0
$\begingroup$

My understanding of a truncated normal distribution $\mathcal{N}(\mu,\sigma;a,b)$ is that it results from scaling the values of a normal distribution within the bounds $[a; b]$ such that the area under the curve within these bounds becomes 1 (please correct me, if I'm wrong).

However, in a case where I want to describe a variable $z$ that results from a latent variable $y$ (with normally distributed noise $\sigma$) such that

$ z = \begin{cases} a & \text{if $y<a$}\\ b & \text{if $y>b$}\\ y & \text{otherwise} \end{cases} $

.. the truncated normal distribution in the above form would not be an appropriate distribution for $z$. Indeed, as $\sigma$ increases, the pdf within $]a;b[$ should not be scaled such that the area under the curve is 1 (hence the title "without scaling"); rather, the probability that $z$ is exactly $a$ or exactly $b$ should increase.

Now let's say, I'm not interested in the density per se, but rather in computing the probability $p(x; \epsilon)$ that the true value of whatever $z$ is measuring is $x$ ($\pm \epsilon$). Assuming that I measured $z=\mu$ and that I know the measurement noise $\sigma$, then I think that $p(x; \epsilon)$ can be described as

$ p(x; \epsilon) = \begin{cases} \int\limits_{-\infty}^{a+\epsilon}\mathcal{N}(\mu,\sigma) & \text{if $x=a$}\\ \int\limits_{b-\epsilon}^{\infty}\mathcal{N}(\mu,\sigma) & \text{if $x=b$}\\ \int\limits_{x-\epsilon}^{x+\epsilon}\mathcal{N}(\mu,\sigma) & \text{otherwise} \end{cases} $

  1. Is this correct?
  2. Is there a name for this form of $p(x;\epsilon)$ or even a name for the underlying probability density, which would seems like a special form of a truncated normal distribution? (-> would help researching on this)
  3. Are there other probability distributions that represent are able to represent a similar scenario as outlined above?
$\endgroup$
3
  • $\begingroup$ As you correctly point out, there's a mix of continuous and discrete densities here, in that values a and b exhibit a point-like non-zero probability, which is an infinite in the density. In the signal processing courses this was represented as a Dirac's delta function, which is always zero except that it is infinite at a point (the origin) and the integral is 1. You can take a look here: en.wikipedia.org/wiki/Dirac_delta_function#Probability_theory $\endgroup$ – polettix Aug 27 '20 at 8:47
  • $\begingroup$ Part of it depends on how exactly a and b are defined. In analytical chemistry a is usually set such that it is the lowest value that can be statistically distinguished from zero. I.e. a usually includes the measurement error of a low value observation plus the error of a blank measurement. It is also common to set values below a to a/2. This means a already includes the $\epsilon$ term. $\endgroup$ – ReneBt Aug 27 '20 at 8:48
  • $\begingroup$ Also, since z is imputed below a and above b you cannot assume that $\mu$=a or b is true for the underlying distribution. The probability of where it would lie beyond those limits depends on the probability characteristics of the population of samples the procedure is applied to.. $\endgroup$ – ReneBt Aug 27 '20 at 8:48
3
$\begingroup$

The observed $z$ is left and right censored (as opposed to truncated), that is, rather than never seeing observations smaller than $a$ and larger than $b$ (truncation), you observe $z=a$ equivalent to the event $y<a$ and $z=b$ equivalent to $y>b$. The contribution to the likelihood from such observations would be $$ \begin{cases} \Phi(\frac{a-\mu}\sigma) & \text{for }z=a \\ \frac1\sigma\phi(\frac{z-\mu}\sigma) &\text{for }a< z <b\\ 1-\Phi(\frac{b-\mu}\sigma) & \text{for }z=b, \end{cases} $$ where $\phi$ and $\Phi$ is the pdf and cdf of the standard normal distribution. Note that the events that $y=a$ and $y=b$ both have probability zero when $y$ is continuously distributed so you don't need to add anything extra to the probabilities in the first and third cases above to account for the fact that $z$ can take the values $a$ or $b$ without $y$ being censored.

Another thread has a detailed discussion on this kind of likelihood involving a mix between discrete and continuous observations.

$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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