# Transformation Of Random Variables

## Transformation of random variables (for monotone mapping)

Consider the probability density function $f_Y(y)$ of $Y$ when transforming a random variable $X$ into

\begin{align*}
Y = g(X)
\end{align*}

through some function $g(\cdot)$, using the probability density function $f_X(x)$ of $X$.

Theorem 1

Let $f_X(x)$ be the probability density function of the random variable $X$ and $Y = g(X)$. If $g(X)$ is a monotone function and $g^{-1}(Y)$ is differentiable, the probability density function of $Y$ is given by

\begin{eqnarray*}
f_Y(y) = f_X\left(g^{-1}(y) \right) \left| \frac{\mathrm{d}}{\mathrm{d}y}g^{-1}(y) \right|
\end{eqnarray*}

First, let’s think through a concrete example. In the following, we denote the cumulative distribution function of $X$ as $F_X(x)$.

Example 1

For $Y = g(X)$, consider the case when $g(X) = a + b X$. That is, $Y = a + bX$. Let $a, b$ be constants and $b \neq 0$. In this case, express $f_Y$ in terms of $f_X$.

\begin{eqnarray*} \definecolor{myblack}{rgb}{0.27,0.27,0.27} \definecolor{myred}{rgb}{0.78,0.24,0.18} \definecolor{myblue}{rgb}{0.0,0.443,0.737} \definecolor{myyellow}{rgb}{1.0,0.82,0.165} \definecolor{mygreen}{rgb}{0.24,0.47,0.44} \end{eqnarray*}

(i) If $b > 0$

\begin{eqnarray*}
F_Y(y) = P(Y \leq y) &=& P(a + bX \leq y) \\
&=& P\left(X \leq \frac{y\,- a}{b}\right) \\
&=& F_X\left(\frac{y-a}{b}\right)
\end{eqnarray*}

Therefore,

\begin{eqnarray*}
f_Y(y) = \frac{\mathrm{d}}{\mathrm{d}y}F_Y(y) &=& \frac{\mathrm{d}}{\mathrm{d}y}F_X\left(\frac{y-a}{b}\right) \\
&=& \frac{\mathrm{d}}{\mathrm{d}x}F_X\left(\frac{y-a}{b}\right) \cdot \frac{\mathrm{d}x}{\mathrm{d}y} \\
&=& \frac{\mathrm{d}}{\mathrm{d}x}F_X\left(\frac{y-a}{b}\right) \cdot \frac{\mathrm{d}}{\mathrm{d}y} \frac{y\,- a}{b}\\
&=& f_X\left(\frac{y-a}{b}\right) \cdot \frac{1}{b}
\end{eqnarray*}

