Is the following probability distribution stationary/constant

Question

For a conservative system, we know that angular momentum, $l$, and total energy, $E$, are constant, i.e. $\dot{l}=\frac{dl}{dt} = 0$ and $\dot{E}=\frac{dE}{dt} = 0$, where $t$ indicates time. Let $l_1,l_2,l_3$ indicate the components of $l$ in $\hat{i},\hat{j},\hat{k}$ directions. \begin{equation} l^2 = l_1^2+l_2^2+l_3^2\\ E = \frac{1}{2}m\dot{x}^2 + mgh \hspace{10pt} \end{equation} Here we consider $l,l_1,l_2,l_3,E$ as random variables. Then we can say that

1. probability distribution of $l$ is stationary with respect to time.
2. probability distribution of $E$ is stationary with respect to time.
3. joint probability distribution of $l$ and $E$ is stationary with respect to time.

But can we say the following?

1. joint probability distribution of $l_1,l_2,l_3$ is stationary with respect to time.
2. joint probability distribution of $l_1,l_2,l_3,E$ is stationary with respect to time.

Answer 1

No, the joint probability distribution does not have to be stationary w.r.t. time. Just imagine a precessing top with no friction. The probability distribution for $l^2$ is just a delta peaked at some particular value. But the joint probability distribution of $(l_1,l_2,l_3)$ is a delta peak that moves around in the angular momentum space.