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I have the continuous random variables, u, v, x, and, y, where x and y are independent. Also, conditional on x and y, u is independent of v; that is, $\pmb{u} \bot \pmb{v} \mid \pmb{x}, \pmb{y}$. My question is whether $\pmb{u} \bot \pmb{v} \mid \pmb{x}+ \pmb{y}$. In words, is u, conditional on $\pmb{x} +\pmb{y}$, independent of v? I guess it boils down to whether the sigma algebra, $\sigma(\pmb{x},\pmb{y})$, generated by $\pmb{x}$ and $\pmb{y}$ the same as the sigma algebra, $\sigma(\pmb{x}+\pmb{y})$, generated by $\pmb{x}+\pmb{y}$. Under what conditions is the following, $\sigma(\pmb{x},\pmb{y}) = \sigma(\pmb{x}+\pmb{y})$, true.

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Generally, no. Consider that $U,V|\theta$ are independent, $U|\theta \sim \text{Bernoulli}(\theta)$, $V|\theta \sim \text{Bernoulli}(\theta)$ and that $P(\theta=0.25)=P(\theta=0.75)=0.5$.

Let $X=\theta$ and $Y=-\theta$. Since $\sigma(X,Y) = \sigma(\theta)$, conclude that $U$ is independent of $V$ given $(X,Y)$.

Also, note that \begin{align*} P(U=1|V=1) &= \frac{P(U=1,V=1)}{P(V=1)} \\ &= \frac{P(U=1,V=1,\theta=0.25)+P(U=1,V=1,\theta=0.75)}{P(V=1,\theta=0.25)+P(V=1,\theta=0.75)} \\ &= \frac{0.5(0.25)^2 + 0.5(0.75)^2}{0.5 \cdot 0.25 + 0.5 \cdot 0.75} \\ &= 0.625 \neq 0.5 = P(U=1) \end{align*} That is, $U$ is not independent of $V$. Since $X+Y=0$, $U$ is not independent of $V$ given $X+Y$.

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    Thank you very much for the answer, but are there general conditions under which $\sigma(X,Y) = \sigma(X +Y)$ and would $X$ and $Y$ being normal qualify. – 2017-01-15
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    For $\sigma(X,Y) = \sigma(X+Y)$ you'd need very strong conditions ($X$ and $Y$ being normal does not qualify. For example, consider that $\Omega=\{(0,0),(0,1),(1,0),(1,1)\}$, $X$ is the first coordinate of $\Omega$ and $Y$ is second coordinate of $\Omega$. Note that $\Omega \subset \sigma(X,Y)$ but, since $X(0,1)+Y(0,1)=X(1,0)+Y(1,0)$, $(0,1) \notin \sigma(X+Y)$ and $(1,0) \notin \sigma(X+Y)$. In other words, you'd need $X+Y$ to be bijective over the support of $(X,Y)$ in order for $\sigma(X+Y)=\sigma(X,Y)$. – 2017-01-16
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    Thank you very much, madprob, for your reply. It was helpful. – 2017-01-17