query: if $X,Y$ i.i.d $\mathcal{N}(0,1)$, then $X+Y$ is independent of $X-Y$
positive_passages.docid: 152495
positive_passages.text: Prove that if $X$ and $Y$ are Normal and independent random variables, $X+Y$ and $X-Y$ are independent
negative_passages.docid: 25654
negative_passages.text: E[X| X>Y] for independent X, Y ~ N(0,1)
negative_passages.docid: 95106
negative_passages.text: If $Y|X \sim \mathcal{N}(0,1)$ then is $Y^2|X \sim \chi^2(1)$?
negative_passages.docid: 109228
negative_passages.text: Prove that $v, Tv, T^2v, ... , T^{m-1}v$ is linearly independent
negative_passages.docid: 19973
negative_passages.text: If $X$ and $Y$ are independent $N(0,\sigma^2)$, then $X^2+Y^2$ and $X/Y$ are independent?
negative_passages.docid: 83873
negative_passages.text: Find $E(|X-Y|^a)$ where $X$ and $Y$ are independent uniform on $(0,1)$
negative_passages.docid: 7775
negative_passages.text: Show that $|x-y|\leq |x|+|y|$.
negative_passages.docid: 57055
negative_passages.text: Are $|X|$ and $\operatorname{sgn}(X)$ independent?
negative_passages.docid: 99105
negative_passages.text: Are the functions $\sin^n(x)$ linearly independent?
negative_passages.docid: 94218
negative_passages.text: Showing $d(x,y) = \frac{|x-y|}{1+|x-y|}$ is a distance.
negative_passages.docid: 142855
negative_passages.text: If $A$, $B$ are both pairwise independent of $C$, do we have that $A \cap B$ is also independent of $C$?
negative_passages.docid: 107641
negative_passages.text: Show that an infinite set of families are linearly independent
negative_passages.docid: 25618
negative_passages.text: $x,y,z$ independent and uniform random on $[0,1]$, $P(x \geq  yz)$?
negative_passages.docid: 126769
negative_passages.text: Linear independence of columns implies non-zero determinant
negative_passages.docid: 26860
negative_passages.text: X,Y are independent exponentially distributed then what is the distribution of X/(X+Y)
negative_passages.docid: 55752
negative_passages.text: Prove that $n+1$ vectors in $\mathbb{R}^n$ cannot be linearly independent
negative_passages.docid: 67454
negative_passages.text: Show that $\sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y}) = \sum_{i=1}^n (x_i - \bar{x})(y_i)$.