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.
Best 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*}