From an urn containing $a$ white and $b$ black balls, balls are drawn one by one at random according to the following rules:
(i) at any drawing, if the ball drawn is white, then it is returned to the urn,
(ii) if it is black, it is replaced by a white ball (from another collection of balls).
After n such operations, a ball is drawn from the urn. let $W_n$.denote the number of white balls in the urn after the foregoing operations has been repeated n times. I have the following recursive equation for the question $$\mathbb{E}W_{n+1}=\left(1-\frac{1}{a+b}\right)\mathbb{E}W_n +1 $$ My question is how can ı use the above equation to prove the following recursive equation ? $$\mathbb{E}W_{n}= a+b-b\left(1-\frac{1}{a+b}\right)^n$$
Do you have any advice? I have tried to write the second equation for $n =1,2,\cdots$ but I could not get any connection between them ?
You want to solve,
$$x_{n+1}=cx_{n}+1$$
Define,
$$x_n=u_n+\frac{1}{1-c}$$
Then we have,
$$u_{n+1}=cu_n$$
This is a much easier recurrence to solve. If we start at $u_0$ to get to the next term we multiply by $c$, and multiplying by $c$ more than once results in a power,
$$u_n=u_0c^n=\left(x_0-\frac{1}{1-c}\right)c^n$$
$$x_n=u_n+\frac{1}{1-c}=\left(x_0-\frac{1}{1-c}\right)c^n+\frac{1}{1-c}$$
In your example we have,
$$1-\frac{1}{a+b}=c$$
$$x_0=a$$
$$\mathbb{E}W_n=x_n$$
Going back to the definition I made, how did I come up with $\frac{1}{1-c}$?
If you are ever faced with a recurrence of the form,
$$x_{n+1}=ax_{n}+b$$
Then pretend or assume that the sequence has a limit $L$ and solve for it, it will help you solve the recurrence. I will show you how with your specific example.
Looking at our original recurrence, we have that if the sequence converges to $L$, then
$$L=cL+1 \qquad (1)$$
But we also have,
$$x_{n+1}=cx_n+1 \qquad (2)$$
Subtracting (1) from (2) we have,
$$x_{n+1}-L=c(x_n-L)$$
And now all of a sudden it becomes apparent what substitution to use:
$$x_n-L=u_n$$
$$x_n=u_n+L$$
Solving for $L$ from equation $(1)$ we get:
$$L=\frac{1}{1-c}$$