distribution of linearly transformed variable Let $X$ be a random variable and it takes values from $\ell_{2}$, which is the space of square-summable sequences. Next, let us define random variable $U=a(X - b)$, for some $a \in \mathbb{R}_{+}$ and $b \in \ell_{2}$, also both $a$ and $b$ are not all zeros.
Next, assume that for some random variables $V$ and $Y$, taking values from $\ell_{2}$, defined on, possibly, another probability space we know that $V = a(Y - b)$ and $V\stackrel{d}{=}U$.
Does this mean that
$$
Y\stackrel{d}{=}X
$$
?
Does the result depend on the space which the variables take values from?
 A: I do not believe that the result is dependent on which spaces we define things.
I believe we can even prove a more general result than what you are looking for. Let $\varphi$ be a measurable transformation and define $U = \varphi (X)$ and $V = \varphi(Y)$ and assume that $U \overset{d}{=} V$. For a measurable set $B$ we have
\begin{align*}
\mathbb{P}_{U}(B) &= \mathbb{P}_{\varphi (X)}(B) = \mathbb{P}(X \in \varphi^{-1}(B)) = \mathbb{P}_{X}(\varphi^{-1}(B)) \\
\mathbb{P}_{V}(B) &= \mathbb{P}_{\varphi (Y)}(B) = \mathbb{P}(Y \in \varphi^{-1}(B)) = \mathbb{P}_{Y}(\varphi^{-1}(B))
\end{align*}
So we may conclude that
\begin{align*}
\mathbb{P}_{X}(\varphi^{-1}(B)) = \mathbb{P}_{Y}(\varphi^{-1}(B))
\end{align*}
Implying that $\mathbb{P}_{X} = \mathbb{P}_{Y}$, meaning $X\overset{d}{=}Y$.
A: Let $X=(X_1,\ldots,X_n,\ldots)$, where $X_i$ are scalar random variables, similarly for $U$, $V$, $Y$ and $a$, $b$.
Then it follows immediately with
$$P(Y_1\le y_1, \ldots, Y_n \le y_n,\ldots)$$
$$=P(\frac 1a V_1 + b_1 \le y_1,\ldots,\frac 1a V_n + b_n \le y_n,\ldots)$$
$$=P(V_1 \le a(y_1-b_1),\ldots, V_n \le a(y_n-b_n),\ldots)$$
$$=P(U_1 \le a(y_1-b_1),\ldots, U_n \le a(y_n-b_n),\ldots)$$
$$=P(\frac 1a U_1 + b_1 \le y_1,\ldots,\frac 1a U_n + b_n \le y_n,\ldots)$$
$$=P(X_1\le y_1, \ldots, X_n \le y_n,\ldots)$$
Hence the distribution of $X$ and $Y$ is the same.
