Let $X$ be a random variable uniformly distributed over a nontrival interval $[c,d]$, and let $Y = aX+b$. For what choice of real constants $a$ and $b$ is $Y$ uniformly distributed over [0,1]?

How could I fully comprehend this question? This is my interpretation so far:

$Y$ is a function of X. Does that mean the interval of $Y \subset X$? If yes, does it the interval of $Y$ MUST ALWAYS be a subset of $X$

From $Y = aX+b$ we know that the CDF of Y, $F_Y(y)$ = $$ P(Y\leq y) \\ P(aX+b \leq y) \\ P(X \leq \frac{y-b}{a}) $$ Hence, $F_Y(y)$ = $F_x(\frac{y-b}{a})$

So, for Y to be uniformly distributed over [0,1], $$a > 0 \\ y-b > 0 \\ b < y $$

Let me know if I have the right approach to the problem!


1 Answer 1


All you need to care about is if $X = c$, then $Y = 0$, and if $X = d$, then $Y = 1$. So in particular, we want the solution to the system $$0 = ac + b, \quad 1 = ad + b$$ for $a, b$ given $c, d$. This gives us $$a = \frac{1}{d-c}, \quad b = -\frac{c}{d-c}.$$ No solution is possible if $c = d$, but this is assumed to not be possible.

Of course, this is not a unique solution because there is no stipulation that the transformation $Y = aX + b$ must also be order-preserving. That is to say, you could also solve the system $$1 = ac + b, \quad 0 = ad + b$$ and the resulting transformation would also give $Y \sim \operatorname{Uniform}(0,1)$.

  • $\begingroup$ thank you for the explanation but is $[0,1] \subset [c,d]$? $\endgroup$
    – misheekoh
    Mar 6, 2016 at 21:25
  • 1
    $\begingroup$ @misheekoh No. The transformation works for any $c < d$, so it is not a requirement that $[0,1] \subset [c,d]$. $\endgroup$
    – heropup
    Mar 6, 2016 at 21:34

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.