I have been given a binomial distribution: $$B(m+n;n,p)=\frac{(m+n)!}{m!n!}p^mq^n.$$

Here $m = 10^3$, $n=10^2$, $p=10^{-2}$, $q=1-p.$

I'm using MATLAB to compute log $B(m+n;n,p)$ and store the value in logB


I get logB as NaN . How can I modify the formula to avoid floating point error and get a valid answer?

  • 1
    $\begingroup$ Taking those factorials is probably giving you infinities, the division of which gives you NaN. You need a factorial division function which preemptively cancels. $\endgroup$ – Richard Jun 2 '20 at 13:53
  • 2
    $\begingroup$ You have a ratio of a product of $n$ terms by a product of $n$ terms. Write this as the product of $n$ ratios, each of which should of moderate size. $\endgroup$ – Wolfgang Bangerth Jun 2 '20 at 14:24
  • $\begingroup$ @Richard I'm trying to simplify but still stuck: ((m+n)(m+n-1)...(m+1))/n! $\endgroup$ – Rhombus Jun 2 '20 at 14:27
  • $\begingroup$ @WolfgangBangerth could you elaborate a bit? Thanks. $\endgroup$ – Rhombus Jun 2 '20 at 14:43
  • $\begingroup$ You want to compute $\frac{abc}{def}$ where both enumerator and denominator end up very large numbers. So compute it as $\frac ad\cdot \frac be \cdot \frac cf$ where now each fraction is of moderate size. $\endgroup$ – Wolfgang Bangerth Jun 2 '20 at 18:31

This is why gammaln exists

logB = gammaln(m+n+1) - gammaln(m+1) - gammaln(n+1) + m*log(p) + n*log(q);


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.