Question: Use LOTUS to show that for X ~ Pois(λ) and any function g(.), E(Xg(X)) = λE(g(X+1)).
The steps I have taken are shown below
It is given that mean is E(X)=λ
The Law of Unconscious Statistician is
$$E(g(X))=\sum_{x} g(x)f_X(x)$$ Then,
$$E(Xg(X))=\sum_{x} Xg(x)f_X(x)$$ $$=E(X)E(g(x)f_X(x))$$
Is this solution correct?
The second step is not correct; you cannot split a sum like that into two expectations.
\begin{align} E[X g(X)] &= \sum_{x=0}^\infty x g(x) f_X(x) \\ &= \sum_{x=1}^\infty x g(x) f_X(x) & \text{addend is zero when $x=0$} \\ &= \sum_{x=1}^\infty x g(x) e^{-\lambda} \frac{\lambda^x}{x!} & \text{plug in PMF of Poisson distr.} \\ &= \lambda\sum_{x=1}^\infty g(x) e^{-\lambda} \frac{\lambda^{x-1}}{(x-1)!} & \text{rearrange terms} \\ &= \lambda \sum_{x=0}^\infty g(x+1) e^{-\lambda} \frac{\lambda^x}{x!}. & \text{shift indexing} \end{align}