Definition: Derived Distribution

X is a continuous random variable. Given the PDF of $X$, and that $Y = g(X)$. The PDF of $Y$ is called a derived distribution.

To calculate the derived distribution, the principal method for doing so is the following two-step approach.

1. Calculate the CDF $F_Y$ of $Y$ using the formula $$F_Y(y) = P(g(X) \leq y) = \int_{ \set{ x \mid g(x) \leq y }} f_X(x) \d x$$

2. Differentiate to obtain the PDF of $Y$: $$f_Y(y) = \frac{ \d F_Y }{ \d y } (y)$$

Example: 4.1

Let $X$ be uniform on $[0, 1]$, and let $Y = \sqrt{ X }$. Want to know PDF of $y$.

1. Calculate the CDF $F_Y$ of $Y$ using the formula \begin{align*} F_Y(y) &= P(\sqrt{ x } \leq y) = P(x \leq y^2) \br &= \int_{ -\infty}^{ y^2 } \sqrt{ x } \d x \br &= y^2, 0 \leq y \leq 1 \end{align*}

2. Differentiate to obtain the PDF of $Y$: $$f_Y(y) = F’_Y(y) = \frac{ \d }{ \d y } y^2 = 2y, 0 \leq y \leq 1$$

Outside the range $[0, 1]$, the CDF $F_Y(y)$ is constant, with $F_Y(y) = 0$ for $y \leq 0$, and $F_Y(y) = 1$ for $y \geq 1$. i.e.

$$F_Y(y) = \begin{cases} 0 & y < 0 \br y^2 & 0 \leq y \leq 1 \br 1 & y > 1 \end{cases}$$

By differentiating, we see that $f_Y(y) = 0$ for $y$ outside $[0,1]$.

$$f_y = \begin{cases} 0 & y < 0 \br 2y & 0 \leq y \leq 1 \br 0 & 1 < y \end{cases}$$

Corollary : The Linear Case

Let $X$ be a continuous random variable with PDF $f_(X)$, and let $Y$ be a linear function of $X$, i.e.

$$Y = aX + b$$

where $a$ and $b$ are scalars, with $a \neq 0$. Then,

$$f_Y(y) = \frac{ 1 }{ \abs{ a }} f_X(\frac{ y - b }{ a })$$

Proof [expand]

Assuming $a > 0$.

1. Calculate the CDF $F_Y$ of $Y$ using the formula \begin{align*} F_Y(y) &= P(Y \leq y) \br &= P\pare{aX + b \leq y} \br &= P\pare{X \leq \frac{ y-b }{ a }} \br &= F_X\pare{ \frac{ y-b }{ a }} \end{align*}

2. Differentiate to obtain the PDF of $Y$ $$f_Y(y) = \frac{ \d F_Y }{ \d y } (y) = \frac{ 1 }{ a } f_X \pare{ \frac{ y-b }{ a }}$$

It’s easy to verify in a similar way that when $a < 0$, $$f_Y(y) = \frac{ \d F_Y }{ \d y } (y) = -\frac{ 1 }{ a } f_X \pare{ \frac{ y-b }{ a }}$$

$\therefore f_Y(y) = \frac{ 1 }{ \abs{ a }} f_X \pare{ \frac{ y - b }{ a }}$

Corollary : The Monotonic Case

Suppose that $g$ is strictly monotonic and that for some function $h$ and all $x$ in the range of $X$ we have

$$y = g(x) \text{ if and only if } x = h(y)$$

Assume that $h$ is differentiable. Then, the PDF of $Y$ in the region where $f_Y(y) > 0$ is given by

$$f_Y(y) = f_X(h(y)) \abs{ \frac{ \d h }{ \d y } (y)}$$