Find E[X] and Var[X] of a particle moving along a graph

Question

A particle moves along the following graph

0 <---> 1 <---> 2

If the particle is at 0, it will next move to 1. If the particle is at 1, it will next move to 0 with probability 0.99 and to 2 with probability 0.01. Starting at 0, let X denote the number of steps needed to reach 2.

I'm having trouble with this question. I tried to use X = X1 + X2 + X3 + ... , and i noticed that it can only reach 2 with even steps so I made a new variable Y = Y1 + Y2 + Y3 + ..., where Y represents the even round that it takes to reach and created a summation of $\sum_{i=1}^\infty 0.99^i(0.01)*2$ but I dont think its correct as it seems to tend to infinte

Answer 1

The first thing to observe is that state $2$ can only be reached after an even number of steps, if starting at state $0$. We can see this by coloring states $0$ and $2$ red, and coloring state $1$ blue. Each transition from one state to another must change the color of the state we are in, either from red to blue, or from blue to red. Since $0$ and $2$ share the same color, it is only possible to reach $2$ after an even number of color changes.

This also implies that if the number of steps is odd, then we are always at state $1$.

So suppose we have taken $2n-1$ steps for some positive integer $n$ but have not reached state $2$. We are at state $1$, and the probability of being at state $2$ and stopping at the next step, is simply $0.01$. In this case, we will have taken $X = 2n$ steps. But with probability $0.99$, we will return to state $0$, after which we will again be at state $1$ after $2(n+1)-1 = 2n+1$ steps. So the number $n$ represents the number of times we have entered state $1$.

To have reached state $1$ upon the $n^{\rm th}$ time without stopping at state $2$, we must have had consecutive $n-1$ failures to transition from state $1$ to state $2$. So the probability of this event is simply $(0.99)^{n-1}$. Then the probability of the next step ending at state $2$ is $(0.01)$ times this probability; hence $$\Pr[X = 2n] = (0.01)(0.99)^{n-1}, \quad n \in \{1,2,\ldots\}.$$ consequently, $X/2$ is a geometric random variable with success parameter $p = 0.01$ and counts the number of times we have entered state $1$ before reaching state $2$, and $X$ is twice a geometric random variable that counts the number of steps to reach state $2$.

The mean and variance of $X$ are straightforward to calculate and are left as an exercise.

Answer 2

If we let $Y_i$ be the event that it reaches point 2 on the graph in $2i$ steps then the equation for the probability of $Y_i$ should be:

$P(Y_i) = (0.99)^{i-1}(.01)$ then the summation is equal to 1,

$\displaystyle\sum_{i=1}^{\infty}P(Y_i) = 1$