Let $x_1,x_2,\ldots,x_n$ and $y_1,y_2,\ldots,y_n$ be uniformly, independent random variables. What can I say about the distribution of the following two variables?

$$S_1 = (x_1,x_2,\ldots,x_n,-\sum(x_i))$$

$$S_2 = (y_1,y_2,\ldots,y_n,-\sum(y_i))$$

My guess is that they have exactly the same distribution, but got stuck on how to argue for it. It would be easier without $\sum(x_i)$ and $\sum(y_i)$, since both $S_1$ and $S_2$ are made of independent identical random variables.


I denote by $U$ the support of your variables. So any variable $z$ among $x_1,x_2, \ldots ,x_n,y_1,y_2, \ldots ,y_n$ satisfies

$$ P(z\in E)=\frac{\mu_1 (E \cap U)}{\mu_1 (U)}, $$

for any measurable subset $E$ of $\mathbb R$, where $\mu_1$ denotes the Lebesgue measure on $\mathbb R$.

Note that $S_1$ and $S_2$ both take their values in the hyperplane $H$ defined by the equation $$t_1+t_2+t_3+ \ldots +t_n+y=0$$

The map $p : H \to {\mathbb R}^{n}, (x_1,x_2, \ldots ,x_n,y) \mapsto (x_1,x_2, \ldots ,x_n)$ is a linear isomorphism. Let $A$ be a measurable subset of ${\mathbb R}^{n+1}$. Then

$$ P(S_1 \in A)=\frac{\mu_n(p(A \cap H))}{\mu_n(U^n)}=P(S_2 \in A) $$

where $\mu_n$ denotes the Lebesgue measure on ${\mathbb R}^n$. So $S_1$ and $S_2$ are identically distributed.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.