Independence of sum of random variables Suppose we have a family $(Y_i, i\in\mathbb{I})$ of independent random variables. Then if we have measurable functions of those random variables, they are also independent. However, my question is: If for $n\in\mathbb{N}$ $(Y_1,\ldots ,Y_{n+1})$ is a finite collection of independent random variables, does that imply that $Y_{n+1}$ is independent of $Y_{1} +\cdots+Y_{n+1}$. And if so, how does one prove this. Thanks in advance!
 A: If I understand your question correctly, the easy answer is: no. As a counterexample, consider $Y_1,Y_2\sim U\left([0,1]\right)$ and let $X=Y_1+Y_2$. Your question is then: are $Y_2$ and $X$ independent? However, easy implications such as
$$X\leq\frac{1}{2}\ \implies\ Y_2\leq\frac{1}{2}$$
$$Y_2\geq\frac{1}{2}\ \implies\ X\geq\frac{1}{2}$$
show that $Y_2$ and $X$ are not independent.

Edit:
Judging from the comments, the question is not: Show that $Y_{n+1}$ is independent of $Y_1+\ldots+Y_{n+1}$, but Show that $Y_{n+1}$ is independent of $Y_1+\ldots+Y_{n}$


*

*$Y_{n+1}$ and $Y_1$ are independent, such that
$$P(Y_{n+1}|Y_1)=P(Y_{n+1})$$

*$Y_{n+1}$ and $Y_2$ are independent, such that
$$P(Y_{n+1}|Y_2)=P(Y_{n+1})$$

*Combining this shows
$$P(Y_{n+1}|Y_1,Y_2)=P(Y_{n+1})$$

*Generalizing this shows
$$P(Y_{n+1}|Y_1,Y_2,\ldots,Y_n)=P(Y_{n+1})$$


Thus $Y_{n+1}$ is independent of $Y_1,Y_2,\ldots,Y_n$. Thus $Y_{n+1}$ is also independent of $Y_1+Y_2+\ldots+Y_n$.
A: Let $Y_2$ have some distribution with expectation $E_2$.
Then $E(Y_2+Y_1|Y_1=a)=E_2+a$, while $E(Y_2+Y_1|Y_1=b)=E_2+b$. This is enough to say that the two conditional distributions differ.
