A fair four-sided die has its sides labeled U, D, L, and R, respectively. A token is placed at (0, 0) on the Cartesian plane and the die is then rolled repeatedly. After each roll, the token is moved as follows:
Token Moves
U (a, b) → (a, b + 1)
D (a, b) → (a, b − 1)
L (a, b) → (a + 1, b)
R (a, b) → (a − 1, b)
Let the random variable Yn be the taxicab distance. Taxicab distance is calculated by |a|+|b|.
What is the expected value of Yn and why?
What is the variance of Yn and why?
What I know so far:
After using a program that I made and tracking the sample space of n=1, n=2, n=3 and n=4 I managed to determine an equation that seems to generate the correct expected value 100% of the time.
E(Yn)=E(Y(n-1))*(2n-1)/(2n-2)
I don't know how I'd go about proving this equation and I lack a conceptual understanding of why it works. I need help with finding why this relation works.
The expected value is to be $0$; if it was not, then then die must be biased toward a particular direction(s).
If each roll is treated as a new instance of a random variable, the variance of each roll with be exactly $1$. For example, with just one roll, you are guaranteed to be one square from the origin. If the new location of the token is set as a new origin, the variance of the next roll must also be $1$, and so on...
One of the most important rules of statistics says that variances of random variables are always additive; regardless of whether you are adding or subtracting the output values of the random variables. Thus, the variance in the question you are asking will simply be equal to the number of rolls performed.