An urn contains$a$white and $b$black balls. After a ball is drawn, it is returned to the urn if it is white; but if it is black, it is replaced by a white hall from another urn. Let $M_n$denote the expected number of white balls in the urn after the foregoing operation has been repeated$n$times.

1. Derive the recursive equation $M_{n+1}=\left(1-\frac{1}{a+b}\right)M_n+1$
   
2. Use part (a) to prove that $M_n=a+b-b{\left(1-\frac{1}{a+b}\right)}^n$
   
3. What is the probability that the $(n+1)$ball drawn is white?

Given in the question that the define random variables $X_n$which counts how many of white balls is in the urn after $n$draws. The recursive equation $M_{n+1}=\left(1-\frac{1}{a+b}\right)M_n+1$

Let's find the conditional expectation of $X_{n+1}$given $X\_\left\{n\right\}.$

Consider what has been drawn in the $n+1$draw. If we have drawn a white ball (probability $X_n/(a+b)$), the expected value $X_{n+1}$is $X_n$since we return back that white ball. If we have drawn a black ball (probability $1-X\_\left\{n\right\}/(a+b)),$we replace it with the white ball, so the expected value of $X_{n+1}$is$X\_\left\{n\right\}+1.$

Hence

$E\left(X_{n+1}\mid X_n\right)=\frac{X_n}{a+b}\times X_n+\left(1-\frac{X_n}{a+b}\right)\times \left(X_n+1\right)$

$=\left(1-\frac{1}{a+b}\right)X_n$

Applying expectation to both sides and using the relation $E\left(E\left(X_{n+1}\mid X_n\right)\right)=E\left(X_{n+1}\right)=M_{n+1}$

We get,$M_{n+1}=\left(1-\frac{1}{a+b}\right)M_n$which has been claimed.

Given in the question to prove that $M_n=a+b-b{\left(1-\frac{1}{a+b}\right)}^n$

Let's prove it by induction.

For $n=0$,

We have that$M_0=a+b-b=a$

which is true since we have $a$white balls at the beginning. Suppose that the expression is true for$M_n$. Consider $M_{n+1}$. Using the relation from part (a), we have that

role="math" $M_{n+1}=\left(1-\frac{1}{a+b}\right)M_n$

$=\left(1-\frac{1}{a+b}\right)\left(a+b-b{\left(1-\frac{1}{a+b}\right)}^n\right)=a+b-b{\left(1-\frac{1}{a+b}\right)}^{n+1}$

The claimed expression hold for every$n\ge 0$

Define indicator random variable $I_{n+1}$that is equal to one if and only if the$n+1$drawn ball is white.

Let's find the conditional expectation of $I_{n+1}$a given$X_n$.

We have that,$E\left(I_{n+1}\mid X_n\right)=\frac{X_n}{a+b}$

since if we know that there are $X_n$white balls, we know that the probability that we draw white ball is $X_n/(a+b)$.

Applying the expectation to both sides, we have that$P\left(I_{n+1}=1\right)=E\left(I_{n+1}\right)=\frac{1}{a+b}{M}_n=\frac{1}{a+b}\left(a+b-b{\left(1-\frac{1}{a+b}\right)}^n\right)$