For the random walk with step sizes: $S_i = \begin{cases} &+1 &\text{probability} &p, \\ &-2 &\text{probability} &q=1-p \end{cases}$

Let $T_n = \sum_{i=1}^mS_i$ be the displacement after a fixed, not random, number of steps $n$.

Find the probability distribution: $P(T_n=t)$

and the mean and variance of $T_n$ in terms of a general $n$ and $p$.

Note that $\frac{S_i+2}{3}$ is distributed as $B(1,p)$, giving that $\frac{T_n+2n}{3}$ is distributed as $B(n,p)$. Thus:

$$\mathbb{E}\left(\frac{T_n+2n}{3}\right)=np, \mathbb{V}\left(\frac{T_n+2n}{3}\right)=np(1-p)$$

from which you can derive:

$$\mathbb{E}(T_n)=n(3p-2), \mathbb{V}(T_n)=9np(1-p)$$

Furthermore, knowledge of the binomial distribution gives that:

$$\mathbb{P}\left(\frac{T_n+2n}{3}=k\right)=\binom{n}{k}p^k(1-p)^{n-k}$$

which can be manipulated to find the distribution of $T_n$ itself.

Let X_n be the random variable corresponding to the number of +1 steps in n turns. $X_n$ will follow Binomial Distribution with parameters $n$ and $p$. If $k$ is the number of +1 steps. Then, $n-k$ is the number of +2 steps. Then, $k+2(n-k) =t$. On solving, we get $k = 2n - t$.

Thus, $P(T_n = t) = P(X_n = 2n-t) = {n \choose 2n-t}p^{2n-t}(1-p)^{t-n}$.

Now, $E[S_i] = p + (1-p)*(-2) = 3p-2$ and, $Var[S_i] = E[S_i^2] -E[S_i]^2 = p + (1-p)(-2)^2 -(3p-2)^2 = 9p(1-p)$.

Using linearity of expectations and the fact that each step of the walk is independent and identically distributed, we get $E[T_n] = nE[S_i] =3np-2n$ and $Var[T_n] = nVar[S_i] =9np[1-p]$.

$P(T_n=t)=P(\sum_{i=1}^n S_i=t)$ so this is essentially the $n$-fold convolution of $S_1,S_2,...,S_n$. Not sure if a neat answer exists.

$E(T_n)=E(\sum_{i=1}^nS_i)=\sum_{i=1}^nE(S_i)=\sum_{i=1}^n[1\times p-2\times(1-p)]=n(3p-2)$

$Var(T_n)=Var(\sum_{i=1}^nS_i)=\sum_{i=1}^nVar(S_i)=nVar(S_1) $ due to iid $S_i$.

Find $Var(S_1)=E(S_1^2)-(E(S_1))^2$. Simplify.

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.