I have read that, in the context of matrix factorization, performing maximum likelihood estimation under the assumption that the entries are Poisson generated is equivalent to minimizing the generalized Kullback-Leibler divergence between the matrix and the product of its factors.

Does anyone know how to show this?

(for an example of this claim, see, e.g. page 2 of Lee and Seung's paper on NMF in Nature.)

I worked it out in the end and I'll link to it here, in case someone else is interested. I wrote it up here http://building-babylon.net/2015/02/17/maximum-likelihood-estimation-for-non-negative-matrix-factorisation-and-the-generalised-kullback-leibler-divergence/

We want to proof that \begin{align} argmin_{W,H} \qquad D_{KL}(\boldsymbol{V} | \boldsymbol {WH}) \quad = \quad argmax_{W,H} \qquad p(\boldsymbol{V} | \boldsymbol{W}\boldsymbol{H}) \end{align} under a Poisson distribution.

The KL divergence (actually the I-divergence) is defined as \begin{align} D_{KL}(\boldsymbol{V} | \boldsymbol{W}, \boldsymbol{H}) = \sum_f \sum_n \left[ v_{fn} \ln \frac{v_{fn}}{\sum_k w_{fk}h_{kn}} + \sum_k w_{fk}h_{kn} - v_{fn} \right] \end{align} And the likelihood can be expressed in terms of the KL divergence: \begin{align} \ln p(\boldsymbol{V} | \boldsymbol{W}\boldsymbol{H}) &= \sum_{f,n} \ln \left[ \exp\left\lbrace -\sum_k w_{fk}h_{kn}\right\rbrace \frac{(\sum w_{fk}h_{kn})^{v_{fn}}}{v_{fn}!} \right] \\ &= \sum_{f,n} \left[ {v_{fn}} \ln\sum w_{fk}h_{kn} -\sum_k w_{fk}h_{kn} - \ln v_{fn}! \right] \\ &= \sum_{f,n} \left[ {v_{fn}} \ln\frac{\sum w_{fk}h_{kn}}{v_{fn}} -\sum_k w_{fk}h_{kn} - \ln v_{fn}! +v_{fn}\ln v_{fn} \right] \\ &= -D_{KL}(\boldsymbol{V} | \boldsymbol{W}, \boldsymbol{H}) - v_{fn} + \left[ v_{fn}\ln v_{fn} - \ln v_{fn}! \right] \end{align} Therefore, both the likelihood and the KL divergence have the same optimum with respect to $\boldsymbol{W}, \boldsymbol{H}$.

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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