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I am working on a problem, in which the prompt indicates that independent random variables X and Y are independent and both follow a poisson distribution, that is X~Poisson$(\lambda)$ and Y~Poisson$(\theta)$. I am asked to compute the MGF of X. Since the different random variables have different means, can I ignore the Y? That is, can I accurately interpret the prompt to being asking me to simply derive MGF of the poisson distribution of X?

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The computation would be the same, just with $\lambda$ instead of $\theta$ or vice versa. For completeness, here is the derivation: \begin{align} M_X(t) &= \mathbb E[e^{tX}]\\ &= \sum_{n=0}^\infty e^{tn} e^{-\lambda}\frac{\lambda^n}{n!}\\ &= e^{-\lambda} \sum_{n=0}^\infty \frac{(\lambda e^t)^n}{n!}\\ &= e^{-\lambda}e^{\lambda e^t}\\ &= e^{\lambda(e^t-1)}. \end{align}

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