Jacobian¶

Obtaining the pdf of a transformed variable (using a one-to-one transformation) is simple using the Jacobian (Jacobian of inverse)

\[ \begin{aligned} Y &= g(X)\newline X &= g^{-1}(Y)\newline f_{Y}(y) &= f_{X}(g^{-1}(y)) \lvert \frac{dx}{dy} \rvert \end{aligned} \]

The modulus ensures that the probability density is positive when the transformation is either increasing or decreasing.

For the two variable case, let \(\mathbf{X} = (X_{1}, X_{2})\) denote the random vector. Suppose we have a one-to-one transform (and its inverse) defined as follows

\[ \begin{aligned} \mathbf{Y} &= T(\mathbf{X})\newline \mathbf{Y} &= (Y_{1}, Y_{2})\newline Y_{1} &= u_{1}(X_{1}, X_{2})\newline Y_{2} &= u_{2}(X_{1}, X_{2})\newline T^{-1} &= (w_{1}(Y_{1}, Y_{2}), w_{2}(Y_{1}, Y_{2}))\newline X_{1} &= w_{1}(Y_{1}, Y_{2}\newline X_{2} &= w_{2}(Y_{1}, Y_{2} \end{aligned} \]

Then, the pdf is given by the expression

\[ \begin{aligned} f_{Y}(y) &= f_{X}(w_{1}(y_{1}, y_{2}), w_{2}(y_{1}, y_{2})) \lvert \mathbf{J} \rvert \newline \mathbf{J} &= \begin{vmatrix} \frac{\partial x_{1}}{y_{1}} &\frac{\partial x_{1}}{y_{2}}\newline \frac{\partial x_{2}}{y_{1}} &\frac{\partial x_{2}}{y_{2}} \end{vmatrix} \end{aligned} \]

If \(A\) denotes the support of \(\mathbf{X}\), then \(B = T(A)\) will denote the support of \(\mathbf{Y}\). Everywhere else, the value of pdf for \(Y\) will be zero. Refer to the exercises section for a demonstration of this method.

This method is powerful and applies to change of variables when doing integrals as well.

\[ \begin{aligned} \int \int f_{X}(x_{1}, x_{2}) dx_{1}dx_{2} = \int \int f_{Y}(w_{1}(y_{1}, y_{2}), w_{2}(y_{1}, y_{2})) \lvert J \rvert dy_{1}dy_{2} \end{aligned} \]

where \(J\) is the same as defined earlier.

The two variable expression extends to multiple variables as well. Suppose the transformations are given by

\[ \begin{aligned} Y_{i} &= u_{i}(X_{1}, \ldots X_{n}) \quad \forall i = 1, 2, \ldots, n\newline X_{i} &= w_{i}(Y_{1}, \ldots Y_{n}) \quad \forall i = 1, 2, \ldots, n\newline \implies f_{\mathbf{Y}}(\mathbf{y}) &= f_{\mathbf{X}}(w_{1}(\mathbf{y}), \ldots, w_{n}(\mathbf{y})) \detm{\mathbf{J}}\newline \text{with}\quad \mathbf{J} &= \begin{vmatrix} \frac{\partial x_{1}}{\partial y_{1}} &\cdots &\frac{\partial x_{1}}{\partial y_{n}}\newline \vdots &\ddots &\vdots\newline \frac{\partial x_{n}}{\partial y_{1}} &\cdots &\frac{\partial x_{n}}{\partial y_{n}}\newline \end{vmatrix} \end{aligned} \]

However, we may not be always have a one-to-one transformation. The simplest case is \(Y = X^{2}\). In such scenarios, we will partition the space so that each has a one-to-one transform.

For instance, in the above example the partitions will be \(-\infty < x < 0\) and \(0 \leq x < \infty\). Then, we individually calculate the pdf for \(Y\) in all the partitions. If the partitions have an overlap (from the view of support of \(y\)), we will sum up the corresponding pmf to get the final pmf.

Continuing with the same example, both the partitions map to \(0 <\leq y < \infty\). Hence, after finding the pmf in each partition, we can simply sum it up (due to symmetry of the transform). Refer to the exercises for an illustration of this principle.