Distribution of the product of an inverse uniform distribution and a constant

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: $$p(z | d)$$?

Thanks

• 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?
– whuber
Commented Jun 9, 2023 at 14:23
• @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! Commented Jun 9, 2023 at 15:07

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.

• Thanks @whuber added more clarifications in the question. Commented Jun 10, 2023 at 1:43
• As usual, if you want an expression for the density, just differentiate the distribution function.
– whuber
Commented Jun 10, 2023 at 15:41
• Got it, so $\cfrac{1}{z^2}$ is actually the density. Commented Jun 10, 2023 at 15:57
• 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 $f(z) = 1/z^2.$
– whuber
Commented Jun 10, 2023 at 16:01