PROBABILITIES Course

Generating function

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When dealing with combinatorics, generating functions (Fonctions génératrices) can make your life easier!

You will write a function, in which having $a_k * x^k$ means that

  • for $n=k$
  • the number of distributions is $a_k$

For instance, let's say your result is $5 x^2 + 4 x^3 + ...$ then that's means that if $n=2$ then the number of distributions is $5$...

It's usually defined like this

\[ G(x) = \sum_{k \in N} a_k * x^k \]

Creating a generating function

If we got $k$ issues and $m$ experiments ($m = Card(X)$ for instance) then we would have

\[ G(x) = \prod_{j=1}^{m} \sum_{i=k_j}^{n_j} x^i \]
  • this is a product of sums
  • for each value of X (indexed by $j$)
  • we calculate a sum,
  • from $k_j$ (the minimum of times you want this value)
  • to $n_j$ (the maximum of times you want this value)
  • once you multiplied and factorized the result, $a$ in $a * x^i$ is the number of distributions for $n=i$.

You may also use the other methods as explained on these websites


Example (1)

Let's say you are rolling two dices. You got $m=2$ and each experiment goes from $1$ to $6$, so you have $k_1=k_2=1$ (min) to $n_1=n_2=6$ (max). We will have

\[ \begin{split} G(x) = (\sum_{i=1}^{6} x^i) * (\sum_{i=1}^{6} x^i) = (x+x^2+x^3+x^4+x^5+x^6)^2 \\ = x^2 + 2 x^3 + 3 x^4 + 4 x^5 + 5 x^6 + 6 x^7 + 5 x^8 + 4 x^9 + 3 x^{10} + 2 x^{11} + x^{12} \end{split} \]

You can develop that easily using websites (😂) like wolframalpha.

If you are asked the number of issues/the cardinal

  • when the sum is $7$ then it's $6$
  • when the sum is $8$ then it's $4$
  • ...

Okay, you could actually find that yourself like for the $sum=7$ then we have $(1,6),\ (2,5),\ (3,4)$ so that's 6 (without the order like $1+6=6+1$), but a generating function is scalable and that was not that hard (aside from developing the function 😬).


Example (2)

You have $n=12$ cakes (chocolate, vanilla, strawberry),

  • each person must have at least two flavors
  • and they can't have chocolate more than 4 times

Since we need "at least two", then $k_1=k_2=k_3=2$. Aside from $n_1$, we don't have restrictions on vanilla/strawberry, so $n_2=n_3=12$. As for chocolate, $n_1=4$ (since "up to" 4).

\[ \sum_{i=2}^{12} x^i * \sum_{i=2}^{12} x^i * \sum_{i=2}^{4} x^i \]

is evaluated (using wolframalpha 😶) as

@ x^{28} + ... + 18 x^{12} + ... + x^6 @

So we have 18 ways of distributing our cakes (since $n=12$). If $n=6$ then it would be 1, etc.