I want to get checked if my attempt is okay. 


First off, let me shortly describe what Polya's urn is: 


A certain urn initially contains a red and a blue ball. We now repeatedly do the following : we (uniformly at random) pick a ball from the urn, and then put it back in together with an additional ball of the same colour. Let $X\_n$ denote the number of red balls and $Y\_n$ denote the fraction of red balls after we’ve done this $n$ times, i.e. $$
Y\_n = \frac{X\_n}{n+2}. $$


And the following is my attempt so far. (let me assume that $Y\_n$ is a martingale without showing it)


\begin{align\*}
X\_{n+1} &= X\_n + 1\_{\{\text{$n+1^{th}$ ball is red}\}} \\
X\_n &= 1 + \sum\_{1\leq i\leq n}1\_{\{\text{$i^{th}$ ball is red}\}} \\
\to EX\_n &= 1 + \sum\_{1\leq i\leq n}E[1\_{\{\text{$i^{th}$ ball is red}\}}] = 1 + nE[1\_{\{\text{$i^{th}$ ball is red}\}}] = 1 + \frac{n}{2}
\end{align\*}
since the indicator function take only values $0$ or $1$ with the same probability here (picking either ball has the same probability).


Thus, $EX\_n = 1 + \frac{n}{2} = \frac{n+2}{2}$, so $EY\_n = \frac{1}{2}$.


So now we know the expectation of $Y\_n$, **but is it enough to say that $Y$ is the uniform distribution on $[0,1]$**? 


(where $Y\_n\xrightarrow[]{a.s.}Y$ (Martingale convergence theorem)) 


Thank you for your help in advance!



Martingale convergence theorem in Polya's urn
The answer is yes, the distribution of $Y\_n$ converges to the uniform distribution on $[0,1]$.  

More generally, if we initially have $r$ red and $b$ blue balls in the urn, then the distribution of the proportion of the red balls in the urn converges to the beta distribution with parameters $r,b$; see e.g. [Section 4](http://sites.science.oregonstate.edu/~math_reu/proceedings/REU_Proceedings/Proceedings2008/2008Drinanepaged.pdf).




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Moreover, in your case of $r=b=1$, letting 
$$p\_{n,k}:=P(X\_n=k),$$
it is easy to check directly by induction on $n$, using the recursion 
$$p\_{n,k}=p\_{n-1,k}\frac{n+1-k}{n+1}+p\_{n-1,k-1}\frac{k-1}{n+1}$$
for natural $n$ and $k$, that 
$$p\_{n,k}=\frac1{n+1}\,1\big\{k\in\{1,\dots,n+1\}\big\}$$
for $n=0,1,\dots$; that is, $X\_n$ has the discrete uniform distribution on the set 
$\{1,\dots,n+1\}$. 


Therefore, it is clear that the distribution of $Y\_n=\frac{X\_n}{n+2}$ converges to the uniform distribution on $[0,1]$. 


