MGF Question (and infinite sum)

Question

I am having trouble with the following.

Let $X$ equal the number of flips of a fair coin that are required to observe heads-tails on consecutive flips.

a) Find the pmf of $X$. Hint: Draw a tree diagram.

b) Show that the mgf of $X$ is $\displaystyle M(t) = \frac{e^{2t}}{(e^t - 2)^2}$

I know that the pmf is $\displaystyle \Pr[X = x] = \frac{x-1}{2^x}$, for $x=2,3,4,5 \ldots$.

so the mgf is $$M(t) = e^{2t}\left(\frac{1}{4} + e^t \cdot \frac{2}{6} + e^{2t} \cdot \frac{3}{8}+ \ldots \right)$$

Can someone show me how this infinite sum is equal to $(e^t-2)^2$?

I believe it is some manipulation of the geometric series but I am not sure.

Thanks

Answer 1

So, you say that $P(X = x) = \frac{x-1}{2^x}$. We thus have that: \begin{align*} f(t) & = E[e^{tX}] = \sum_{x = 2}^\infty e^{tx}\frac{x-1}{2^x} \\ & = \sum_{x = 2}^\infty x\frac{e^{tx}}{2^x}-\sum_{x = 2}^\infty \frac{e^{tx}}{2^x} \\ & = \sum_{x = 2}^\infty \frac{\partial}{\partial t}\left(\frac{e^{tx}}{2^x}\right)-\sum_{x = 2}^\infty \left(\frac{e^t}{2}\right)^x \\ & = \frac{\partial}{\partial t}\sum_{x = 2}^\infty \left(\frac{e^t}{2}\right)^x-\sum_{x = 2}^\infty \left(\frac{e^t}{2}\right)^x \end{align*} Here, I "interchanged" differentiation and summation. We really should be careful about convergence (as it is NOT true in general), but this usually isn't as big of a concern in probability.

Now, note that: $$\sum_{x = 2}^\infty \left(\frac{e^t}{2}\right)^x$$ is a geometric sum. We can thus evaluate it as: $$\sum_{x = 2}^\infty\left(\frac{e^t}{2}\right)^x = \frac{e^{2t}}{4}\frac{1}{1-e^{t}/2} = \frac{e^{2t}}{4-2e^t}$$ Now, we can take the $t$ derivative of this to get that: $$\frac{\partial}{\partial t} \sum_{x = 2}^\infty \left(\frac{e^t}{2}\right)^x = \frac{(4-2e^t)2e^{2t}+2e^{2t}e^t}{(4-2e^t)^2} = \frac{8e^{2t}-2e^{3t}}{(4-2e^t)^2} = \frac{2e^{2t}(4-2e^t)}{(4-2e^t)^2} = \frac{e^{2t}}{2-e^t}$$ Now, we can preform the subtraction to get that: $$f(t) = \frac{e^{2t}}{2-e^t}-\frac{1}{2}\frac{e^{2t}}{2-e^t} = \frac{1}{2}\frac{e^{2t}}{2-e^t}$$

It appears I've made some mistake somewhere, but the important concepts here are:

1. Recognize that: $$x\frac{e^{xt}}{2^x} = \frac{\partial}{\partial t} \left(\frac{e^{t}}{2}\right)^x$$
2. "Commute" the partial derivative and infinite sum
3. Apply the geometric sum formula.

Answer 2

Let $z = e^t/2$. Then

$$\begin{align*} M_X(t) &= \operatorname{E}[e^{tX}] \\ &= \sum_{x=2}^\infty e^{tx} \frac{x-1}{2^x} \\ &= \sum_{x=2}^\infty (x-1) \left(\frac{e^t}{2}\right)^x \\ &= \sum_{x=1}^\infty x z^{x+1} \\ &= z^2 \sum_{x=1}^\infty x z^{x-1} \\ &= z^2 \frac{d}{dz} \left[\sum_{x=1}^\infty z^x \right] \\ &= z^2 \frac{d}{dz} \left[ \frac{z}{1-z} \right], \quad |z| < 1 \\ &= z^2 \cdot \frac{1}{(1-z)^2} \\ &= \frac{e^{2t}/4}{(e^{t}/2 - 1)^2}, \quad t < \log 2 \\ &= \frac{e^{2t}}{(e^{t} - 2)^2} \end{align*}$$ as claimed.