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Let me preface this by saying that I'm not familiar with differential equations, other than basic "separable" differential equations. This problem has come up in a Probability problem that I am doing.

Consider a random variable $X$ with the PDF given by $$f(x) = \dfrac{x-1}{2}\text{, } x \in (1, 3)$$ and $0$ elsewhere. I am asked to find a monotonic transformation $Y = u(X)$ such that $Y$ is uniformly distributed in $(0, 1)$.

I am wondering about a more explicit way to attempt this problem, rather than trial-and-error.

By the Method of Transformations (if we further assume that $u^{-1}$ is differentiable), I know that $$f_Y(y) = f_X(u^{-1}(y))\left|\dfrac{\text{d}}{\text{d}y}[u^{-1}(y)] \right| = \dfrac{u^{-1}(y)-1}{2}\cdot\left|\dfrac{\text{d}}{\text{d}y}[u^{-1}(y)] \right| = 1\text{.}$$

If we assume that the derivative of $u^{-1}$ is always positive, setting $h = u^{-1}$, we have the differential equation $$(h-1)h^{\prime} = 2\text{.}$$

WolframAlpha tells me this is a first-order non-linear differential equation, with solutions $$\begin{align} h(y) &= 1-\sqrt{c_1+4y+1}\\ h(y) &= 1+\sqrt{c_1+4y+1}\text{.} \end{align}$$ Which one do I choose? Also, does it matter what initial condition I impose on this?

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3 Answers 3

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A variant of this problem comes up a lot when you're trying to simulate something using a Monte Carlo code. Here's how I would obtain $u(x)$, and it doesn't require the solution of any differential equations:

The CDF $F(x)$ is given by

$$ F(x) = \int_1^x f(w) dw = \frac{(x-1)^2}{4} $$

$F(x)$ is uniformly distributed between 0 and 1 (my wording here may not be correct, but basically it runs from 0 to 1) for $x \in (1,3)$. If we set this equal to $y$, then we obtain:

$$ y = \frac{(x-1)^2}{4} $$

Thus, we can just simply choose

$$ u(x) = F(x) = \frac{(x-1)^2}{4} $$

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  • $\begingroup$ However, in my experience, the more useful result is usually $u^{-1}(x)$, which is given by: $ u^{-1}(y) = \sqrt{4y}+1 $. I think you were kind of getting at this with your differential equation stuff -- this matches $h(y) = 1 + \sqrt{c_1 + 4y + 1}$ with $c_1 = -1$. The reason $u^{-1}$ is useful is because computer codes generally have a function that generates uniformly distributed numbers between (0,1) -- i.e., computers can generate $y$ easily. But, if you want $x$, you need a function that maps $y$ to $x$. That function is $u^{-1}$. $\endgroup$
    – nukeguy
    Jul 19, 2016 at 20:14
  • $\begingroup$ OH, I see what you did - so you exploited the fact that $F(X)$ is uniformly distributed in $(0, 1)$. Clever! $\endgroup$ Jul 19, 2016 at 20:15
  • $\begingroup$ Yep! From the wording of your problem, however, it sounds like you may be more interested in $u^{-1}$. (It's weird that they don't just ask you for a CDF.) See the comment above for my thoughts on it. $\endgroup$
    – nukeguy
    Jul 19, 2016 at 20:16
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The easy way for this problem, as is the case for many pdf problems, is to work with CDF's instead. Here, since $f(x) = \frac{x-1}{2}$ on $(1,3)$, $$ F(x) = \left\{ \array{0 & x\leq1\\ \frac{(x-1)^2}{4} & 1< x < 3 \\ 1 & x\geq 3 }\right. $$ And this needs to match the CDF of the uniform distribution ojn $(0,1)$ $$F(y) = y$$

So $$y= \frac{(x-1)^2}{4} \implies x = 1+\sqrt{4y}$$ and this is the needed transformation.

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$$(h-1)h'=2 $$ is a separable differential equation: it can be simply re-written as $$ \frac{d}{dt}\left(\frac{h^2}{2}-h\right) = 2 $$ from which: $$ (h-1)^2=(4t+C) $$ and $$ h = 1\pm\sqrt{C+4t} $$ readily follow. Have also a look at the generating function for Catalan numbers.

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