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I have to sets of r.v. $X_1,X_2,\ldots X_n$, and $Y_1,Y_2,\ldots Y_n$. I have that the increments in both collections are independent. That is: $X_1,X_2-X_1,\ldots, X_n-X_{n-1}$ are independent, and $Y_1,Y_2-Y_1,\ldots, Y_n-Y_{n-1}$ are independent. I also have that the corresponding increments between X and Y have the same distribution. That is, $X_1$ and $Y_1$ have the same distribution, and $X_i-X_{i-1}$ have the same distribution as $Y_i-Y_{i-1}$, for $1< i\le n$.

I need to show that the vectors $(X_1,X_2,\ldots, X_n)$, and $(Y_1,Y_2,\ldots,Y_n)$ have the same distribution.

I think I almost had it, but I got a problem with measurability.

my attempt:

$P(X_1\in B_1,X_2\in B_2,\ldots,X_n \in B_n)\\=\int_{\mathbb{R}^n}I_{z_1\in B_1,z_2+z_1\in B_2,\ldots z_1+z_2+z_3+\cdots+z_{n-1}+z_n\in B_n}(z_1,z_2,\ldots,z_n)d\mu_{(X_1,X_2-X_1,\ldots ,X_n-X_{n-1})}$

indpendence:

$\\=\int_{\mathbb{R}^n}I_{z_1\in B_1,z_2+z_1\in B_2,\ldots z_1+z_2+z_3+\cdots+z_{n-1}+z_n\in B_n}(z_1,z_2,\ldots,z_n)d\mu_{X_1}d\mu_{X_2-X_1}\ldots d\mu_{X_n-X_{n-1}}$

same distribution of increments:

$\\=\int_{\mathbb{R}^n}I_{z_1\in B_1,z_2+z_1\in B_2,\ldots z_1+z_2+z_3+\cdots+z_{n-1}+z_n\in B_n}(z_1,z_2,\ldots,z_n)d\mu_{Y_1}d\mu_{Y_2-Y_1}\ldots d\mu_{Y_n-Y_{n-1}}$

$=P(Y_1\in B_1,Y_2\in B_2,\ldots,Y_n \in B_n)$

But in order for this to work I need that $\{(z_1,z_2,\ldots,z_n):z_1\in B_1, z_1+z_2 \in B_2, \ldots, z_1+z_2+\cdots+z_n \in B_n\} \in \mathcal{B}(\mathbb{R}^n)$.

But is it clear that this set is Borel-measurable? If not, must I use another approach to solve the problem? Any hints?

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Clearly,

$$\{(z_1,\ldots,z_n); z_1 \in B_1, \ldots, z_1+\ldots+z_n \in B_n\} = \bigcap_{k=1}^n \left\{(z_1,\ldots,z_n); \sum_{j=1}^k z_j \in B_k \right\}$$

and therefore it suffices to show that each set

$$\left\{(z_1,\ldots,z_n); \sum_{j=1}^k z_j \in B_k \right\}$$

is Borel measurable. This, however, follows from the fact that the mapping

$$(z_1,\ldots,z_n) \mapsto f_j(z) := \sum_{j=1}^k z_j$$

is Borel measurable (as it is continuous);

$$\left\{(z_1,\ldots,z_n); \sum_{j=1}^k z_j \in B_k \right\} = f_j^{-1}(B_k) \in \mathcal{B}(\mathbb{R}^n)$$

for any $B_k \in \mathcal{B}(\mathbb{R})$.

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