Suppose that X1 and X2 are independent and identically distributed discrete random variables. The moment generating function of X1 + X2 is:

M(t)= 0.01e^(-2t) + 0.15e^(-t) +0.5925 + 0.225e^(t) + 0.0225e^(2t)

Find P[X<0]

I know that since X1 and X2 are independent that M(t) = Mx(t)^2, but I don't know where to go from there.

Any suggestions/hints?


$$M(t)=\frac{1}{100}e^{-2t}+\frac{3}{20}e^{-t}+\frac{237}{400}+\frac{9}{40}e^{t}+\frac{9}{400}e^{-2t}\\=\frac{e^{-2t}}{400}\left(4+60e^{t}+237e^{2t}+90e^{3t}+9e^{4t}\right)\\=\frac{e^{-2t}}{400}\left(2+15e^{t}+3e^{2t}\right)^{2}$$ So $$M_{X}(t)=\frac{1}{20}(2e^{-t}+15+3e^{t})$$ And by definition, $$M_{X}(t)=\mathbb{E}(e^{tX})=\sum p_{i}e^{x_{i}t}$$ So $P(X<0)=P(X=-1)=\frac{2}{20}=0.1$


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