Question

conditional expectations in Polya's urn scheme. An urn contains 1 black and 2 white balils. One ball is drawn at random and its color noted. The ball is replaced in the um, together with n additional ball of its color. There are now four balls in the urn. Again, one ball is drawn at random from the urn, then replaced along with an additional bal of its color. The process continues in this way. Let B be the number of black balls in the urn just before the nth ball is drawn.(thus B₁ Is 1) for n ≥ 1, find E(Bₙ₊₁ I Bₙ)

Answer 1

In Polya's urn scheme, the conditional expectation E(Bₙ₊₁ | Bₙ) for n ≥ 1 can be found by considering the possible outcomes for Bₙ. If Bₙ = 1, the probability is 2/3 and the expectation is ⅔. If Bₙ = 2, the probability is 1/3 and the expectation is 2⅓.

Step-by-step explanation:

In Polya's urn scheme, an urn contains 1 black and 2 white balls. The process starts with drawing one ball at random and noting its color. The ball is then replaced along with an additional ball of its color, resulting in four balls in the urn. This process continues, with one ball being drawn at random and replaced with an additional ball of its color.

To find the conditional expectation E(Bₙ₊₁ | Bₙ) for n ≥ 1, we need to find the expected number of black balls in the urn just before the nth ball is drawn.

For n ≥ 1, Bₙ represents the number of black balls in the urn just before the nth ball is drawn. Since we know that B₁ is 1, we can find E(Bₙ₊₁ | Bₙ) by considering the possible outcomes for Bₙ.

If Bₙ = 1, then the nth ball drawn must be white, so Bₙ₊₁ will still be 1. The probability of Bₙ = 1 is 2/3 (since there are 2 white balls for the first draw).

If Bₙ = 2, then the nth ball drawn must be black, so Bₙ₊₁ will be 3 (since we replace the drawn black ball and add an additional black ball). The probability of Bₙ = 2 is 1/3 (since there is 1 black ball for the first draw).

Therefore, the conditional expectation E(Bₙ₊₁ | Bₙ) is given by:

E(Bₙ₊₁ | Bₙ) = (1/3) * 2 + (2/3) * 3 = 2⅓