# Why constrain mean and standard deviation when proving Gaussian is maximum differential entropy pdf?

I'm reading Bishop's Pattern Recognition and Machine Learning. In chapter 1.6: Information Theory (page 53) when trying to derive the maximum differential entropy pdf from the definition of continuous entropy, the author states:

"Let us now consider the maximum entropy configuration for a continuous variable. In order for this maximum to be well defined, it will be necessary to constrain the first and second moments of p(x) as well as preserving the normalization constraint" (where p(x) is the pdf we wish to learn the form of)

My two questions:

1. What does it mean for a maximum to be well-defined? Intuition + math please!
2. Why does constraining the first and second moments of p(x) ensure this maximum is well defined - i.e., why do we not need to constrained all moments.

Thanks!

• It is instructive to find a maximum entropy distribution subject to a constraint only on its mean--or even subject to no constraints at all. What would it look like?
– whuber
Commented Sep 24, 2014 at 21:20
• Hm - i suspect you are hinting that the resulting pdf would just be an inproper distribution (so a horizontal line if x is one dimensional)? However surely this could be avoided by constraining integral over p(x) = 1? For example, cauchy distribution has an undefined mean if I recall correctly? Commented Sep 25, 2014 at 7:02