\vert_{ {\hat{x}^{m}(y)} } = 0 \\ -\frac{1}{2} \Big( -H^{T} R^{-1} Ty + H^{T} R^{-1} H \hat{x}^{m} \Big) = 0 \\ H^{T} R^{-1} H \hat{x}^{m} = H^{T} R^{-1} Ty \\ \hat{x}^{m} = H^{-1} Ty \\ \text{with $\tilde{y}=Ty$} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \\ \rightarrow \hat{x}^{m} = H^{-1} \tilde{y} \\ \text{or} \qquad \hat{x}^{m} = (H^{T} R^{-1} H)^{-1} H^{T} R^{-1} \tilde{y} \end{align*}
It is a bit unclear because results for equation 1 and 2 are not the same. Because: $\hat x^{m}_1 - \hat x^{m}_2 = \left(H^TR^{-1}H\right)^{-1}H^TR^{-1}\left(y- Ty\right) \neq 0.$
How can I show that the estimate does not depend on linear transformation ? I suppose here is a little skill applied.