Derivation of moment generating function of a mixture of random variables

moment-generating-functionsprobabilityrandom variablesstatistics

Question

Let $Z$, $X$ and $Y$ be random variables, where $Z$ is formed by choosing a sample from $X$ with probability $q$ or a sample from $Y$ with probability $1 – q$. If $X$ and $Y$ are independent and have moment generating functions $M_X(t)$ and $M_Y(t)$ respectively, then show that the moment generating function of $Z$ is $$M_Z(t) = qM_X(t) + (1 – q)M_Y(t).$$

My working

$$\begin{aligned}
M_Z(t) & = \mathbb{E}\left(e^{tZ}\right)\\[2 mm]
& = \mathbb{E}\left\{e^{t[qX + (1 – q)Y]}\right\}\\[2 mm]
& = \mathbb{E}\left(e^{tqX}\right)\mathbb{E}\left(e^{t(1 – q)Y}\right)\\[2 mm]
& = M_X(tq)M_Y[t(1 – q)]
\end{aligned}$$
Here, I am stuck. If my working is correct so far, then how should I proceed to show the desired expression for $M_Z(t)$?

Any intuitive explanations or suggestions will be greatly appreciated 🙂

Best Answer

Requested from comments:

  • If we were looking for $\mathbb E\left[Z \right]$ with independence of $X$ and $Y$ from which is chosen, we would say $\mathbb E\left[Z\right] = q\mathbb E\left[X\right] + (1-q)\mathbb E\left[Y\right]$
  • Whenever $Z=X$ you have $e^{tZ}=e^{tX}$. Similarly with $Z=Y$
  • So with probability $q$ we set $e^{tZ}=e^{tX}$ and $=e^{tY}$ with probability $1−q$.
  • With independence of $X$ and $Y$ from which is chosen, this give the desired expectation $\mathbb E\left[e^{tZ}\right] = q\mathbb E\left[e^{tX}\right] + (1-q)\mathbb E\left[e^{tY}\right]$
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