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!

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