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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

  1. Did I correctly justify to only focus on closed rectangles in $f[\mathbb{R}^2]$?
  2. Did I calculate the pushforward measure correctly?
  3. 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*}

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2 Answers 2

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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 $$g(y) = \sum_{x \in f^{-1}(\{y\})}\frac{1}{Jf(x)}.$$

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  • $\begingroup$ I have added a solution based on your answer. Is it correct? $\endgroup$ Commented Apr 5, 2022 at 23:49
  • $\begingroup$ @WilfredMontoya It looks correct. I would check it using Mathematica. $\endgroup$
    – Mason
    Commented Apr 6, 2022 at 1:25
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Hint: First note that $f$ is a diffeomorphism onto its image (i.e. it is continuously differentiable, has an inverse, and its inverse is also continuously differentiable). Thus we have the $C^1$ change of variables formula involving the Jacobian of $f$ (or $f^{-1}$). Now the Radon-Nikodym derivative (or density) is involved in the measurable change of variables formula, and it is unique a.e.. So it suffices to identify it with a Jacobian. One needs to be mindful of the directions; measures/ RN derivatives are homological (they are pushed forward) while forms/ Jacobians are cohomological (they are pulled back).

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