Polya's urn (martingale) Suppose you have an urn containing one red ball and one green ball. You draw one at random; if the ball is red, put it back in the urn with an additional red ball , otherwise put it back and add a green ball . Repeat this procedure and let the random variable $X_n$ be the number of red balls in the urn after n draws.
Let $Y_n=\frac{1}{n+2}X_n$. Find $\mathbb{\mathbb{\textrm{E}}}\left(Y_{n}\right)$ and prove that $Y_n$ is a martingale with respect to $X_n$.

MY ATTEMPT:
We have $\mathbb{\mathbb{\textrm{E}}}\left(\left.X_{n+1}\right|X_{n}\right)=X_{n}+\dfrac{X_{n}}{n+2}=\dfrac{n+3}{n+2}X_{n}$, so
$\mathbb{\mathbb{\textrm{E}}}\left(\left.Y_{n+1}\right|X_{n}\right)=\dfrac{1}{n+3}\mathbb{\mathbb{\textrm{E}}}\left(\left.X_{n+1}\right|X_{n}\right)=\dfrac{1}{n+3}\cdot\dfrac{n+3}{n+2}X_{n}=\dfrac{1}{n+2}\cdot X_{n}=Y_{n}$
It's ok?
And, can you help me to find $\mathbb{\mathbb{\textrm{E}}}\left(Y_{n}\right)$?
 A: Looks good, as indeed
$$ X_{n+1} = X_n + R_{n+1},$$
where $R_{i}$ denotes the indicator variable that takes value $1$ if color of the $i$-th ball extracted is red, and $0$ if green. By definition we have that the urn contains $X_n$ red and $n+2-X_n$ green balls after $n$ extractions. Then the conditional probability given $X_n$ of a red ball on the $n+1$-th extraction (equal to the conditional expectation given $X_n$ of $R_{n+1}$ that we need) is $$\frac{X_n}{n+2}=Y_n.$$
We also observe that
$$ X_n = 1+\sum_{i=1}^n R_i. $$
Taking expectation we get:
$$ \mathbf{E}\left[ X_n\right] = 1+\sum_{i=1}^n \mathbf{E}\left[ R_i\right].$$
As all $R_i$ have the same distribution as $R_1$, we get:
$$ \mathbf{E}\left[ R_i\right] = \mathbf{E}\left[ R_1\right] =\frac{1}{2},$$
for all $i\in \{1,\ldots , n\}$.
Our indicator variables have the same distribution due to the fact that the sequence of variables $R_1,\ldots, R_n$ is exchangeable, as its joint distribution
$$\mathbf{P}\left(R_1=c_1,\ldots, R_n=c_n\right)   $$ $$= \mathbf{P}\left(R_1=c_1\right)\mathbf{P}\left(R_2=c_2 | R_1=c_1\right) \ldots\mathbf{P}\left(R_n=c_n | R_1=c_1,\ldots, R_{n-1}=c_{n-1}\right) $$ 
$$ = \frac{c!(n-c)!}{(n+1)!}$$ depends on $c_1,\ldots,c_n$ only through the number of red balls $c$, $c=c_1+\ldots + c_n$.
To conclude, we have $\mathbf{E}[X_n]=(n+2)/2$, so
$$\mathbf{E}[Y_n]=1/2.$$
This can also be seen directly as we already know that $Y_n$ is a martingale, so (proof here)
$$\mathbf{E}[Y_n]=\mathbf{E}[Y_{n-1}]=\ldots = \mathbf{E}[Y_1]=1/2.$$
