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?

Question

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.

Answer 1

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\ $.