Solution.

The density is a sum of products of a function of \(x\) and a function of \(y\), so the double integral splits into single integrals. With \(t_1, t_2\ne 0\), integration by parts gives \[\int _0^1e^{xt_1}\,dx = \frac {e^{t_1} - 1}{t_1},\qquad \int _0^1xe^{xt_1}\,dx = \frac {(t_1 - 1)e^{t_1} + 1}{t_1^2},\] \[\int _0^2e^{yt_2}\,dy = \frac {e^{2t_2} - 1}{t_2},\qquad \int _0^2ye^{yt_2}\,dy = \frac {(2t_2 - 1)e^{2t_2} + 1}{t_2^2} .\] Hence \begin {align*} M_{X,Y}(t_1,t_2) & = \frac 13\int ^2_0\int ^1_0e^{xt_1+yt_2}(x+y)\, dx\, dy\\ & = \frac 13\left [\int _0^1xe^{xt_1}\,dx\int _0^2e^{yt_2}\,dy + \int _0^1e^{xt_1}\,dx\int _0^2ye^{yt_2}\,dy\right ]\\ & = \frac 13\left [\frac {(t_1 - 1)e^{t_1} + 1}{t_1^2}\cdot \frac {e^{2t_2} - 1}{t_2} + \frac {e^{t_1} - 1}{t_1}\cdot \frac {(2t_2 - 1)e^{2t_2} + 1}{t_2^2}\right ], \end {align*}

with the value at \(t_1 = 0\) or \(t_2 = 0\) given by the limit. The marginal m.g.f.s are \(M_X(t_1) = \lim _{t_2\to 0}M_{X,Y}(t_1, t_2)\) and \(M_Y(t_2) = \lim _{t_1\to 0}M_{X,Y}(t_1, t_2)\); for example, since \(\frac {e^{2t_2} - 1}{t_2}\to 2\) and \(\frac {(2t_2 - 1)e^{2t_2} + 1}{t_2^2}\to 2\), \[M_X(t_1) = \frac 23\left [\frac {(t_1 - 1)e^{t_1} + 1}{t_1^2} + \frac {e^{t_1} - 1}{t_1}\right ].\] □