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?

  • $\begingroup$ Got something from the answer below? $\endgroup$ – Did Aug 30 '13 at 6:53
  • $\begingroup$ @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. $\endgroup$ – yyzhang Aug 30 '13 at 8:10
  • $\begingroup$ Then it is time to "close" the question, no? $\endgroup$ – Did Aug 30 '13 at 9:18
  • $\begingroup$ @Did Ok, it is time. $\endgroup$ – 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 $1\leqslant k\leqslant n$.

Not sure we can be more specific than that, unfortunately.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.