This might be a very basic question but I am wondering how to understand the following point rigorously. Let $X_1,\dots, X_n \sim N(\mu_i, \sigma^2)$, $Y_1, Y_2,\dots, Y_n \sim N(\mu_i, \sigma^2)$ with the two sequences being independent of each other. Let $T_i = (X_i + Y_i)/2$. Why does the following hold true:
The conditional distribution of $(X_i, Y_i|T_i)$ is readily determined from the conditional distributions of $X_i|T_i$.
What does this statement mean formally? We know that $(X_i, Y_i|T_i)$ does not have a density with respect to Lebesgue measure, but that $X_i|T_i$ does. What does this claim imply for the latter density ?
We have that $$ \begin{align*} \Pr [(X,Y)\in A|T=t]&=\Pr [(X,2T-X)\in A|T=t]\\&=\Pr [f(X,T)\in A|T=t]\\&=\Pr [(X,T)\in f^{-1}(A)|T=t]\\ &=\Pr [(X,t)\in f^{-1}(A)|T=t]\\ &=\Pr [X\in A_t|T=t] \end{align*} $$ for $A_t:=\{x\in \mathbb{R}:(x,2t-x)\in A\}$. Therefore the conditional distribution $Q(A,t):=\Pr [(X,Y)\in A|T=t]$ is determined by the conditional distribution $R(B,t):=\Pr [X\in B|T=t]$, that is, we have the identity $Q(A,t)=R(A_t,t)$, so we can use $R$ to compute $Q$, or in other words $$ Q(A,t)=\int_{A}Q(d\mathbf{r},t)=\int_{A_t}R(dx,t)=\int_{A_t}f_{X|T}(x,t)\,d x $$ where $\mathbf{r}:=(x,y)$ and the latter integral is in the case that $R(dx,t)$ have a conditional density (respect to the Lebesgue measure).