Quiero hacer que la respuesta de @ Glen_b sea más explícita, aquí hay una respuesta adicional solo porque no cabe como un comentario.
f(x)∝N(x|−1/2λ1/λ2,−1/(2λ2))
For the unbounded variable, you can explicitly solve for the Lagrange multipliers
λ1 and
λ2 in terms of the constraint values (
a1,a2 in the Wikipedia article). With
a1=μ,a2=μ2+σ2, you then get
λ1=μ/σ2,λ2=−0.5σ2, so the standard Gaussian
N(x|μ,σ2).
For the bounded variable x>xmin, I (and mathematica) cannot solve for λ1,2 explicitly anymore because of the error function term that appears when computing the partition function (1/c in wikipedia). This means that that the μ and σ2 parameters of the truncated Gaussian are not the mean and variance of the continuous variable you started with. It can even happen that for xmin=0, the mode of the Gaussian is negative! Of course the numbers all agree again when you take xmin→−∞.
If you have concrete values for a1,a2, you can still solve for λ1,2 numerically and plug in the solutions into the general equation and you are done! The values of λ1,2 from the unbounded case may be a good starting point for the numerical solver.
This question is a duplicate of /math/598608/what-is-the-maximum-entropy-distribution-for-a-continuous-random-variable-on-0