The entropy of a Poisson distribution has no closed form. The entropy, in nats, is:
H(λ)=−∑k=0..∞PrThese notes evaluate some numerical approximations to the entropy.
For high firing rates, the Poisson distribution becomes approximately Gaussian with \mu=\sigma^2=\lambda. Using the forumula for the entropy of the Gaussian, this implies
H(\lambda) = \tfrac 1 2 \log(2 \pi e \lambda) + \mathcal{O}\left(\tfrac 1 \lambda \right)Compare this to the fist few terms of the series expression, which is increasingly accurate for large \lambda, but diverges for small \lambda:
H(\lambda) = \frac{1}{2}\log(2 \pi e \lambda) - \frac{1}{12 \lambda} - \frac{1}{24 \lambda^2}-\frac{19}{360 \lambda^3} + O\left(\frac{1}{\lambda^4}\right)Another way to calculate the entropy for low rates is calculate it using a finite number of terms in the series expansion. For low rates, the probability of k> \lambda + 4\sqrt{\lambda} events is neglegible, so we only need to sum a small number of terms.
H(\lambda) \approx - \textstyle\sum_{k=0}^{\lceil \lambda + 4\sqrt{\lambda} \rceil} \frac {\lambda^k e^{-\lambda}}{k!} \left[ k \log(\lambda) - \lambda - \log(k!) \right]Numerically, you can use the above calculation for \lambda<1.78, taking terms out to k<\lceil \lambda + 4\sqrt{\lambda} \rceil. For \lambda\ge 1.78, the third-order approximation for large \lambda becomes accurate.
H(\lambda) \approx \begin{cases} - \textstyle\sum_{k=0}^{\lceil \lambda + 4\sqrt{\lambda} \rceil} \frac {\lambda^k e^{-\lambda}}{k!} \left[k \log(\lambda) - \lambda - \log(k!)\right] & \lambda<1.78\\ \frac{1}{2}\log(2 \pi e \lambda) - \frac{1}{12 \lambda} - \frac{1}{24 \lambda^2}-\frac{19}{360 \lambda^3} & \lambda\ge 1.78 \\ \end{cases}The relative error in this approximation is |\hat H-H|/|H|<0.0016