Transformation of Random Vector

Definition

$\mathbf{Y}=g(\mathbf{X})$

Discrete

Let $\mathbf{X}$ be a Random Vector with joint pmf $p_{\mathbf{X}}(\mathbf{x})$ and support $S$, and $g:\mathbf{X}\to \mathbf{Y}$ be a vector Bijective transformation $p_{\mathbf{Y}}(\mathbf{y})=P(\mathbf{Y}=\mathbf{y})=p(g(\mathbf{X})=\mathbf{y})=P(\mathbf{X}=g^{-1}(\mathbf{y}))=P_{\mathbf{X}}(g^{-1}(\mathbf{y}))$

Continuous

Let $\mathbf{X}$ be a Random Vector with joint pdf $f_{\mathbf{X}}(\mathbf{x})$ and support $S$, $g:\mathbf{X}\to \mathbf{Y}$ be a vector Bijective transformation, and $\mathbf{x}=g^{-1}(\mathbf{y})$

One-to-One Transformation Case

$f_{\mathbf{Y}}(\mathbf{y})=f_{\mathbf{X}}(g^{-1}(\mathbf{y}))\cdot |J|,\mathbf{y}\in S_{\mathbf{Y}}$ where $J=\frac{\partial \mathbf{x}}{\partial \mathbf{y}}=\left[\begin{array}{ccc}\frac{\partial x_1}{\partial y_1}& \cdots & \frac{\partial x_1}{\partial y_n}\\ ⋮& \ddots & ⋮\\ \frac{\partial x_m}{\partial y_1}& \cdots & \frac{\partial x_m}{\partial y_n}\end{array}\right]$ is a Jacobian Matrix, and $S_{\mathbf{Y}}=\{\mathbf{y}=g(\mathbf{x})|\mathbf{x}\in S_{\mathbf{X}}\}$

Many-to-One Transformation Case

Let $g:\mathbf{X}\to \mathbf{Y}$ be a many-to-one transformation, $\mathbf{x}=g^{-1}(\mathbf{y})$, $A_1,\dots ,A_k$ be a partition of a set $S_X={⨄}_{i=1}^k{A}_i$ whose each partition’s transformation result is one-to-one Then, $f_{\mathbf{Y}}(\mathbf{y})=\sum _{i=1}^k|J_i|f_{\mathbf{X}}(g_i^{-1}(\mathbf{y})),\mathbf{y}\in S_{\mathbf{Y}}$ where $J_i=\frac{\partial {\mathbf{x}}_i}{\partial \mathbf{y}}=\left[\begin{array}{ccc}\frac{\partial x_{1i}}{\partial y_1}& \cdots & \frac{\partial x_{1i}}{\partial y_n}\\ ⋮& \ddots & ⋮\\ \frac{\partial x_{mi}}{\partial y_1}& \cdots & \frac{\partial x_{mi}}{\partial y_n}\end{array}\right]$ is a Jacobian Matrix for each partition, and $S_{\mathbf{Y}}=\{\mathbf{y}=g(\mathbf{x})|\mathbf{x}\in S_{\mathbf{X}}\}$, and $\uplus$ is a union of the pairwise disjoint sets

CDF Technique

If $g$ is increasing, then $F_{\mathbf{Y}}(\mathbf{y})=P(\mathbf{Y}\le \mathbf{y})=P(g(\mathbf{X})\le \mathbf{y})=P(\mathbf{X}\le g^{-1}(\mathbf{y}))=F_{\mathbf{X}}(g^{-1}(\mathbf{y}))$

If $g$ is decreasing, then $F_{\mathbf{Y}}(\mathbf{y})=P(\mathbf{Y}\le \mathbf{y})=P(g(\mathbf{X})\le \mathbf{y})=P(\mathbf{X}\ge g^{-1}(\mathbf{y}))=1-F_{\mathbf{X}}(g^{-1}(\mathbf{y}))$