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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

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