# Determining one distribution from another

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).