A proof regarding the moment generating function.

Question

We are required to prove that two random variables $(X,Y)$ are independent if and only if $m_{X,Y}(t_1,t_2)=m_X(t_1)m_Y(t_2)$ where $m(\_)$ is the moment generating function.

Supposing $(X,Y)$ are independent and then proving $m_{X,Y}(t_1,t_2)=m_X(t_1)m_Y(t_2)$ is trivial since for two independent variables $X,Y$, $E(g_1(X)g_2(Y))=E(g_1(X))E(g_2(Y))$.

Now conversely, we say, $m_{X,Y}(t_1,t_2)=m_X(t_1)m_Y(t_2)$, i.e

$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{t_1x+t_2y}f_{X,Y}(x,y)dydx=\int_{-\infty}^{\infty}e^{t_1x}f_X(x)dx\int_{-\infty}^{\infty}e^{t_2y}f_Y(y)dy$=$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{t_1x+t_2y}f_X(x)f_Y(y)dydx$

Hence,

$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{t_1x+t_2y}f_{X,Y}(x,y)dydx-\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{t_1x+t_2y}f_X(x)f_Y(y)dydx=0$

that is,

$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{t_1x+t_2y}(f_{X,Y}(x,y)-f_X(x)f_Y(y))dy\:dx=0$

Now, if $(f_{X,Y}(x,y)-f_X(x)f_Y(y))\geq 0$ we have what we want, but I am not sure whether that is true in general, is it ?

Can anyone help ?

Answer 1

Consider a joint distribution $(\tilde{X}, \tilde{Y})$ where $\tilde{X}$ and $\tilde{Y}$ are independent, and the marginal distributions are the same as $X$ and $Y$ respectively. By the direction you have proved, $m_{\tilde{X}, \tilde{Y}}(t) = m_X(t) m_Y(t)$. Further, by the assumption $m_X(t) m_Y(t) = m_{X,Y}(t)$, we have $m_{\tilde{X},\tilde{Y}}(t) = m_{X,Y}(t)$. Finally, use the uniqueness of MGFs to conclude that $(\tilde{X},\tilde{Y})$ and $(X,Y)$ must have the same distribution, and thus $X$ and $Y$ are independent.