Normally distributed random walk

Question

Consider a random walk on the real line with discrete steps taken once each second. Suppose that each step $\Delta X_i$ is an independent normally distributed random variable with mean 0 and variance $\sigma^2$. So, if at t=0, you start at position $X_0$, then after one second you will be at $X_{1} = X_0 + \Delta X_1$, then after two seconds $X_{2} = X_1 + \Delta X_2$, etc.

If $X_0 = 25$, and $\sigma = 2$, what is the probability that between 12 and 13 seconds, the random walker is standing at a position greater than 32? This is equivalent to $P({X_{12} > 32})$.

I know the root mean square translation distance is $\sigma \sqrt{n}$ after $n$ steps, which is $2 \sqrt{12}$ in our case, but I do not know how to transform this into the desired probability.

Answer 1

Guide:

We have $X_{12} = X_0 +\sum_{i=1}^{12}\Delta X_i=25 + \sum_{i=1}^{12}\Delta X_i$.

Hence you just have to solve for $P(25 + \sum_{i=1}^{12}\Delta X_i>32)$.

It might help if you use a notation $W=\sum_{i=1}^{12}\Delta X_i$, you might first want to find the mean and the variance of $W$ and note that $W$ follows a normal distrbution.

Answer 2

first you can easily verify that $$ X_{12} = X_{11} + \Delta X_{12} = X_0 + \sum_{t = 1}^{12} \Delta X_t = 25 + \sum_{t = 1}^{12} \Delta X_t. $$

Since all the $\Delta X_t$ have a normal distribution and are independent we get that also the sum $\sum_{t = 1}^{12} \Delta X_t$ has a normal distribution and therefore $X_{12}$ has a normal distribution.

A simple calculation reveals that \begin{align} \mathbb{E}[X_{12}] &= X_0 = 25, \\ \text{Var}(X_{12}) &= \sum_{t = 1}^{12} \text{Var}(\Delta X_{t}) = 12 \cdot 4. \end{align}

Therefore $X_{12} \sim \mathcal{N} (25, 48)$. Can you solve it from here?