Problem
Let $f: \mathbb{R}^2 \to \mathbb{R}^2$ with $f(x,y) = (e^{2x+y}, e^{x+y})$.
Compute the density function $\frac{df[\lambda_2]}{d\lambda_2}$ of the pushforward measure $f[\lambda_2]$, where $\lambda_2$ is the Lebesgue measure on $\mathcal{B}(\mathbb{R}^2)$.
My approach
For $f[\lambda_2]: \mathcal{B}(\mathbb{R}^2) \to [0,\infty]$ we have $f[\lambda_2](A) = \lambda_2(f^{-1}[A])$. We just need to compute $f[\lambda_2](Q)$ for a closed rectangle $Q = [a_1,b_1]\times[a_2,b_2] \subset f\left[\mathbb{R}^2\right] = \left( \mathbb{R}^+ \right)^2$, since the set of all closed rectangles generates $\mathcal{B}(\mathbb{R}^2)$.
\begin{align*}
(x,y) \in f^{-1}\big[ [a_1,b_1]\times[a_2,b_2] \big] \Leftrightarrow\ &\begin{cases}
a_1 \leq e^{2x+y} \leq b_1\\
a_2 \leq e^{x+y} \leq b_2
\end{cases}\\
\Leftrightarrow\ &\begin{cases}
\ln a_1 \leq 2x+y \leq \ln b_1\\
\ln a_2 \leq x+y \leq \ln b_2
\end{cases}\\
\Leftrightarrow\ &\begin{cases}
\ln a_1 – \ln a_2 \leq x \leq \ln b_1 – \ln b_2\\
2\ln a_2 – \ln a_1 \leq y \leq 2\ln b_2 – \ln b_1
\end{cases}\\
\Leftrightarrow\ &\begin{cases}
\ln \frac{a_1}{a_2} \leq x \leq \ln \frac{b_1}{b_2}\\
\ln \frac{a_2^2}{a_1} \leq y \leq \ln \frac{b_2^2}{b_1}
\end{cases}\\
f^{-1}\big[ [a_1,b_1]\times[a_2,b_2] \big] =\ &\left[\ln\frac{a_1}{a_2}, \ln \frac{b_1}{b_2}\right]\times\left[\ln\frac{a_2^2}{a_1}, \ln \frac{b_2^2}{b_1}\right]\\
\lambda_2\left( f^{-1}\big[ [a_1,b_1]\times[a_2,b_2] \big] \right) =\ &\left( \ln \frac{b_1}{b_2} – \ln\frac{a_1}{a_2} \right)\left( \ln \frac{b_2^2}{b_1} – \ln\frac{a_2^2}{a_1} \right)\\
=\ &\ln\frac{a_2b_1}{a_1b_2} \ln \frac{a_1b_2^2}{a_2^2b_1}
\end{align*}
For $\frac{df[\lambda_2]}{d\lambda_2}: \mathbb{R}^2 \to \mathbb{R}$ we have:
$$\int_{a_2}^{b_2} \int_{a_1}^{b_1} \frac{df[\lambda_2]}{d\lambda_2}(x,y) dx dy \overset{!}{=} \lambda_2\left( f^{-1}\big[ [a_1,b_1]\times[a_2,b_2] \big] \right)$$
I am unable to find such function $\frac{df[\lambda_2]}{d\lambda_2}(x,y)$.
Questions
- Did I correctly justify to only focus on closed rectangles in $f[\mathbb{R}^2]$?
- Did I calculate the pushforward measure correctly?
- How can I calculate the density function?
Edit: Solution
\begin{align*}
(x_1,x_2)\in f^{-1}[\{(y_1,y_2)\}]\Leftrightarrow&
\begin{cases}
e^{2x_1+x_2}=y_1\\
e^{x_1+x_2}=y_2
\end{cases}\\
\Leftrightarrow&
\begin{cases}
2x_1+x_2=\ln y_1\\
x_1+x_2=\ln y_2
\end{cases}\\
\Leftrightarrow&
\begin{cases}
x_1=\ln y_1-\ln y_2=\ln\frac{y_1}{y_2}\\
x_2=2\ln y_2-\ln y_1=\ln\frac{y_2^2}{y_1}
\end{cases}\\
f^{-1}[\{(y_1,y_2)\}] = &
\begin{cases}
\left\{\left(\ln\frac{y_1}{y_2},\ln\frac{y_2^2}{y_1}\right)\right\} & \text{if } y_1,y_2 > 0\\
\emptyset & \text{else}
\end{cases}
\end{align*}
Let $y_1,y_2 > 0$:
\begin{align*}
Df(x_1,x_2) = &
\begin{bmatrix}
2e^{2x_1+x_2}&e^{2x_1+x_2}\\
e^{x_1+x_2}&e^{x_1+x_2}
\end{bmatrix}\\
Df\left(\ln\frac{y_1}{y_2},\ln\frac{y_2^2}{y_1}\right) = &
\begin{bmatrix}
2y_1&y_1\\
y_2&y_2
\end{bmatrix}\\
\left| \det Df\left(\ln\frac{y_1}{y_2},\ln\frac{y_2^2}{y_1}\right)\right| = &\left| 2y_1y_2 – y_1y_2 \right| = \left| y_1y_2 \right| = y_1y_2\\
\end{align*}
For $\frac{df[\lambda_2]}{d\lambda_2}:\mathbb{R}^2 \to \mathbb{R}$ we have:
\begin{align*}
\frac{df[\lambda_2]}{d\lambda_2}(y_1,y_2) = &\sum_{(x_1,x_2)\in f^{-1}[\{(y_1,y_2)\}]} \frac{1}{\left| \det Df\left(x_1,x_2\right)\right|}\\
= &
\begin{cases}
\frac{1}{y_1y_2} & \text{if } y_1,y_2>0\\
0 & \text{else}
\end{cases}
\end{align*}
Best Answer
I'll give a method that works for any $C^1$ function $f$ whose derivative is invertible almost everywhere (notably, $f$ does not have to be one-to-one). By the change of variables theorem (see theorem 3.9 on page 122 of "Measure Theory and Fine Properties of Functions" by Evans and Gariepy ), for any Lebesgue-measurable $A \subset \mathbb{R}^2$, \begin{align} \lambda_2(f^{-1}(A)) &= \int_{\mathbb{R}^2}1_{A}(f(x))\,dx \\ &= \int_{\mathbb{R}^2}1_{A}(f(x))\frac{1}{Jf(x)}Jf(x)\,dx \\ &= \int_{\mathbb{R}^2}\sum_{x \in f^{-1}(\{y\})}1_{A}(f(x))\frac{1}{Jf(x)}\,dy \\ &= \int_{A}\sum_{x \in f^{-1}(\{y\})}\frac{1}{Jf(x)}\,dy, \\ \end{align} where $$Jf(x) = |\det Df(x)|.$$ Hence the density of $\lambda \circ f^{-1}$ is $$g(y) = \sum_{x \in f^{-1}(\{y\})}\frac{1}{Jf(x)}.$$