(ii) If $b \begin{eqnarray*} F_Y(y) = P(Y \leq y) &=& P(a + bX \leq y) \\ &=& P\left(X \textcolor{myred}{\geq} \frac{y\,- a}{b}\right) \\ &=& 1\, – P\left(X \leq \frac{y\,- a}{b}\right) \\ &=& 1- F_X\left(\frac{y-a}{b}\right) \end{eqnarray*} Therefore, \begin{eqnarray*} f_Y(y) = \frac{\mathrm{d}}{\mathrm{d}y}F_Y(y) &=& \frac{\mathrm{d}}{\mathrm{d}y}\left[ 1 \,- F_X\left(\frac{y-a}{b}\right) \right]\\ &=& – \frac{\mathrm{d}}{\mathrm{d}y} F_X\left(\frac{y-a}{b}\right) \\ &\vdots&\\ &=& – f_X\left(\frac{y-a}{b}\right) \cdot \frac{1}{b} \end{eqnarray*} From the above, it follows that \begin{eqnarray*} f_Y(y) = f_X\left(\frac{y-a}{b}\right) \cdot \frac{1}{|b|} \end{eqnarray*} You can see that Example 1$f_Y(y)$is in the form of Theorem 1. Now let us prove Theorem 1. For$Y = g(X)$, (i) When$g(x)$is a monotonically increasing function \begin{eqnarray*} F_{Y}(y)=P(Y \leq y) &=& P(g(X) \leq y)\\ &=&P\left(X \leq g^{-1}(y)\right) \\ &=& F_{X}\left(g^{-1}(y)\right) \end{eqnarray*} \begin{eqnarray*} f_{Y}(y)=\frac{\mathrm{d}}{\mathrm{d} y} F_{Y}(y) &=& \frac{\mathrm{d}}{\mathrm{d} y} F_{X}\left(g^{-1}(y)\right) \\ &=&\frac{\mathrm{d}}{\mathrm{d} x} F_{X}\left(g^{-1}(y)\right) \cdot \frac{\mathrm{d}x}{\mathrm{d}y} \\ &=& f_{X}\left(g^{-1}(y)\right) \frac{d}{d y} g^{-1}(y)\tag{1} \end{eqnarray*} (ii) When$g(x)$is a monotonically decreasing function \begin{eqnarray*} F_{Y}(y)=P(Y \leq y)&=& P(g(X) \leq y)\\ &=&P\left(X \textcolor{myred}{\geq} g^{-1}(y)\right) \\ &=&1- P\left(X \leq g^{-1}(y)\right)\\ &=&1- F_{X}\left(g^{-1}(y)\right) \end{eqnarray*} \begin{eqnarray*} f_{Y}(y)=\frac{\mathrm{d}}{\mathrm{d} y} F_{Y}(y) &=& \frac{\mathrm{d}}{\mathrm{d} y} \left[ 1- F_{X}\left(g^{-1}(y)\right) \right]\\ &=&\textcolor{myred}{-} \frac{\mathrm{d}}{\mathrm{d} x} F_{X}\left(g^{-1}(y)\right) \cdot \frac{\mathrm{d}x}{\mathrm{d}y} \\ &=& f_{X}\left(g^{-1}(y)\right) \left( \textcolor{myred}{-} \frac{d}{d y} g^{-1}(y) \right)\tag{2} \end{eqnarray*} Therefore, from (1) and (2) \begin{eqnarray*} f_Y(y) = f_X\left(g^{-1}(y) \right) \left| \frac{\mathrm{d}}{\mathrm{d}y}g^{-1}(y) \right| \tag{☆} \end{eqnarray*} Another expression of (☆) is the following table expression. \begin{eqnarray*} g\left( g^{-1}(y)\right) = y \end{eqnarray*} Then, by differentiating both sides by$y$, we obtain \begin{eqnarray*} 1 &=& \frac{\mathrm{d}}{\mathrm{d}y}g\left( g^{-1}(y)\right) \\ &=& \frac{\mathrm{d}}{\mathrm{d}g^{-1}}g\left( g^{-1}(y)\right) \cdot \frac{\mathrm{d}g^{-1}}{\mathrm{d}y} \\ &=& g^{\prime}\left( g^{-1}(y)\right) \cdot \frac{\mathrm{d}g^{-1}}{\mathrm{d}y} \\ \\ &\therefore& \frac{\mathrm{d}g^{-1}}{\mathrm{d}y} = \frac{1}{g^{\prime}\left( g^{-1}(y)\right)} \end{eqnarray*} Substituting the above expression into (☆), we obtain \begin{eqnarray*} f_Y(y) = f_X\left(g^{-1}(y) \right) \left| \frac{1}{g^{\prime}\left( g^{-1}(y)\right)} \right| \end{eqnarray*} ## Transformation of random variables (if not monotone