My question is about Entropy rate of a dog looking for a bone. According to this problem, A dog walks on the integers, possibly reversing direction at each step with probability $p=0.1$ . Let $X_0 = 0$. The first step is equally likely to be positive or negative. A typical walk might look like this: $$(X_0,X_1,...)=(0,−1,−2,−3,−4,−3,−2,−1,0,1,...).$$
c) What is the expected number of steps that the dog takes before reversing direction.
I'm at a loss as to how $E(S) = 10$. (shown below)
$$\ E(S) = \sum_{s=1}^{\infty} s(0.9)^{s-1}(0.1)$$ $$\ = 10$$
The probability the dog reverses on the first step is 0.1. The probability it reverses on the second step is (0.9)(0.1) since it must NOT reverse on the first step. The probability it reverses on the third step is (0.9)^2(0.1). In the same way, the probability it reverses on the "nth" step is (0.9)^{n-1}(0.1). (You can check that thw probability it reverses on some step is 0.1+ (0.9)(0.1)+ (0.9)^2(0.1)+ …= 0.1(1+ 0.9+ 0.9^2+ …)= 0.1(1/(1-.9))= 0.1/0.1= 1, as it must be, since the sum is a "geometric series".)
The expected value is 1(0.1)+ 2(0.9)(0.1)+ 3(0.9)^2+ …= 0.1( 1+ 2(0.9)+ 3(0.9)^2+ …).
To do that, recall that the derivative of x^n is nx^(n-1). The sum above is 0.1 times the derivative of 1+ x+ x^2+ x^3+... for x= 0.9. That is again a geometric series so has sum 1/(1- x)= (1- x)^(-1). The derivative of that is (-1)(1- x)^(-2)(-1)= (1- x)^(-2). When x= 0.9 that is (.1)^{-2}= 1/(10^2)= 100 and multiplying by 0.1 we have 10.