Ant Movement

Solution

Answer: \( (-2.5,\,-2.5). \)

Let the ant’s position after one step be the random vector \[ (X,Y). \] Each second the ant moves in one of four directions with equal probability \( \frac14 \): Right \(1\): \((1,0)\) Up \(2\): \((0,2)\) Left \(3\): \((-3,0)\) Down \(4\): \((0,-4)\) First compute the expected displacement in the \(x\)-direction during one step. \[ E(X) = \frac14(1) + \frac14(0) + \frac14(-3) + \frac14(0) \] \[ E(X) = -\frac12. \] Now compute the expected displacement in the \(y\)-direction during one step. \[ E(Y) = \frac14(0) + \frac14(2) + \frac14(0) + \frac14(-4) \] \[ E(Y) = -\frac12. \] Thus the expected movement per second is \[ (-0.5, -0.5). \] Over \(5\) seconds, expected displacements add linearly. Therefore \[ E(\text{position after 5 steps}) = 5(-0.5,-0.5). \] \[ = (-2.5,-2.5). \]