# Finding Probability Generating function for $P\left\{ X > n+1\right\}$

I am trying to find probability generating function for $P\left\{ X > n+1\right\}$.

Let X be a random variable assuming the values $0, 1, 2, ...$. The notation both for the distribution of $X$ and for it's tails are $P\left\{ X = j\right\} = p_j$, $P\left\{ X > j\right\} = q_j$. So the generating functions of the sequences $\{p_j\}$ and $\{q_j\}$ are $$P(s) = p_0 + p_1s+ p_2s^2 + p_3s^3+...$$ and $$Q(s) = q_0 + q_1s+ q_2s^2 + q_3s^3+...$$

As $P(1) = 1$, the series for $P(s)$ converges absolutely at least for $-1 \leq s \leq 1$. The coefficients of $Q(s)$ are less than unity, and so the series for $Q(s)$ converges at least in the open interval $-1 < s < 1$.

Also for $-1 < s < 1$ there is a known identity which provides the relation $$Q\left( s\right) =\dfrac {1-P\left( s\right) } {1-s}$$

In most of similar problems the approach i have been taking is to convert the desired probability into some form of an algebraic equation of $P\left\{ X = j\right\}$, $P\left\{ X > j\right\}$ and may be 1. Then substitute in one of the known generating function and solve for an expression of the generating function of the desired probability.

I am unsure this time this idea is working as i can n't seem to shake off of the tail due to +1 on the n.

$$P\left\{ X > n+1\right\} = 1-P\left\{ X \leq n+1\right\}$$ $$=1 -P\left\{ X = n+1\right\}-P\left\{ X < n+1\right\}$$ I am unsure if should further expand $P\left\{ X < n+1\right\}$ part as it would keep unfolding recursively.

Any help would be much appreciated.

-
The GF is usually as a sum over non-negative integers, because it's assumed that a sequence is regarded as a function with domain in $n=0, 1\cdots$. If we call $s_n = P(X > n+1)$ , there is an ambiguity (what about $n=-1$?). So, you should make clear of what GF you are referring to. –  leonbloy Aug 4 '12 at 21:15
I am afraid I do not understand what is the question. –  Did Aug 4 '12 at 21:19
@did I am trying to find a generating function for $P(X>n+1)$ so the probability is such that X takes values higher than n. In the form of $P(s)$ and $Q(s)$ described above. –  Hardy Aug 4 '12 at 21:23
@leonbloy i think like you said n can never be -1. It only value for natural numbers. –  Hardy Aug 4 '12 at 21:25
Since the GF for $P(X\gt n)$ is $Q(s)$, are you not simply after $(Q(s)-q_0)/s$? –  Did Aug 4 '12 at 21:25

Let $H(s)=\sum\limits_{n=0}^{+\infty}\mathrm P(X\gt n+1)\,s^n$, then $$sH(s)=\sum_{n=1}^{+\infty}\mathrm P(X\gt n)\,s^{n}=\sum_{n=0}^{+\infty}\mathrm P(X\gt n)\,s^{n}-\mathrm P(X\gt0)\cdot1=Q(s)-q_0$$
Using the identity in the question and $q_0= 1-p_0$. We can state $Q(s) - q_0 = P_{0}+\dfrac {1-\dfrac {P\left( s\right) } {s}} {1-s}$. The answer provided is $\dfrac{P_{0}}{s}+\dfrac {1-\dfrac {P\left( s\right) } {s}} {1-s}$. I am guessing there must be a typo in the answer provided. –  Hardy Aug 5 '12 at 0:11
@Hardy: No, the answer is fine; you seem to have made a mistake. Using the identity in the question and $q_0=1-p_0$ yields $Q(s)-q_0=p_0+\frac{s-P(s)}{1-s}$, and then dividing through by $s$ gives the answer provided. –  joriki Aug 5 '12 at 6:51