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Year 2 of the mathematics course of the University of Oxford

Options I course of the University of Oxford

Past exam of the options I course of the University of Oxford

2008

The inverse transformation is $x=2u+v$, $y=v$, with absolute <Jacobian determinant> two. The original positivity constraints become $v>0$ and $2u+v>0$. Thus
$$
\boxed{A=\{(u,v):v>\max(0,-2u)\}},\qquad
f_{U,V}(u,v)=\tfrac12e^{-u-v}\mathbf1_A.
$$
Integrating out $v$ gives
$$
f_U(u)=\frac12e^{-u}\int_{\max(0,-2u)}^\infty e^{-v}\,dv
=\frac12e^{-u-\max(0,-2u)}
=\boxed{\frac12e^{-|u|}}.
$$
This is the centred <Laplace distribution> with scale one.

Solution

Past exam of the options I course of the University of Oxford / 2008 / M / M1 / Solution

Past exam of the options I course of the University of Oxford / 2008 / M / M1 / Solution

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Past exam of the options I course of the University of Oxford / 2008 / M / M1 / Solution

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