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If X is Poisson then I know the m.g.f is exp(λ[$e^t$−1]). And i need to find out the m.g.f of X-2. I am not able to reach the answer given. I am thinking X-2 is also Poisson with mean λ-2 but I am not sure whether I am right. Please help me. Thankyou.

I have proceeded in this way: Since X~Poisson(λ) So m.g.f is given by $M_x$(t)=$e^λ$[$e^t$−1]

Now, Y=X-2~Poisson(λ-2) So m.g.f is given by $M_y$(t)=exp(λ-2)[$e^t$-1] =exp(λ$e^t$-2$e^t$-λ+2)

But the correct answer given in my book is $M_y$(t)=exp(λ$e^t$-2t-λ)

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Azka
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1 Answers1

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I proceeded in the following way.

We know that $M_{a+bx}$$(t)$= $E$$[e^{(a+bx)t}]$ = $e^{at}$$E$$[e^{btx}]$ = $e^{at}$ $M_x(bt)$.

Now X~P($\lambda$)

This implies $M_x(t)$=$e^{\lambda(e^t -1)}$

Y=X-2 $\Rightarrow$ X=Y+2 $$\Rightarrow M_X(t)=M_{Y+2}(t) = E(e^{(y+2)t})$$ $$=e^{2t}E(e^{yt})$$ $$e^{2t} M_Y(t)$$

$$\Rightarrow e^{\lambda(e^t-1)}=e^{2t} M_Y(t)$$ $$\Rightarrow M_Y(t)=\frac{e^{(\lambda e^t - \lambda)}}{e^{2t}}$$ $$=e^{\lambda e^t - 2t - \lambda}$$

Azka
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