J
jatin.agrawal
Member
By the property of MGF, M(0) = 1 always and this result can be used to crosscheck the MGF obtained for any PDF...
But for PDF, f(x) = 0.5(1-x), -1<x<1
the MGF is e^(-t) * 1/t^2 * {e^2t - 2t - 1}
But this MGF does'nt follow above rule
Pls tell me where i am wrong????
But for PDF, f(x) = 0.5(1-x), -1<x<1
the MGF is e^(-t) * 1/t^2 * {e^2t - 2t - 1}
But this MGF does'nt follow above rule
Pls tell me where i am wrong????