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

  • 1
    $\begingroup$ With any continuous distribution the probability of any single number is zero, so truncated normal, theoretically, fulfills this requirement $\endgroup$
    – Firebug
    Commented Oct 12, 2021 at 23:51
  • $\begingroup$ Right, but probability 0 does not mean impossible, which temperature 0 physically is. I would hope the density would go to zero so that it is not only a probability 0 event, but literally an impossible measurement $\endgroup$
    – Craig
    Commented Oct 13, 2021 at 1:17
  • 1
    $\begingroup$ If for physical reasons the probability of temperature doesn't have to go smoothly to zero as the temperature approaches zero you can just define the domain to be $(0, \infty)$ and still use the truncated Gaussian. $\endgroup$
    – jbowman
    Commented Oct 13, 2021 at 1:29
  • 2
    $\begingroup$ The MaxEnt distribution which is constrained to be positive and have a given mean and variance is the truncated normal, and as far as I can tell there is no obvious way to get around this fact. For example, if you constrain the problem so that you require the density to be smooth and differentiable at 0 then there won't be a MaxEnt distribution: you could maximize the entropy up-to some tolerance $\epsilon$, but you would just get a density which approximates the truncated normal you are trying to avoid. $\endgroup$
    – guy
    Commented Oct 13, 2021 at 2:16
  • $\begingroup$ This question confuses probability with probability density. The formula given for $P$ is a density. It needn't go to zero as $t_i\to 0.$ Indeed, it's even OK for it to diverge. $\endgroup$
    – whuber
    Commented Oct 18, 2021 at 16:44

1 Answer 1


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.

  • $\begingroup$ Thank you for taking the time to help me $\endgroup$
    – Craig
    Commented Oct 19, 2021 at 17:25

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