This is Example 2(i) of Sheldon Ross' A First Course in Probability.
I have a question about a variation of this problem:
Consider a particle initially located at a given point in the plane, and suppose that it undergoes a sequence of steps of fixed length, but in a completely random direction. Specifically, suppose that the new position after each step is one unit of distance from the previous position and at an angle of orientation from the previous position that is uniformly distributed over (0, 2π).
The variation I have would be to find the expected value of the final point instead. My question is whether this is equal to $0$ as the expected value of the X-coordinate can be written as,
$$E(X) = \sum_{i = 1}^{n} E(X_i)$$
and $X_i$ would just be equal to $\cos{\theta_i}$, where $\theta_i$ is the angle randomly chosen at each turn. Then,
$$E(X_i) = \frac{1}{2\pi} \int_{0}^{2\pi} \cos{\theta_i} dx = 0$$
which means that the expected value of $X$ would be $0$. The same argument can be used for $Y$ to show that the expected coordinate would be $(0, 0)$
Is this correct or am I missing something?