I am dealing with temperature measurements, and normally we assume the probability of getting a measurement $t_i$ with a certain uncertainty $\sigma_t$ given the model ('true' value) $M(x_i)$ (where $x_i$ is some independent variable, like position) is:
$$P(t_i| M) = N \exp[-\frac{(t_i-M(x_i))^2}{2\sigma_{t}^2}]$$
This is a great approximation but technically incorrect, since temperatures can never be negative, and nor can they ever be 0, so the probability must go to 0 at 0.
I know the usual method for support on the half line with mean and variance gives a truncated Gaussian distribution, but that does not work for me, because it does not go to zero at zero. My intuition leads me to a log-normal, but I haven't been able to prove it to myself.
Any help is appreciated
Best Answer
The lognormal need not be the maximum-entropy distribution with those properties.
Consider $LN(0.0500, 0.1700)$. In Mathematica, I calculate that it has an entropy of $-0.3030183$.
We can also a find a mixture distribution with the same mean, same variance, and same limiting density at 0, by mixing $w$ of $LN(s,0.1701)$ and $1-w$ of $LN(0.0500, 0.1699)$ for appropriate $s$ and $w$. Solving for the variables gives $s=0.0499998$, $w=.502259$, and then the mixture has an entropy of $-0.3030156$. So for that choice of mean and variance, the mixture has higher entropy than the pure lognormal.
The lognormal distribution probably comes close to maximizing entropy, but the exact maximum may be something unfamiliar instead.