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Is it possible to estimate the average end point of a 1-dimensional random walk of n steps where the probability of going "left" is p and going "right" is 1-p?

Thanks.

dabd
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  • Yes: the mean of the endpoint is exactly $(1-2p)n$, the endpoint itself is $(1-2p)n+O(\sqrt{n})$. – Did Apr 28 '13 at 23:07

1 Answers1

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Let $X_t$ be the position at time $t$.

We can think of the random walk as being the sum of a set of random variables, $Z_1$, $Z_2$, $Z_3$, ..., each taking the value 1 with probability $p$ and the value -1 with probability $1-p$.

Hence:

$$X_t = \sum\limits_{i=1}^t Z_i$$

So by the linearity of expectation:

$$\mathbb{E}[X_t] = \sum\limits_{i=1}^t\mathbb{E}[Z_i] = \sum\limits_{i=1}^tp -(1-p) = (2p - 1)t$$

user73445
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  • I was thinking about applying this to estimating the average bankroll after t games of playing a game where we have probability p of winning (1-p) of losing. If we win we gain 1-2*r units where r is a game fee that is typically very small comparing to 1, (for example 2% of 1 unit). If we lose a game we lose 1 unit. In this case the expectation would be ((1-2r)p-(1-p))t. – dabd Apr 29 '13 at 00:05