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For a discrete random variable $X$, we have known the exact expression of its PGF $G(z)=E[z^X]$.

The question is how can I get $Pr\{X>k\}$ from this PGF.

I want to have an explicit expression of the probability in terms of $G(z)$.

The Z-Transform is too complex to use. And an explicit expression is preferred rather than a upper bound. Is there some other way?


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I found in'Univariate Discrete Distributions, 3ed' another possible solution to this problem. $\Pr\{X\geq k\}=\sum_{j=k}^\infty (-1)^{j+k} C_{j-1}^{k-1} \frac{G^{(j)}(z)|_{z=1}}{j!}. – Severals-user45972 Oct 26 '12 at 4:07
up vote 2 down vote accepted

Using complex integration over the direct unit circle $S^1$, one gets $$\mathbb P(X\gt k)=\frac1{2\mathrm i\pi}\oint_{S^1}\frac{1-G(z)}{1-z}\cdot\frac{\mathrm dz}{z^{k+1}}. $$

To prove this, one can rely on two ingredients:

(1) A distribution transform

For every nonnegative integer valued random variable $X$ integrable and not almost surely zero, there exists a random variable $Y$ such that, for every $k\geqslant0$, $$ \mathbb P(Y=k)=\frac{\mathbb P(X\gt k)}{\mathbb E(X)}. $$ This is because each RHS is nonnegative and their sum is $1$ thanks to the well-known formula $$ \sum_{k\geqslant0}\mathbb P(X\gt k)=\mathbb E(X). $$ If the PGF of $X$ is the function $G$ defined by $G(z)=\mathbb E(z^X)$, then $\mathbb E(X)=G'(1)$ and the PGF of $Y$ is the function $F$ defined by $$ F(z)=\mathbb E(z^Y)=\frac{1-G(z)}{\mathbb E(X)\cdot(1-z)}. $$ (2) The "extraction" of coefficients of an entire series through Cauchy formula

For every nonnegative integer valued random variable $Y$ with PGF $F$ and every $k\geqslant0$, $$ \mathbb P(Y=k)=\frac1{2\mathrm i\pi}\oint_{S^1}F(z)\cdot\frac{\mathrm dz}{z^{k+1}}. $$ To see this, expand $F$ as $F(z)=\sum\limits_{n\geqslant0}\mathbb P(Y=n)z^n$ and use the fact that, for every $n\geqslant0$, $$ \oint_{S^1}z^n\cdot\frac{\mathrm dz}{z^{k+1}}=\left\{\begin{array}{ccc}2\mathrm i\pi & \text{if} & n=k,\\ 0 & \text{if} & n\ne k.\end{array}\right. $$ Using (1) and (2) together, one gets $$ \frac{\mathbb P(X\gt k)}{\mathbb E(X)}=\frac1{2\mathrm i\pi}\oint_{S^1}\frac{1-G(z)}{\mathbb E(X)\cdot(1-z)}\cdot\frac{\mathrm dz}{z^{k+1}}, $$ which is equivalent to the desired formula.

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Is this from the Z-Transform thoery? Why the image unit $i$ is not involved? can you give me some reference on this? Thanks. – Severals-user45972 Oct 25 '12 at 9:33
See edited version. – Did Oct 25 '12 at 10:12
Excellent! Thanks! – Severals-user45972 Oct 25 '12 at 10:27
However, I think a factor of $\frac{1}{2\pi i}$ is missed in the formula. – Severals-user45972 Oct 26 '12 at 4:04
Indeed. Thanks. – Did Oct 26 '12 at 4:52

If $$G(z) = \operatorname{E} (z^X) = \sum_{x=0}^{\infty}\Pr(X=x)z^x$$ then to identify $\Pr(X=x)$ for a particular (non-negative integer) $x$, you take the $x$th derivative of $G(z)$ at $0$ and divide by $x!$.

Or use standard generating function techniques, such as those described in generatingfunctionology.

So for $\Pr(X \gt k)=1 -\Pr(X \le k)$ you can use $$1-\sum_{x=0}^{k}\Pr(X=x) =1- \sum_{x=0}^{k}\frac{G^{(x)}(0)}{x!}$$

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Thanks. But this gives only a particular value. But I want to get $\Pr\{X>k\}$, which is a sum of infinite $p(s)$s. What can I do then? – Severals-user45972 Oct 25 '12 at 9:21
@Severals-user45972: bit added – Henry Oct 25 '12 at 9:26

This solution is for the positive RV, i.e. RV that takes only positive values. In this case one can show that $$ \sum_{k \geq 0}\mathbf{P}(X \geq k)=\mathbf{E}X=G'_{X}(0) $$

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Yes, you are right. – Severals-user45972 Oct 25 '12 at 9:37
What is $G'_0(X)$? – Did Oct 25 '12 at 10:11
Edited:$\left. \frac{dG_{X}(z)}{dz} \right|_{z=0}$ – Alex Oct 25 '12 at 10:16

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