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Let X and Y be independent and exponentially distributed both. with parameter $\lambda = 1$. Let $U = 2X + Y$ and $V=Y$. Find the joint pdf of $(U,V)$.

My attempt I found the Jacobian and plug in the joint pdf of $(X,Y)$, which is $(1/2\pi)e^{(-1/2)(x^2 + y^2)}$ but this is incorrect.

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This is one of the examples where the use of Dirac delta to compute pdfs and joint pdfs proves invaluable. Enforce the two constraints with delta functions $$ f_{U,V}(u,v)=\int_0^\infty dx\int_0^\infty dy e^{-x-y}\delta(u-(2x+y))\delta(v-y)\ , $$ and kill first the $y$-integral with one delta $$ f_{U,V}(u,v)=\Theta(v)\int_0^\infty dx e^{-x-v}\delta(u-(2x+v))=\Theta(v)e^{-v}\frac{1}{2}e^{-(u-v)/2}\Theta(u-v)\ , $$ where $\Theta(z)=1$ if $z>0$ and $0$ otherwise. So the final answer is $$ f_{U,V}(u,v)=\frac{1}{2}e^{-(u+v)/2}\,\qquad\text{for }u>v>0, $$ and $0$ otherwise. You can check that it is correctly normalized: $$ \int_0^\infty dv\int_v^\infty du \frac{1}{2}e^{-(u+v)/2}=1\ , $$ as it should. Note that the Theta constraints arise from the need to make sure that the values for which the argument of each delta is equal to zero lie within the integration range: for example $\delta(v-y)$ implies $y=v$, but $y$ must be $>0$, so also $v$ must be $>0$.

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