If this equation is an ensemble average over phase space
$$\langle A\rangle=∫_Γ ∏^{3N} \ {\rm d}q_i \ {\rm d}p_i\ A({q_i},\,{p_i})ρ({q_i},\,{p_i})$$
what is the time average of $A$ and how that can be calculated?
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Using an abbreviated notation, the time average is
$$\bar A \equiv \lim_{T\to \infty} \frac 1 T \int_0^T A(p(t),q(t)) \ dpdq$$
Also notice that your expression is wrong. It should be, if you want to write it explicitly,
$$\langle A \rangle = \int_{\Gamma} \prod_{i=1}^{3N} dp_i dq_i \ A(p_1,q_1,\dots p_{3N}, q_{3N}) \ \rho (p_1,q_1,\dots p_{3N}, q_{3N})$$
or, in abbreviated notation,
$$\langle A \rangle = \int A(p,q) \rho(p,q) \ dp dq$$
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$\begingroup$ Daer Valerio, Thank you so much. Is this type of questions generally solve in say MD simulation or they are just in theory? $\endgroup$ Commented Feb 23, 2018 at 23:43
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$\begingroup$ @Robertharry I am sorry, I don't understand the question. You want to know how to apply such relations to MD simulations? $\endgroup$– valerioCommented Feb 23, 2018 at 23:46
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$\begingroup$ Like I wanted to have an example that this has been solved $\endgroup$ Commented Feb 24, 2018 at 0:05
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$\begingroup$ @Robertharry You want to know if it is possible to explicitly calculate time averages in realistic situations? In MD simulations you can, of course, since you have all the $p$ and $q$ at every time step, but for a number of particles (max$\approx 10^{10}$) much smaller than what you have in real-life systems ($\approx 10^{24}$). However, usually it works pretty well. $\endgroup$– valerioCommented Feb 24, 2018 at 0:12
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$\begingroup$ You are missing a factor $1/T$ in your definition of $\bar A$. $\endgroup$ Commented Feb 24, 2018 at 10:04