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Question

Suppose I have a particle at position $0$ that, every second, moves either 1 step right w.p. $p$ or 1 step left w.p. $1-p$. The particle stops when it reaches position $1$. If $T$ is the time it takes until the particle stops, what is $Var[T]$?

Thoughts

I know that $E[T]$ can be calculated by noting a simple linear relationship between when the particle is at $0$ vs when it is at $-1$; but I do not see how to approach finding the variance. Of course, I would like to be able to find an approach which easily generalizes (i.e. particle starts at $0$ and ends when it reaches $n>0$.

Also, this was in Ross Stochastic Processes chapter 1 (not homework, just piqued my interest) and so it should have an elementary solution.

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    You can say $E[T^2]=E[T^2|\mbox{first move right}]p + E[T^2|\mbox{first move left}](1-p)$ and note that, given we first move left, the stochastics of $T$ can be understood in terms of itself. – 2017-01-13
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    I suspected as much, but how do we relate $E[T^2|\text{first move left}]$ to $E[T^2]$? I think theres a simple relationship I am overlooking... – 2017-01-13
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    Similar to how you did it for the computation of $E[T]$. How did you do that one? – 2017-01-13
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    Note that $E[T|\text{you're at -1}]=2*E[T|\text{you're at 0}] = 2*E[T]$ and then compute the total probability. – 2017-01-13
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    And why is that relationship true? (Actually it is not: $E[T|\mbox{first move left}] = 1 + 2E[T]$). But, what relationship does the remaining time $R$ (not including the first step) have with the original $T$ variable? – 2017-01-13
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    I reasoned intuitively- because it takes $E[T]$ moves to get from -1 to 0, and then another $E[T] $ moves to get from 0 to 1. I'm not seeing how this generalized to $E[T^2]$- sorry if I'm being unreasonably dense! – 2017-01-13
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/51725/discussion-between-michael-and-user3000877). – 2017-01-13

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As we discussed, you can note that $$ T = \left\{ \begin{array}{ll} 1 &\mbox{ , if first move right} \\ 1 + A + B & \mbox{ , if first move left} \end{array} \right.$$ where $A$ and $B$ are independent and have the same distribution as $T$. You can work out the details and perhaps create your own full answer to mark as "best answer" if you like.