Why is the characteristic function of the sum of R.V define in the joint space the same as the characteristic function of the sum in the sum space? Suppose I have a set of independent random variables $\{X_1,X_2,\dots,X_n\}$ and I calculate a new variable $Y(X_1,\dots,X_n)=\sum^{n}_{i=1}X_i$. In order to calculate the characteristic function of Y, $\phi_Y(t)$, I saw in many books, that the correct procedure is the following:
\begin{equation}\phi_Y(t)=\phi_{\sum^{n}_{i=1}X_i}(t)=E[e^{i\, \sum^{n}_{j=1}X_j\,t}]=\int_{\mathbb{R}^{n}}e^{i\, \sum^{n}_{j=1}X_j\, t}\, f_{X_1,X_2,\dots,X_n}(x_1,x_2,\dots,x_n)\,\text{d}^{n}x=\prod^{n}_{j}\int^{\infty}_{-\infty}e^{i\, X_j\, t}\, f_{X_j}(x_j)\, \text{d}x_j=\prod^{n}_{j}\phi_{X_j}(t).
\end{equation}
What I can't understand is why or even if this procedure is the same as doing this
\begin{equation}\phi_Y(t)=E[e^{i\,Y\,t}]=\int^{\infty}_{-\infty}e^{i\,y\,t}\, f_{Y}(y)\, \text{d}y.
\end{equation}
I get that you can calculate the expected value of any function of the variables where you define your probability space, for example, the variance: $Var(X)=E[(X-E(X))^{2}]$. But I can´t understand why $\phi_Y(t)=\phi_{\sum^{n}_{i=1}X_i}(t)$ is a valid step.
 A: To see that the two expressions
$$
\int_{\mathbb{R}^n}e^{it\sum_\limits{j=1}^nx_j}f_X(x)d^nx
$$
and
$$
\int_\mathbb{R}e^{ity}f_Y(y)dy\ ,
$$
reduce to the same thing, you shouldn't really need to do anything more than observe that both are equal to
$$
E\Bigg(e^{it\sum_\limits{j=1}^nX_j}\Bigg)=E\big(e^{itY}\big)\ .
$$
However, if you're curious to see how the second integral can be reduced to the first, you simply need to replace $\ f_Y(y)\ $ in the second integral by its expression in terms of $\ f_X\ $, which it must have, because $\ Y=\sum_\limits{j=1}^nX_j $. One equation (of several) connecting the two is
$$
f_Y(y)=\int_{\mathbb{R}^{n-1}}f_X\Big(x_1,x_2,\dots,y-\sum_{j=1}^{n-1}x_j\Big)dx_1dx_2\dots dx_{n-1}\ .
$$
Substituting the expression on the right of this identity for $\ f_Y(y)\ $ in the integral $\ \int_\mathbb{R}e^{ity}f_Y(y)dy\ $ gives
\begin{align}
\int_\mathbb{R}e^{ity}f_Y(y)dy=&\int_\mathbb{R}e^{ity}\int_{\mathbb{R}^{n-1}}f_X\Big(x_1,x_2,\dots,y-\sum_\limits{j=1}^{n-1}x_j\Big)dx_1dx_2\dots dx_{n-1}dy\\
=&\int_{\mathbb{R}^{n-1}}\int_\mathbb{R}e^{ity}f_X\Big(x_1,x_2,\dots,y-\sum_{j=1}^{n-1}x_j\Big)dy\,dx_1dx_2\dots dx_{n-1}\\
=&\int_{\mathbb{R}^{n-1}}\int_\mathbb{R}e^{it\sum_\limits{j=1}^nx_j}f_X\big(x_1,x_2,\dots,x_n\big)dx_ndx_1dx_2\dots dx_{n-1}\ ,
\end{align}
by changing variables in the innermost integral from $\ y\ $ to $\ x_n=y-\sum_\limits{j=1}^{n-1}x_j\ $.
