Let's assume I have a measurement $d + \epsilon_d$ where the true value of the measurement is $d$ and the error associated with that measurement $\epsilon_d$ has a uniform distribution $\epsilon_d \sim U(-0.5, 0.5)$.

I have a random variable that is a function of that measurement and error: $Z = \cfrac{B * f}{d + \epsilon_d}$ where $B$ and $f$ are constants.

  • What is the distribution of $Z$?
  • What is the posterior probability $p(z|d)$?

I looked at the wikipedia article on inverse uniform distributions, but wasn't sure how to apply that to this problem.

More clarifications:

$(B * f) > 0$, and the support of $d + \epsilon_d$ in the interval $[d - \frac{1}{2}, d + \frac{1}{2}]$ is also always positive.

When I try to workout the algebra using the wikipedia notation, I end up with something like:

Define random variable $X = \frac{d + \epsilon_d}{B * f}$.

Define reciprocal of random variable $Z = \frac{1}{X} = \frac{B * f}{d + \epsilon_d}$.

Define support $a = d - \frac{1}{2}$ and $b = d + \frac{1}{2}$.

Probability density function is (per wiki): $g(z) = \frac{1}{z^2} * \frac{1}{b - a} = \frac{1}{z^2}$

Therefore, $g(z) = \cfrac{(d + \epsilon_d)^2}{B^2 * f^2} = \cfrac{1}{z^2}$

  • Does this look like the correct distribution?
  • And if so, how do I go from the expression for this distribution to the conditional distribution expression: $p(z | d)$?


  • $\begingroup$ This is strange for two reasons. First, is $d$ the measurement or, as suggested by your formula for $Z,$ the true value being measured? If $\delta$ is the true value then the measured value is $d=\delta+\epsilon_d,$ whence $d+\epsilon_d = \delta + 2\epsilon_d,$ so the distinction is important. Second, why use the complex expression "$Bf$" for the numerator unless there are some unstated special relationships among $B,$ $f,$ $d,$ and $\epsilon_d$? If there aren't, then where is the difficulty in applying the Wikipedia article? $\endgroup$
    – whuber
    Commented Jun 9, 2023 at 14:23
  • 1
    $\begingroup$ @whuber, yes, you are correct. Edited for clarity. measurement is $d + \epsilon_d$. $B$ and $f$ are simply constants. But I'm still not sure how to apply the wikipedia article. If you have an answer, open to it. Thank you! $\endgroup$ Commented Jun 9, 2023 at 15:07

1 Answer 1


I am going to take some liberties interpreting this question (and its helpful title) as asking,

For constants $C$ and $d$ and uniformly distributed random variable $\epsilon$ supported on $[-1/2,1/2],$ what is the distribution of $Z = C/(d+\epsilon)$?

Before we go on, the Wikipedia article derives the distribution of $1/U$ where $U$ has a uniform distribution on the interval $[0, t]$ with $t\gt 0.$ This is done by computing

$$\Pr\left(\frac{1}{U}\le y\right) = \Pr\left(U \ge \frac{1}{y}\right) = \frac{1}{t}\max\left(0, t - \frac{1}{y}\right) = \max\left(0, 1 - \frac{1}{yt}\right)\tag{*}$$

for all numbers $y\ge 0$ (since obviously $\Pr(1/U\le y)=0$ for all negative $y$). We'll capitalize on this at the very end.

Returning to the problem at hand, applying the same basic definition works best: the value of the distribution function of $Z$ at any number $z$ is

$$F_Z(z) = \Pr(Z \le z) = \Pr\left(\frac{C}{d+\epsilon}\le z\right).$$

To simplify the algebra it would be nice to clear the denominator, but that's made tricky because when it's negative, the inequality is reversed. So, to that end, consider the following:

  • When $C=0,$ $Z=0$ constantly, solving the problem. From now on then assume $C\ne 0.$

  • When $C\lt 0,$ the event $C/(d+\epsilon)\le z$ is equivalent to $1/(d+\epsilon)\ge z/C.$ Otherwise, when $C\gt 0,$ the equivalent event is $1/(d+\epsilon)\le z/C.$

  • The support of the denominator $d+\epsilon$ is the interval $[d-1/2, d+1/2].$ If this is entirely positive or entirely negative, things are relatively simple; but otherwise we need to break the interval into its negative and positive parts. An elegant way to handle this, with a minimum of fuss, is to express the distribution of $d+\epsilon$ as a mixture of a positively supported and negatively supported distribution. (When a variable $X$ has a distribution function $F_X$ with weight $q$ and $Y$ has a distribution function $F_Y$ with weight $p,$ the distribution function of their weighted mixture is $qF_X + pF_Y.$)

Specifically, let $l = \min(0, d-1/2)$ and $u=\max(0, d+1/2)$, so that $[l,u]$ covers all the numbers of possible relevance in this analysis. For any two numbers $a$ and $b,$ let $U(a,b)$ designate a uniform distribution on the interval $(\min(a,b),\max(a,b))$ (so we don't care what order $a$ and $b$ might be in). By checking the three cases $u\lt 0,$ $l\le 0 \le u,$ and $l\gt 0,$ you can readily verify that a weighted mixture of a $U(l,0)$ variable with weight $q = 1/2-d$ and a $U(0,u)$ variable with weight $p=d+1/2$ has the same distribution as $d+\epsilon.$

(Any experts who have been following along might object that $p$ or $q$ can be negative. That's correct -- but the math still works out, because $p+q=1$ and, because we know we're computing a valid distribution function (that of $Z$), we are guaranteed the expressions we are about to derive will give a legitimate distribution function.)

To deal with the uniform distributions on $(l,0]$ when $l\lt 0,$ negate the random variable. This will convert the event $1/X \le z/C$ to $1/(-X)\ge -z/C$ (notice the change in the direction of the inequality) and $1/X \ge z/C$ becomes $1/(-X)\le -z/C.$

This analysis has reduced the problem to finding some probabilities that $1/X\le \pm z/C$ or $1/X\ge \pm z/C$ where $X$ has a uniform distribution on a positive interval anchored at zero -- either $(0,u)$ or $[0,-l].$ Apply the results $(*)$ in the Wikipedia article directly in either case and combine those results using the definition of a weighted mixture. The only challenge is to keep track of the $\pm$ signs and the directions of the inequalities.

If any of this looks tricky or abstract, I recommend plotting the distributions involved in each of the three cases.

  • $\begingroup$ Thanks @whuber added more clarifications in the question. $\endgroup$ Commented Jun 10, 2023 at 1:43
  • $\begingroup$ As usual, if you want an expression for the density, just differentiate the distribution function. $\endgroup$
    – whuber
    Commented Jun 10, 2023 at 15:41
  • $\begingroup$ Got it, so $\cfrac{1}{z^2}$ is actually the density. $\endgroup$ Commented Jun 10, 2023 at 15:57
  • $\begingroup$ It's the right idea but not quite correct: you have to be specific about the support of the function. A defining fact about any density $f$ is that $\int_{\infty}^{+\infty}f(x)\,\mathrm dx = 1,$ but that's not the case for $f(z) = 1/z^2.$ $\endgroup$
    – whuber
    Commented Jun 10, 2023 at 16:01

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