Question 3.8: The joint density of X and Y is given by f(x,y) = {1/2ye^-xy......

The conditional density of X, given that Y = 1, is given by

f_{X\mid Y}(x\mid1)={\frac{f(x,1)}{f_{Y}(1)}}

{}=\frac{\frac{1}{2}e^{-x}}{\int_{0}^{\infty}\frac{1}{2}e^{-x}\,d x}=e^{-x}

Hence, by Proposition 2.1,

E{\Bigl[}e^{X/2}|Y=1{\Bigr]}=\int_{0}^{\infty}e^{x/2}f_{X|Y}(x|1)\;d x

=\int_{0}^{\infty}e^{x/2}e^{-x}\,d x

=2