Let $X$ be a geometric random variable with parameter $\theta$ find $m_X(t)$

Question

Let $X$ be a geometric random variable with parameter $\theta$ find $M_X(t)$, moment generating function.

$m_X(t)=E(e^{tX})=\sum_x e^{tx} p_X(x)$

$$\sum_x e^{tx} p_X(x)= \sum_x e^{tx}(1-\theta)^x\theta$$ $$=\theta\sum_x[e^t(1-\theta)]^x$$

I need to simplify $\sum_x[e^t(1-\theta)]^x$

To do this I could use $\sum_{k=1}^{\infty} q^k = \frac{q}{1-q}.$

However, how to show that $e^t(1-\theta)<1$?

But I'm not sure how to do this.

Answer 1

Let $\mathbb P(X = k) = (1-\theta)^{k-1}\theta, k=1,2,\ldots$. Then \begin{align} M_X(t) &= \mathbb E[e^{tX}]\\ &= \sum_{n=1}^\infty e^{tn} \mathbb P(X=n)\\ &= \sum_{n=1}^\infty e^{tn} (1-\theta)^{n-1}\theta\\ &= \frac\theta{1-\theta}\sum_{n=1}^\infty((1-\theta)e^t)^n\\ &= \frac\theta{1-\theta} \cdot\frac{(1-\theta) e^t}{\theta e^t-e^t+1}\\ &= \frac{\theta e^t}{1-(1-\theta)e^t}. \end{align} Note that the series only converges for $t<\log\left(\frac1{1-\theta}\right)$.