# Joint probability density function of N composite random variables

If I got some random independent variables $$X_1,X_2,...,X_n$$ with all with identical PDFs, for example $$f_{X_i}(x_i) = e^{-x_i}$$. Now I have some new set of random variables, where $$S_k=\sum_i^k X_i, k=1,2,...n$$, how do I find the joint PDF of these new variables, like the PDF of $$S_1$$?

I would think because $$X_i$$ are all independent that I should just multiple all the PDFs in order to find the joint PDF: \begin{align} f_{S_1}(x_1) &= f_{X_1}(x_1)\\ f_{S_2}(x_1,x_2) &= f_{X_1}(x_1)\cdot f_{X_2}(x_2)\\ &...\\ f_{S_n}(x_1,...,x_n) &= f_{X_1}(x_1)\cdot ...\cdot f_{X_n}(x_n) \end{align} or maybe because the way the random variable is defined they should be added: \begin{align} f_{S_1}(x_1) &= f_{X_1}(x_1)\\ f_{S_2}(x_1,x_2) &= f_{X_1}(x_1)+f_{X_2}(x_2)\\ &...\\ f_{S_n}(x_1,...,x_n) &= f_{X_1}(x_1)+...+f_{X_n}(x_n) \end{align}

Which one of these is correct or am I way off?

• The density of $S_n$ depends on the 1-dimensional variable $s$: $f_{S_n}(s)$. So $f_{S_n}(x_1, ..., x_n)$ does not make sense. We have $f_{S_n}(s) = d/ds P[S_n\leq s]$. Aug 31, 2020 at 4:35

$$\newcommand{\bbx}{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}{\mathcal{#1}} \newcommand{\mrm}{\mathrm{#1}} \newcommand{\pars}{\left(\,{#1}\,\right)} \newcommand{\partiald}[]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}{\left\vert\,{#1}\,\right\vert}$$ \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty}\ldots\int_{0}^{\infty}\mrm{f}_{X_{1}}\pars{x_{1}}\ldots \mrm{f}_{X_{n}}\pars{x_{n}}\delta\pars{S - \sum_{\ell = 1}^{n}x_{\ell}}\dd x_{1}\ldots\dd x_{n}} \\[5mm] = &\ \int_{0}^{\infty}\ldots\int_{0}^{\infty}\mrm{f}_{X_{1}}\pars{x_{1}}\ldots \mrm{f}_{X_{n}}\pars{x_{n}}\ \times \\[2mm] &\ \bracks{\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic} \expo{\pars{S - \sum_{\ell = 1}^{n}x_{\ell}}s}{\dd s \over 2\pi\ic}} \dd x_{1}\ldots\dd x_{n} \\[5mm] = &\ \bbx{\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic}\expo{Ss} \bracks{\int_{0}^{\infty}\mrm{f}_{X}\pars{x}\expo{-sx}\dd x}^{n} {\dd s \over 2\pi\ic}}\label{1}\tag{1} \\ & \end{align} For example, with $$\ds{\mrm{f}_{X}\pars{x} = \expo{-x}}$$; (\ref{1}) is reduced to \begin{align} & \mbox{} \\ &\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic}{\expo{Ss} \over \pars{x + 1}^{n}}\,{\dd s \over 2\pi\ic} = {1 \over \pars{n - 1}!}\lim_{s \to -1}\totald[n - 1]{\expo{Ss}}{s} = \bbx{S^{n - 1}\expo{-S} \over \pars{n - 1}!} \\ & \end{align}