9231 · 4.5
Probability generating functions flashcards
Revision flashcards for Cambridge 9231 Probability generating functions (syllabus 4.5). Flip, recall, then mark a real past-paper question.
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What is the definition of the Probability Generating Function (PGF) for a discrete random variable $X$?
$G_X(t) = E(t^X) = \sum_{x=0}^{\infty} P(X=x)t^x$. It's a power series in a dummy variable $t$, where the coefficient of $t^x$ is the probability $P(X=x)$.
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What is the value of $G_X(1)$ for any PGF?
$G_X(1) = \sum P(X=x) = 1$, since the sum of all probabilities in a distribution is 1.
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How do you find the mean, $E(X)$, using a PGF?
Differentiate the PGF with respect to $t$ and evaluate at $t=1$. Formula: $E(X) = G_X'(1)$.
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How do you find the variance, $Var(X)$, using a PGF?
$Var(X) = G_X''(1) + G_X'(1) - [G_X'(1)]^2$. This is a standard formula to memorise.
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How can you find the probability $P(X=r)$ from its PGF, $G_X(t)$?
Method 1: Find the coefficient of $t^r$ in the series expansion of $G_X(t)$. Method 2: Use the formula $P(X=r) = \frac{G_X^{(r)}(0)}{r!}$, where you evaluate the $r$-th derivative at $t=0$.
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What is the PGF of a Bernoulli distribution, $X \sim B(1, p)$?
$G_X(t) = q + pt$, where $q = 1-p$. This is because $P(X=0)=q$ and $P(X=1)=p$.
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What is the PGF of a Binomial distribution, $X \sim B(n, p)$?
$G_X(t) = (q+pt)^n$, where $q=1-p$. This comes from the fact that a Binomial RV is the sum of $n$ independent Bernoulli trials.
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What is the PGF of a Geometric distribution, $X \sim Geo(p)$ (number of trials)?
$G_X(t) = \frac{pt}{1-qt}$, where $q=1-p$. (Note: for number of failures before first success, it's $\frac{p}{1-qt}$).
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What is the PGF of a Poisson distribution, $X \sim Po(\lambda)$?
$G_X(t) = e^{\lambda(t-1)}$. This is derived from the Poisson probability formula and the series for $e^x$.
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If $X$ and $Y$ are independent random variables, what is the PGF of their sum, $Z=X+Y$?
The PGF of the sum is the product of the individual PGFs: $G_Z(t) = G_{X+Y}(t) = G_X(t)G_Y(t)$.
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Common Trap: What is the difference between $G_X'(0)$ and $G_X'(1)$?
$G_X'(0) = P(X=1)$, the probability that the variable takes the value 1. In contrast, $G_X'(1) = E(X)$, the mean of the distribution. Do not mix them up.
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Common Trap: Forgetting a term in the variance formula.
The variance is $Var(X) = G_X''(1) + G_X'(1) - [G_X'(1)]^2$. A frequent error is to forget the $+G_X'(1)$ term. Remember it as $E(X(X-1)) + E(X) - [E(X)]^2$.