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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.

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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 $0$.

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  • $\begingroup$ "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
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    $\begingroup$ @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
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    $\begingroup$ @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

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