Consider a random variable $Y$ with density function $f_Y(y)$ and moment generating function $m_Y(t)$ and cumulant generating function $\kappa_Y(t)$. Then a random variable $X$ derived from $Y$ by the so-called "exponential tilting" has the following density. For some nonzero $\lambda$,
$$ f_X(x; \lambda) := \frac{e^{\lambda x}f_Y(x)}{m_Y(\lambda)} = e^{\lambda x - \kappa_Y(\lambda)}f_Y(x). $$
Then the question is to verify that the density function for $S_{n,X}:=X_1 +\cdots+X_n$ is $$ f_{S_{n,X}}(s; \lambda) = e^{\lambda s - n\kappa_Y(\lambda)}f_{S_{n,Y}}(s), $$ where $f_{S_{n,Y}}(s)$ means the density function for $S_{n,Y}:=Y_1+\cdots+Y_n$.
This result is only listed in my reading since the authors think it is obvious. However, I do not see why it is true. Could anyone help me, please? Thank you!
Update: Using the convolution argument as suggested in the answer, by induction, I have:
\begin{align*} f_{S_{n,X}}(s; \lambda) &= \int\cdots\int_{\{(x_1, \dots, x_n): x_1+\cdots+x_n =s\}} f(x_1, \dots, x_n) dx_1\cdots dx_n \\ &= \int\cdots\int_{\{(x_1, \dots, x_n): x_1+\cdots+x_n =s\}} f_X(x_1)\dots, f_X(x_n) dx_1\cdots dx_n \\ &= \dots \end{align*}
I do not know how to reduce from this to the integration given in the answer. Or how to understand the integrand given in the answer.
The probability density function for $S_{n,X}$ is the convolution of the probability density functions of $X$ (assuming the $(X_i)_{i=1,\dots,n}$ are i.i.d.): $$ f_{S_n,X}(s)=\int\cdots\int f_X(s-x_1;\lambda)\cdots f_X(x_{n-2}-x_{n-1};\lambda)\,f_X(x_{n-1};\lambda)\,\mathrm dx_1\cdots\mathrm dx_{n-1}, $$ so \begin{align*} f_{S_n,X}&(s)\\ &=\int\cdots\int {\mathrm e}^{\lambda\left((s-x_1)+(x_1-x_2)+\dots+(x_{n-2}-x_{n-1})+x_{n-1}\right)-n\kappa_Y(\lambda)}\,f_Y(s-x_1)\cdots f_Y(x_{n-1})\,\mathrm dx_1\cdots\mathrm dx_{n-1}\\ &={\mathrm e}^{\lambda s-n\kappa_Y(\lambda)}\,(\underbrace{f_Y*\cdots*f_Y}_{n\text{ times}})(s)\\ &={\mathrm e}^{\lambda s-n\kappa_Y(\lambda)}\,f_{S_n,Y}(s). \end{align*}