Given the following probability distribution $$f(x) = \frac{\theta^x e^{-\theta}}{x!}, x=0,1,2,3,.. $$
Find the moment generating function, MGF, then find mean of random variable $Y=(X+1)^3$
To my knowledge $$M_x(t) = \sum{e^{tx}f(x)}$$
$$\sum{\frac{\theta^x e^{tx-\theta}}{x!}}, x=0,1,2,..$$
How to solve this summation ? then I am supposed to get MGF for $Y$, but I don't know what to do with power ?
Best Answer
$$M_X(t) = \mathbb{E}[e^{tX}] = \sum_{x=0}^{+\infty} e^{tx} \frac{\theta^x e^{-\theta}}{x!} = e^{-\theta}\sum_{x=0}^{+\infty} \frac{\theta^x e^{tx}}{x!} = e^{-\theta}\sum_{x=0}^{+\infty} \frac{(\theta e^{t})^x}{x!}.$$
As suggested by @Imaosome:
$$\sum_{k=0}^\infty\frac{a^k}{k!} = e^a.$$
Hence:
$$M_X(t) = e^{-\theta}e^{\theta e^{t}} = e^{\theta(e^t - 1)}.$$
For the mean of $Y = (X+1)^3$....
$$\mathbb{E}[Y] = \mathbb{E}[(X+1)^3] = \mathbb{E}[X^3 + 3X^2 + 3X + 1] = \mathbb{E}[X^3] + 3\mathbb{E}[X^2] + 3\mathbb{E}[X] + \mathbb{E}[1] = \\ = \left(\frac{d^3}{dt^3}M_X(t)\right)_{t=0} + 3\left(\frac{d^2}{dt^2}M_X(t)\right)_{t=0} + 3\left(\frac{d}{dt}M_X(t)\right)_{t=0} + 1 = ...\\$$