# Distribution of sum of mixed random variables

Suppose there are $n$ i.i.d mixed random variables $U_1,U_2,\cdots,U_n$. Each has a mass of probability $e^{-\tau}$ at $0$ and a pdf $$f(u)=\left\{\begin{matrix} e^{-u}, & 0<u\leq \tau\\ 0, & u>\tau \end{matrix}\right..$$

Define $X_n=\sum_{i=1}^{n}U_i$. I want to find out the probability distribution of $X_n$. I have determined the distribution for $n=2$ ,which has a probability $e^{-2\tau}$ at $0$ and a pdf as follows$$f_{X_2}(x)=\left\{\begin{matrix} (x+2e^{-\tau})e^{-x}, & 0<x\leq \tau\\ (2\tau-x)e^{-x}, & \tau<x\leq 2\tau \\ 0,& otherwise \end{matrix}\right..$$

I am trying to find out the distribution of $X_n$ recursively according to $X_n=X_{n-1}+U_n$. But this makes it even more difficult. Can anyone help me?

• Got something from the answer below? – Did Aug 30 '13 at 6:53
• @Did, yeah, I think it is the same idea as mine. I am afraid I cannot go further. I am thinking about another way, using the Laplace transform and inverse Laplace transform. – yyzhang Aug 30 '13 at 8:10
• Then it is time to "close" the question, no? – Did Aug 30 '13 at 9:18
• @Did Ok, it is time. – yyzhang Aug 31 '13 at 3:46

The distribution $X_n$ has weight $\mathrm e^{-n\tau}$ at $0$ and density $x\mapsto p_n(x)\mathrm e^{-x}$ elsewhere, where $p_1=\mathbf 1_{(0,\tau)}$ and, for every $n\geqslant1$, $$p_{n+1}=\mathrm e^{-\tau}p_n+\mathrm e^{-n\tau}p_1+p_1\ast p_n=\mathrm e^{-n\tau}\mathbf 1_{(0,\tau)}+\mathrm e^{-\tau}p_n+\mathbf 1_{(0,\tau)}\ast p_n,$$ where $\ast$ denotes convolution. For example, the case $n=1$ reads $$p_{2}=2\mathrm e^{-\tau}\mathbf 1_{(0,\tau)}+\mathbf 1_{(0,\tau)}\ast\mathbf 1_{(0,\tau)},$$ which coincides with the distribution of $X_2$ you computed.
In general, the function $p_n$ is nonzero on $(0,n)$ only and coincides with a given polynomial function of degree at most $n-1$ on each interval $(k-1,k)$ for $1\leqslant k\leqslant n$.