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Suppose that at time 0 I have a urn with a red ball and a white ball. At time $n$ I take a random ball with uniform probability and return this ball with one extra ball of the same colour to the urn. Let $R_n$, $W_n$ be the number of red, white balls at time $n$. Then, \begin{align} E[R_{n+1}\mid F_n]=\frac{n+3}{n+2}R_n \end{align} (the expected number of balls at time n+1 knowing the number of ball extracted at time n)

I can't understand how to compute that value.

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Hints: First, at the beginning of time $n+1$, there are exactly $n+2$ balls (why?). We then sample a ball: with probability $R_n/(n+2)$, we draw a red ball (why?), in which case $R_{n+1}=R_n+1$, while with probability $(n+2-R_n)/(n+2)$, we draw a white ball, so we are stuck with $R_{n+1}=R_n$. Combine these values and probabilities (how?) to get your answer.

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  • $\begingroup$ Yes, at time $n$ we have $R_n+ W_n=n+2$ balls. That's clear to me. The probability to draw a red ball is $\frac{R_n}{n+2}$ because we have (at the beginning of time $n+1$) exactly $R_n$ red balls. After that we have $R_n+1$ red balls at time $n+1$. In the other case, we draw a white ball with probability $1- \frac{R_n}{n+2}$, and we will not add any red ball, so we have $R_{n+1}=R_n$ red balls. All in all, we have by definition of expectation: $\frac{R_n}{n+2} (R_n+1) + (1-\frac{R_n}{n+2})R_n= \frac{n+3}{n+2} R_n$ $\endgroup$ Oct 6 '19 at 20:31
  • $\begingroup$ It's clear. Just one more question about the definiton of the random variable R_n. How is it "formally defined"? I mean, it goes from $(\Omega, \mathcal{F})$ into $(R, \mathcal{R})$, but in which way? I just have the interpretation: "R_n is the number of red ball at time $n$" $\endgroup$ Oct 6 '19 at 20:35
  • $\begingroup$ That's a tricky question; in general, people tend to ignore the underlying space $(\Omega,\mathcal{F})$ and simply assume it is sufficiently large to support all the randomness in the stochastic process (here, all the uniform selections). $R_n$ is just the counting function for the number of red balls selected up to time $n$, which requires measurability of the preimages $R_n^{-1}(k)$ for all $k\in \mathbb{N}$, but again, this is usually swept under the rug. $\endgroup$
    – J.G
    Oct 6 '19 at 20:39
  • $\begingroup$ Thanks. I asked this because in the computation of the expected value I used the "formula" $ \sum_k k P(X=k)$. But in this case I just considered $k \in {n,n+1}$ , this is because I was conditioned to $F_n$, right? $\endgroup$ Oct 6 '19 at 20:46
  • $\begingroup$ Sorry for bothering you, but in this example what would be the right probability space? Could you try to be more precise? $\endgroup$ Oct 9 '19 at 14:14

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