Computing the cdf of some Discrete Distribution...

Question

I'm trying to find the cdf of the following discrete distribution:

$$f(k)=\frac{(1-\rho )^2 \rho ^{k-1}}{\left(1-\rho ^k\right) \left(1-\rho ^{k+1}\right)}+\frac{(1-\rho ) \rho ^{M-1}}{1-\rho ^M}$$

with domain of support

$$k\in \{1,...,M-1\}, k < M, \mbox{and}\ M \in Z.\ \mbox{Also}, 0<\rho<1.$$

I know its a proper pmf because, when $M=10$, I can compute

$$\sum _{k=1}^9 \left(\frac{(1-\rho )^2 \rho ^{k-1}}{\left(1-\rho ^k\right) \left(1-\rho ^{k+1}\right)}\right)+\frac{(1-\rho ) \rho ^{9}}{1-\rho ^{10}} = 1.$$

I'd like to compute the cdf, but I simply don't know how to do this. I can compute the pmf from the cdf for other distributions, but I can't do it in reverse.

Any help would be greatly appreciated.

Answer 1

Consider the sequences:

$$g_n=(1-\rho)\frac{\rho^{n-1}}{1-\rho^n}~~,~~f_n=(1-\rho)^2\frac{\rho^{n-1}}{(1-\rho^n)(1-\rho^{n+1})}$$

It is easy to see that the sequence $g$ is an "integral" to the sequence $f$, in other words:

$$g_{n+1}-g_{n}=-f_n$$

Then we see that the sum telescopes

$$\sum_{n=1}^{M-1}f_n=g_1-g_M=\frac{1-\rho^{M-1}}{1-\rho^M}=1-(1-\rho)\frac{\rho^{M-1}}{1-\rho^M}$$

This shows us that if the sequence g which is the "integral" is to be considered as a CDF then the corresponding PMF meaning must be assigned to $$t(n;M)=\begin{Bmatrix}f_n&~~0\leq n\leq M-1\\\frac{\rho^{M-1}}{1-\rho^M}&~~n=M\\ 0& \text{otherwise}\end{Bmatrix}$$

The corresponding CDF then is:

$$T(n;M)=\sum_{k=1}^{n}t(k;M)=\begin{Bmatrix}-g_{n+1}+g_1&~~1\leq n\leq M-1\\1&~~n=M\\ \end{Bmatrix}$$

Answer 2

Given

$f(x)= \begin{array}{lc} & \left\{ \begin{array}{cc} \frac{(1-\rho )^2 \rho ^{x-1}}{\left(1-\rho ^x\right) \left(1-\rho ^{x+1}\right)} & x\leq M-1 \\ \frac{(1-\rho ) \rho ^{M-1}}{1-\rho ^M} & x=M \\ \end{array} \\ \right. \end{array} $

We can calculate the cdf by:

$ F(k)=\sum _{x=1}^k f(x) = $

$$ \begin{array}{cc} & \left\{ \begin{array}{lc} 1 & (k=1\land M=1)\lor (M=2\land k\geq 2) \\ \frac{1}{1+\rho} & (k=1\land M\geq 2)\lor (M=2\land k>1\land k<2) \\ \frac{1-\rho ^k}{1-\rho ^{k+1}} & k>1\land k+1\leq M \\ \frac{\rho -\rho ^{ M }}{\rho -\rho ^{ M +1}} & k<M\land k+1>M\land M>2 \\ \frac{(1-\rho) ( M -2) \rho ^{M-1}}{1-\rho ^M} & k>1\land k\geq M\land M\geq 1\land M<2 \\ \frac{(1-\rho) \rho ^M \left(1-\rho ^{ M }\right)+\left(1-\rho ^{M+1}\right) \rho ^{ M }-2 \rho ^{M+1}+\rho ^M+\rho }{\rho \left(1-\rho ^M\right) \left(1-\rho ^{ M }\right)} & k\geq M\land M>2 \\ \end{array} \\ \right. \end{array} $$