My textbook, Introduction to Probability, first edition, by Blitzstein and Hwang, says the following:

In a location-scale transformation, starting with $X \sim \text{Unif}(a, b)$ and transforming it to $Y = cX + d$ where $c$ and $d$ are constants with $c > 0$, $Y$ is a linear function of $X$ and Uniformity is preserved: $Y \sim \text{Unif}(ca + d, cb + d)$. But if $Y$ is defined as a nonlinear transformation of $X$, then $Y$ will not be linear in general. For example, for $X \sim \text{Unif}(a, b)$ with $0 \le a < b$, the transformed r.v. $Y = X^2$ has support $(a^2, b^2)$ but is not Uniform on that interval.

Firstly, isn't what the authors describe here as a linear function actually an affine function - not linear (that is, isn't $Y = cX + d$ an affine function -- not a linear function)?

And lastly, how is $Y = X^2$ not Uniform on the interval $0 \le a < b$? I'm having difficulty understanding how this is the case.

I would greatly appreciate it if people could please take the time to clarify this.


Yes, the authors are saying 'linear' for 'affine'.

If $X$ has uniform distribution on $(0,1)$ then $Y=X^{2}$ does not have uniform distribution on $(0,1)$: $P(Y \leq y)= P(X \leq \sqrt y)=\sqrt y$ for $0<y<1$. Note that $X^{2} <X$. $X^{2}$ assigns higher probabilites than $X$ for values near $0$.

  • $\begingroup$ "Note that $X^{2} <X$. $X^2$ assigns higher probabilities than $X$ for values near $0$"; can you please explain this? $\endgroup$ – The Pointer Oct 1 '19 at 12:24
  • 1
    $\begingroup$ @ThePointer Unform distribution assigns same probability for intervals of equal length. But $X^{2}$ doe not do this. For example consider $(0,r)$ and $(1-r ,r)$. The probability that $X^{2}$ lies in the first interval is $\sqrt r$ whereas he probability that $X^{2}$ lies in the second interval is $1- \sqrt (1-r)$. You can easily verify that $1- \sqrt (1-r) <\sqrt r$ $\endgroup$ – Kavi Rama Murthy Oct 1 '19 at 12:28
  • $\begingroup$ How did you get that the probability that $X^2$ lies in the second interval is $1- \sqrt (1-r)$? $\endgroup$ – The Pointer Oct 1 '19 at 12:45
  • 1
    $\begingroup$ @ThePointer $P(a<X<b)=b-a$ for $0\leq a<b\leq 1$. So P(1-r <X^{2} <1)=P(\sqrt {1-r} <X <1)=1-(1-\sqrt {1-r})$. $\endgroup$ – Kavi Rama Murthy Oct 1 '19 at 12:47
  • $\begingroup$ Ahh, ok, I understand now. Thank you for taking the time to clarify this. $\endgroup$ – The Pointer Oct 1 '19 at 12:49

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.