This video at 16:20 presents a quick way to compute the expectation of a geometric random variable N with parameter 0.5 http://www.youtube.com/watch?v=xUPdKqGiy84
The author explains that the expectation should verify E[N] = 1 + 0.5 * E[N]
It is then immediate that E[N] = 2
Despite the explanations given in the video, I don't get it. Any help to understand what is going on is welcome.
Let $A=[N=1]=[\text{success at first try}]$. Then $P[A]=\frac12$ and, on $A$, $N=1$. On $A^c$, $N=1+M$ where $M$ is the number of tries before a success starting at the second try, hence the distributions of $M$ and $N$ coincide and $M$ is independent on $A$. Furthermore, $E[N]=E[N:A]+E[N:A^c]$, $E[N:A]=P[A]=\frac12$, $E[N:A^c]=E[1+M:A^c]=E[1+M]P[A^c]=(1+E[N])\frac12$. Thus, $$ E[N]=\tfrac12+\tfrac12(E[N]+1). $$ Solve for $E[N]$.