mapping) Be careful when$g(\cdot)$is not a monotonic function. For example,$Y = X^2$is not a monotonic function on$\mathbb{R}$. However, it is a monotonic function on$\mathbb{R}^{+}_0$and on$\mathbb{R}^{-}$. Thus, let us divide$g(\cdot)$into intervals where it is a monotonic function. Example 2 For$Y = g(X)$, consider the case when$g(X) = X^2$. That is,$Y = X^2$. In this case, express$f_Y$in terms of$f_X$. Let$\Omega$be the sample space. \begin{eqnarray*} F_Y(y) = P(Y \leq y) &=& P\left(X^2 \leq y \right) \\ \\ &=& P\left(X^2 \leq y\, |\, X(\omega) \geq 0,\, \omega \in \Omega \right) \\ &&\ \ + P\left(X^2 \leq y\, |\, X(\omega) \right) \\ &=& P\left(0 \leq X \leq \sqrt{y} \right) + P\left( -\sqrt{y} \leq X \right) \\ &=& P\left(-\sqrt{y} \leq X \leq \sqrt{y} \right)\\ \\ &=& F_X(\sqrt{y} )\, – F_X(-\sqrt{y}) \end{eqnarray*} \begin{eqnarray*} f_Y(y) = \frac{\mathrm{d}}{\mathrm{d}y}F_Y(y) &=& \frac{\mathrm{d}}{\mathrm{d}y}\left[ F_X(\sqrt{y})\, – F_X(-\sqrt{y}) \right] \\ \\ &=& \frac{\mathrm{d}}{\mathrm{d}x}F_X(\sqrt{y}) \cdot \frac{\mathrm{d}}{\mathrm{d}y}\sqrt{y}\\ &&\ \ – \frac{\mathrm{d}}{\mathrm{d}x}F_X(-\sqrt{y}) \cdot \frac{\mathrm{d}}{\mathrm{d}y}( -\sqrt{y}) \\ \\ &=&\frac{1}{2 \sqrt{y}} f_{X}(\sqrt{y})+\frac{1}{2 \sqrt{y}} f_{X}(-\sqrt{y}) \end{eqnarray*} In light of this example, consider the general$Y = g(X)$case. First, let’s introduce$\mathcal{X} = \{ x | \ f_X(x) > 0 \}$. The$\mathcal{X}$is called the support of$X$. Then the following theorem holds. Theorem 2 Let$f_X(x)$be the probability density function of the random variable$X$and$Y = g(X)$. Let$\{\mathcal{X}_i\}$be a partition of$\mathcal{X}$and$f_X(x)$be continuous in each$\mathcal{X}_i\ (i= 1, \dots, k)$. If the monotone function$g_i(x)$defined on each$\mathcal{X}_i$is$g(x) = g_i(x)$for$x \in \mathcal{X}_i$and$g^{-1}_i(y)$is differentiable, the probability density function of$Y$is given by \begin{eqnarray*} f_Y(y) = \sum_{i=1}^{k} f_{X}\left(g_{i}^{-1}(y)\right)\left|\frac{d}{d y} g_{i}^{-1}(y)\right| \end{eqnarray*} In essence, we are considering dividing into intervals where$g(x)$is a monotonic function. Finally, a simple proof of Theorem 2. \begin{eqnarray*} F_{Y}(y)=P(Y \leq y) &=& \sum_{i=1}^{k} P\left(g(X) \leq y\ | \ X \in \mathcal{X}_{i}\right) \\ &=& \sum_{i=1}^{k} P\left(g_i(X) \leq y \right) \\ \end{eqnarray*} where, for each$g_i(x)\$

\begin{eqnarray*}
P\left(g_i(X) \leq y \right) =
\begin{cases}
P\left(X \leq g^{-1}_i(y) \right) & ({\rm When\ } g_i(x) {\rm \ is\ a\ monotonically\ increasing\ function}) \\
\\
P\left(X \geq g^{-1}_i(y) \right) & ({\rm When\ } g_i(x) {\rm \ is\ a\ monotonically\ decreasing\ function})
\end{cases}
\end{eqnarray*}

Then, by applying the same arguments as in Theorem 1, we obtain

\begin{eqnarray*}
f_Y(y) = \sum_{i=1}^{k} f_{X}\left(g_{i}^{-1}(y)\right)\left|\frac{d}{d y} g_{i}^{-1}(y)\right|
\end{eqnarray*}

.

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