3.4 Joint Moment Generating Function
Definition 3.4.1. Let \(X\) and \(Y\) have a joint probability function \(f_{X,Y}(x,y)\) then the joint moment generating function of \(X\) and \(Y\) denoted by \[M_{X,Y}(t_1,t_2)=E\left (e^{t_1X +t_2Y}\right )\] given by \[E\left (e^{\displaystyle {t_1X + t_2Y}}\right )= \begin {cases} \sum \limits _X\sum \limits _Ye^{\displaystyle {t_1X + t_2Y}}\quad P (X=x,Y=y) & \text {discrete}\\\\ \int \limits _X\int \limits _Ye^{\displaystyle {t_1X + t_2Y}}\quad f_{X,Y}(x,y)\, dx\, dy & \text {continuous}\\ \end {cases} \]
- 1.
- If \(M(t_1,t_2)\) exits then the m.g.f of \(X\) is given by \[M_X(t_1) = E\left (e^{t_1X}\right ) = M(t_1,0)\] and the m.g.f of \(Y\) is given by \[M_Y(t_1) = E\left (e^{t_2Y}\right ) = M(0,t_2).\]
- 2.
- If \(X\) and \(Y\) are independent the \(M_{X,Y}(t_1,t_2)=M_X(t_1)M_Y(t_2)\).
Proof. If \(X\) and \(Y\) are independent then \(f_{X,Y}(x,y)=f_X(x)f_Y(y)\). \begin {align*} E\left (e^{t_1X + t_2Y}\right ) & = \int \limits _X\int \limits _Ye^{t_1X}e^{t_2Y}f_X(x)f_Y(y)dxdy\\ & = \int \limits _Xe^{t_1X}f_X(x)dx\int \limits _Ye^{t_2Y}f_Y(y)\,dy\\ & = M_X(t_1)\, M_Y(t_2). \end {align*} □
Example 3.4.3. Find the joint moment generating function for \(X\) and \(Y\) with joint p.d.f. \[f_{X,Y}(x,y)=\frac {x+y}{3},\quad 0<x<1,\quad 0<y<2.\]
Solution.
\begin {align*} M_{X,Y}(t_1,t_2) & = E\left (e^{t_1X+t_2Y}\right )\\ & = \int ^2_0\int ^1_0e^{xt_1+yt_2}\left (\frac {x+y}{3}\right )\, dx\, dy\\ & = \frac {1}{3}\int ^2_0e^{yt_2}\int ^1_0e^{xt_1}\, (x+y)\, dx\, dy\\ & = \frac {1}{3}\int ^2_0e^{yt_2}\int ^1_0\left (xe^{xt_1} + ye^{xt_1}\right )\, dx\, dy\\ & = \frac {1}{3}\int ^2_0e^{yt_2}\left [\frac {xe^{xt_1}}{t_1}\Bigg |^1_0-\int ^1_0\frac {e^{xt_1}}{t_1}dx+\frac {y}{t_1}e^{xt_1}\Bigg |^1_0\right ]\, dy\\ & =\frac {1}{3}\int ^2_0e^{yt_2}\left [\frac {e^{t_1}}{t_1^2}-\frac {e^{xt_1}}{t_1^2}\Bigg |^1_0+\frac {ye^{t_1}}{t_1}-\frac {y}{t_1}\right ]dy\\ & =\frac {1}{3}\int ^2_0e^{yt_2}\left [\frac {e^{t_1}}{t_1}+ \frac {ye^{t_1}}{t_1}-\frac {y}{t_1}-\frac {e^{t_1}}{t_1^2}+\frac {1}{t^2_1}\right ]\, dy\\ & = \frac {1}{3t_1^2}\int ^2_0e^{yt_2}\left [t_1e^{t_1}-e^{\displaystyle {t_1}}+1+yt_1e^{t_1}-yt_1\right ]\, dy\\ & =\frac {1}{3t_1^2}\left [\left (1+t_1e^{t_1}-e^{t_1}\right )\int ^2_0e^{yt^2}\, dy+t_1\left (e^{t_1}-1\right )\int ^2_0ye^{yt_2}dy\right ]\\ & = \frac {1}{3t^2_1}\left [\left (1+t_1e^{t_1}-e^{t_1}\right )\frac {e^{yt_2}}{t_2}\Big |^2_0+t_1\left (e^{t_1}-1\right )\left [\frac {ye^{yt_2}}{t_2}\right ]^2_0-\int ^2_0\frac {e^{yt_2}}{t_2}\, dy\right ]\\ & = \frac {1}{3t_1^2}\left [\left (1+t_1e^{t_1}-e^{t_1}\right )\left (\frac {e^{2t_2}}{t_2}-\frac {1}{t_2}\right )+t_1\left (e^{t_1}-1\right )\left (\frac {2e^{2t_2}}{t_2}-\frac {e^{yt_2}}{t^2_2}\Big |^2_0\right )\right ]\\ & = \frac {1}{3t^2_1}\left [\left (1+t_1e^{t_1}-e^{t_1}\right )\left (\frac {e^{2t_2}-1}{t_2}\right )+t_1\left (e^{t_1}-1\right )\left (\frac {2e^{2t_2}}{t_2}-\frac {e^{2t_2}}{t^2_2}+\frac {1}{t^2_2}\right )\right ]. \end {align*}
Therefore \[ M_{X,Y} (t_1,t_2) = E\left (e^{Xt_1+Yt_2}\right ) = E\left (e^{Xt_1}\, e^{Yt_2}\right ) = E\left (\sum ^{\infty }_{k=0} \frac {(Xt_1)^k}{k!}\cdot \sum ^{\infty }_{m=0}\frac {(Yt_2)^m}{m!}\right ).\] The marginals can be obtained by \begin {align*} M_X(t_1) & = \lim _{t_2\longrightarrow 0}M_{X,Y}(t_1,t_2)\\ M_Y(t_2) & = \lim _{t_1\longrightarrow 0} M_{X,Y}(t_1,t_2). \end {align*} □
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