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.


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.