Transformations of Random Variable - 2
“확률과 통계(MATH230)” 수업에서 배운 것과 공부한 것을 정리한 포스트입니다. 전체 포스트는 Probability and Statistics에서 확인하실 수 있습니다 🎲
Theorem.
Let $X$ and $Y$ be independent RVs.
1. Discrete Case
If $X$ and $Y$ are discrete with pmfs $f_X(x)$ and $f_Y(y)$,
then $X+Y$ has the pmf
\[f_{X+Y} (z) = \sum_u f_X(u) f_Y (z-u)\]2. Continuous Case
If $X$ and $Y$ are continuous with pdfs $f_X(x)$ and $f_Y(y)$,
then $X+Y$ has the pdf
\[f_{X+Y}(z) = \int_u f_X(u) f_Y(z-u) du\][2]번 명제의 경우, 증명하려면 상당한 엄밀성이 요구된다고 정규 수업에서는 [2]번에 대한 증명을 다루진 않았다.
Proof. [1]
따라서,
\[f_V (v) = \sum_u f_{U, V} (u, v) = \sum_u f_X(u) f_Y(v-u)\]$\blacksquare$
Example.
Let $X \sim \text{Gamma}(n, \beta)$, $Y \sim \text{Gamma}(m, \beta)$, and $X \perp Y$.
Find the pdf of $X+Y$.
이때, $u = zy$로 치환적분해보자.
\[\begin{aligned} f_{X+Y} (Z) &= \frac{1}{\Gamma(n) \cdot \beta^n} \frac{1}{\Gamma(m) \cdot \beta^m} \cdot e^{-z/\beta} \cdot \int^z_0 u^{n-1} (z-u)^{m-1} \, du \\ &= \frac{1}{\Gamma(n) \cdot \beta^n} \frac{1}{\Gamma(m) \cdot \beta^m} \cdot e^{-z/\beta} \cdot \int^1_0 (zy)^{n-1} (z-zy)^{m-1} \, z dy \\ &= \frac{1}{\Gamma(n) \cdot \beta^n} \frac{1}{\Gamma(m) \cdot \beta^m} \cdot e^{-z/\beta} \cdot z^{n-1} z^{m-1} z \cdot \int^1_0 y^{n-1} (1-y)^{m-1} \, dy\\ &= \frac{1}{\Gamma(n) \cdot \beta^n} \frac{1}{\Gamma(m) \cdot \beta^m} \cdot e^{-z/\beta} \cdot z^{n+m-1} B(n, m) \\ &= \frac{1}{\cancel{\Gamma(n)} \cdot \beta^n} \frac{1}{\cancel{\Gamma(m)} \cdot \beta^m} \cdot e^{-z/\beta} \cdot z^{n-m-1} \frac{\cancel{\Gamma(n)} \cancel{\Gamma(m)}}{\Gamma(n+m)} \\ &= \frac{1}{\Gamma(n+m) \beta^{n+m}} \cdot z^{n+m - 1} e^{-z\beta} \end{aligned}\]따라서, $X+Y \sim \text{Gamma}(n+m, \, \beta)$인 것이다! $\blacksquare$
위의 예제를 활용하면, $\text{Gamma}(n, \beta)$가 $\text{Exp}(\beta)$의 generalization임을 쉽게 확인할 수 있다.
Let $X_1, \dots, X_n$ follows $\text{Exp}(\beta)$ and they are mutually independent.
We know $\text{Gamma}(1, \beta) = \text{Exp}(\beta)$.
Therefore,
\[(X_1 + \cdots + X_n) \sim \text{Gamma}(n, \beta)\]not 1-1 Transform
$y = g(X)$ and $g$ is not 1-1
Example.
Let $X$ be a discrete RV with $P(X=0) = P(X=2) = P(X=-2) = 1/3$.
Let $Y := X^2$. Find $f_Y(y)$.
Since $Y = X^2$, $Y = \{ 0, 4\}$
따라서,
(1) If $y=4$, then $x=2$ or $x=-2$
\[P(Y = 4) = P(X^2 = 4) = P(X=2) + P(X = -2) = \sum_{x: g(x)=4} f_X (x) = 2/3\](2) If $y=0$, then $x=0$
\[P(Y = 0) = P(X^2 = 0) = P(X=0) = \sum_{x: g(x) = 0} f_X (x) = 1/3\]Example.
Let $X \sim N(0, 1)$. Find the pdf of $Y := X^2$.
\[\begin{aligned} P(Y \le y) &= P(X^2 \le y) \\ &= P(-\sqrt{y} \le X \le \sqrt{y}) \\ &= P(X \le \sqrt{y}) - P(X \le -\sqrt{y}) \\ &= F_X (\sqrt{y}) - F_X (-\sqrt{y}) \end{aligned}\]위의 cdf 식에서 pdf를 유도해보면,
\[\begin{aligned} f_Y(y) &= \frac{d}{dy} P(Y \le y) \\ &= \frac{d}{dy} F_X (\sqrt{y}) - \frac{d}{dy} F_X (-\sqrt{y}) \\ &= f_X (\sqrt{y}) \frac{1}{2\sqrt{y}} + f_X (-\sqrt{y}) \frac{1}{2\sqrt{y}} \\ &= \frac{1}{\sqrt{y}} \cdot f_X (\sqrt{y}) \qquad (\text{$f_X$가 우함수}) \end{aligned}\]Theorem.
Supp. $X$ has pmf or pdf $f_X (x)$.
Let $Y = g(X)$ where $g$ may not be 1-1.
Assume that the support for $f_X(x)$ can be partitioned into $k$ segments s.t., in each segment, $x_i = w_i(y)$ is 1-1 for $i=1, \dots, k$.
Then,
(1) when $X$ is discrete, $\displaystyle f_Y (y) = \sum^k_{i=1} f_X (w_i (y))$
(2) when $X$ ic continuous, $\displaystyle f_Y (y) = \sum^k_{i=1} f_X (w_i(y)) \cdot \left| w’_i (y) \right|$
이어지는 포스트에서는 RV의 momentum인 $E[X]$, $E[X^2]$를 생성하는 함수인 <MGF; Momemtum Generating Function>에 대해 다룬다. 🤩