An urn U contains $r_0$ red balls and $g_0$ green balls. At each stage a ball is selected at random from the urn, we observe its color, we return it to the urn and then we add another ball of the same color. We denote by $b_n$ the total number of balls in U at stage n, by $R_n$ the number of red balls and by $G_n$ the number of green balls at stage n. Finally, we denote by $C_n$ the "concentration" of red balls at stage n,
$C_n$ = $\frac{R_n}{b_n}$ = $\frac{R_n}{R_n + G_n}$
Show that E[$C_n$] = $\frac{r_0}{r_0 + g_0}$ for all n ∈ N
I'm not sure where to begin with this problem. If anyone could walk me through it, or at least give me a place to start, that would be greatly appreciated.
If it helps, the first part of this problem (this is the second part) showed that E[$C_{n+1}|R_n = i$] = $\frac{i}{b_n}$. I already solved that part of the problem.
HINT:
We are adding a ball of the same color based on their initial probability of being picked up.
Take an example,
Concentration of red balls to start with, $C_0 = \displaystyle \frac{r_0}{r_0 + g_0}$
After first draw and addition of an extra ball, expected number of red balls
$ E_1 = \displaystyle \frac{r_0}{r_0 + g_0} (r_0 + 1) + \frac{g_0}{r_0 + g_0} (r_0) = \frac{r_0(r_0 + g_0 + 1)}{r_0 + g_0}$
(If we draw red with the probability of $r_0 / (r_0 + g_0)$, we have $r_0 + 1$ red balls otherwise if we draw green, we continue to have $r_0$ red balls.
Total number of balls after first draw $ = r_0 + g_0 + 1$
So, $C_1 = \displaystyle \frac{r_0}{r_0 + g_0}$ which is same as we started.