The moment generating function of two jointly distributed random variables X1 and X2 is M(t1, t2) = e ^− 0.5 G where G = (7.51t 2 1 + 7.9t 2 2 + 3.8574t1t2 + 135.4t1 + 137.2t2) Using this function, find the correlation coefficient of of X1 and X2
M(t1,t2)=exp(−7.51t12+7.9t22+3.8574t1t2+135.4t1+137.2t22)EX1=∂M(t1,t2)∂t1=−135.42=−67.7EX12=∂2M(t1,t2)∂t12∣t1=t2=0=4575.78Var(X1)=4575.78−(−67.7)2=−7.51<0Since we obtained a negative variance,the problem is incorrectM\left( t_1,t_2 \right) =\exp \left( -\frac{7.51{t_1}^2+7.9{t_2}^2+3.8574t_1t_2+135.4t_1+137.2t_2}{2} \right) \\EX_1=\frac{\partial M\left( t_1,t_2 \right)}{\partial t_1}=-\frac{135.4}{2}=-67.7\\{EX_1}^2=\frac{\partial ^2M\left( t_1,t_2 \right)}{\partial t_1^2}|_{t_1=t_2=0}=4575.78\\Var\left( X_1 \right) =4575.78-\left( -67.7 \right) ^2=-7.51<0\\Since\,\,we\,\,obtained\,\,a\,\,negative\,\,variance, the\,\,problem\,\,is\,\,incorrectM(t1,t2)=exp(−27.51t12+7.9t22+3.8574t1t2+135.4t1+137.2t2)EX1=∂t1∂M(t1,t2)=−2135.4=−67.7EX12=∂t12∂2M(t1,t2)∣t1=t2=0=4575.78Var(X1)=4575.78−(−67.7)2=−7.51<0Sinceweobtainedanegativevariance,theproblemisincorrect
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