I'm sure this should be straight forward but somehow i can't find a similar example online.
How do i calculate expectation and variance in this case:
A random walker takes 1 step backwards (-1) with p= 0.2, 1 step forward (+1) with p= 0.5, and stays at the same position with p = 0.3.
How do i find expectation and variance after n steps?
Best Answer
Here are some of the concepts that you need to understand to solve this problem:
(1) The final location can be viewed as the sum of steps from the initial location: $$X(n) = X(0)+S(1)+S(2)+\dots+S(n)$$
Here $X(0)$ is the initial location, which also contributes to the mean and variance of $X(n)$. Often $X(0)$ is given as the origin (deterministic initial condition), in which case it has a mean and variance of zero, so can be ignored. I do not see an initial condition given in your problem. You would need that to determine the answer.
(2) The mean of a sum of random variables is equal to the sum of the means of the random variables. This is ALWAYS true (when each random variable has a mean), since the expected value is a linear operator:
When $a$ and $b$ are nonrandom: $$\mathbb{E}[aX+bY] = a\mathbb{E}[X]+b\mathbb{E}[Y]$$
(3) When random variables are INDEPENDENT (as in this problem), then the variance of their sum is equal to the sum of their variances.
Without independence, the formula is: $$\operatorname{Var}[aX+bY] = a^2\operatorname{Var}[X]+b^2\operatorname{Var}[Y] + 2ab \operatorname{Cov}[X,Y]$$ Independence implies that $\operatorname{Cov}[X,Y]=0$ (independence is a stronger condition)
(4) You need to know how to compute the mean and variance of each step. Your computation of the mean in the comments is correct (you should edit your original post to include that, rather than have it in the comments). Your computation of the variance is not correct, your computation is actually for $\mathbb{E}[S_1^2] = 0.2(-1)^2 + 0.5(1)^2 + 0.3(0)^2$. In order to compute the variance, you could use that computation, together with the identity: $$\operatorname{Var}[X] = \mathbb{E}[X^2] - \left( \mathbb{E}[X]\right)^2$$