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Suppose I have some continuous random variable $X$. Further, suppose I am interested about a transformed random variable $Y = g(X)$ where $g$ is some increasing function. If I know the CDF of $X$, I can find that the CDF of $Y$ is $F_{Y}(y) = F_{X}(g^{-1}(y))$. To find the PDF of $Y$, I just have to find the derivative of $F_{Y}$.

However, the resulting PDF does not (usually?) integrate to one. Instead, I have to determine some constant $c$ to make it such. Why is it that the constant is lost in the process?

Also, suppose $X$ was defined over some interval $[a, b]$. The random variable $Y$ might be non-zero over some other interval $[c, d]$. So, when taking the above-mentioned normalizing integral, should I take it as $\int_{a}^{b}$ or $\int_{c}^{d}$? Why?


So suppose $X$ is uniformly distributed over the interval $[-1, 1]$. Let $g(x) = 10x + 5$ and $Y = g(X)$. So $$F_{Y}(y) = P(Y\leq y) = P(g(X)\leq y) = P(10X+5\leq y) = P(10X \leq y-5) \\ = P\Big(X\leq \frac{y-5}{10}\Big) = F_{X}\Big(\frac{y-5}{10}\Big).$$

The PDF is then $$f_{Y}(y) = \frac{d}{dy} F_{Y}(y) = F_{X}\Big(\frac{y-5}{10}\Big) = f_{X}\Big(\frac{y-5}{10}\Big)\frac{d}{dy}\Big(\frac{y-5}{10}\Big) \\ = \frac{1}{2}\cdot\frac{1}{10} = \frac{1}{20}.$$

Now $$\int_{-1}^{1} \frac{1}{20} dy = \frac{y}{20}\Bigg|_{-1}^{1} = \frac{1}{10}$$ which is not one. Why am I missing a factor of ten?

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    $\begingroup$ Try working out a specific example e.g. X uniform on [0, 1], and Y_1 = 10X + 5. Or try Y_2 = X^2. $\endgroup$ – Adrian Oct 19 '15 at 17:57
  • $\begingroup$ What exactly is unclear for you in the change of variable method? I assume $g^{-1}$ is well-defined for your transform, ie that $g$ is increasing to satisfy the relation $F_{Y}(y) = F_{X}(g^{-1}(y))$. $\endgroup$ – Xi'an Oct 19 '15 at 18:51
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    $\begingroup$ If the derivative of a CDF does not integrate to unity, then you have made a mistake in differentiation, because (by the Fundamental Theorem of Calculus) the integral will equal $$\lim_{y\to\infty}F_X(g^{-1}(y)) - \lim_{y\to -\infty}F_X(g^{-1}(y)) = 1 - 0 = 1.$$ $\endgroup$ – whuber Oct 19 '15 at 19:06
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    $\begingroup$ Hint: the range of $y$ is not the same as the range of $x$. $\endgroup$ – EdM Oct 19 '15 at 20:37
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    $\begingroup$ If $x$ is defined on [-1,1], and $y=10x+5$, then what are the limits on $y$ for your integral? $\endgroup$ – EdM Oct 19 '15 at 20:42

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