Cramér's theorem - Legendre transformation

Question

Consider $X_1, X_2,\dots $ i.i.d. random variables that are geometrically distributed with parameter $\frac 1 2$. With the moment generating function of $X_1$ (from Wikipedia) one gets $$ \Lambda(t) = \log E[e^{tX_1}] = \log \frac{pe^t}{1-(1-p)e^t}=-\log(2) + t-\log\left(1-\frac 1 2 e^t\right). $$ I'd like to know what the rate function $I(\cdot)$ is though. In my textbook, I have this formula $$ I(x)=\Lambda^*(x) = \sup_{t\in\mathbb R}\left\{ tx - \Lambda(t) \right\} = \sup_{t\in\mathbb R}\left\{ tx +\log(2) - t+\log\left(1-\frac 1 2 e^t\right) \right\}. $$ The solution is supposed to be (but have not found a way to get there): $$ I(x)=x\log x-(1+x) \log\left( \frac{1+x}{2} \right). $$ If you could provide one or two hints, I'd be really glad!

Answer 1

Let me give you few steps

1. Show that $\Lambda(t)$ is convex and therefore $tx-\Lambda(t)$ is concave. Thus there will be just one maximum of $tx-\Lambda(t)$.

2. Now differentiate $tx-\Lambda(t)$ with respect to $t$ and express $e^t$ in terms of $x$.

Response to the comment: your calculation is correct but the confusion occurred because of the definition of geometric distribution that you have used. Use the distribution that has support $\{0,1,2,\ldots\}$ and probability mass function $p_k$, where $$ p_k = \frac{1}{2^{k+1}} \,\text{ for }\,k=0,1,\ldots. $$
Let me know if you can complete it from here.