MGF of Random Variables

Question

The MGF of random variable W is

$$\mathbf{M}_{w}(t) = \frac{9e^2t}{9-t^2}$$

$\mathbf{E}[W]=2$

Find the Mgf of Z=3(W-2). Simplify your result.

To $\mathbf{M}_{w}(t)$ do you take $\mathbf{E}[e^t*z]$ or would you have to take $\mathbf{E}[Z]$ in terms of W?

Answer 1

Neither. The definition of $M_{\mathbf Z}(t)$ is $$M_{\mathbf Z}(t) = \mathbb E[e^{t \mathbf Z}].$$ You are given $M_{\mathbf W}(t) = \mathbb E[e^{t\mathbf W}]$, and so your goal should be to express $\mathbb E[e^{t \mathbf Z}] = \mathbb E[e^{t \cdot 3(\mathbf W-2)}]$ in terms of $\mathbb E[e^{t\mathbf W}]$.

Answer 2

What you might be looking for is a property that relates MGFs to affine transformations.

\begin{align*} M_{aX + b}(t) &= \mathbb{E}[e^{t \cdot (aX + b)}] \\ &= \mathbb{E}[e^{taX + bt}] \\ &= e^{bt} \mathbb{E}[e^{taX}] \\ &= e^{bt} M_{X}(at) \end{align*}