# Uniform distributions: location-scale transformation

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.

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 $$0. Note that $$X^{2} . $$X^{2}$$ assigns higher probabilites than $$X$$ for values near $$0$$.
• "Note that $X^{2} <X$. $X^2$ assigns higher probabilities than $X$ for values near $0$"; can you please explain this? – The Pointer Oct 1 '19 at 12:24
• @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 $1- \sqrt (1-r) <\sqrt r$ – Kavi Rama Murthy Oct 1 '19 at 12:28
• How did you get that the probability that $X^2$ lies in the second interval is $1- \sqrt (1-r)$? – The Pointer Oct 1 '19 at 12:45
• @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})\$. – Kavi Rama Murthy Oct 1 '19 at 12:47