Question

Emily regularly runs a course that starts at the front of her house, goes to point P (which is on the route she takes to go to work), and comes back. She would like to know the total distance. She will use the odometer in her car. This odometer gives a reading that has mean equal to the distance driven and a standard deviation of .05. Consider the following two options.

A -- She drives her car from her house to point P , comes back, and records the odometer reading XA.

B -- (Let’s not waste gas.) She drives her car from her house to point P, records the odometer reading X1, and continues to work. On her way back from work, she starts the odometer when she reaches point P , continues to drive all the way home, and records the odometer reading X2 for the distance from P to home. The course distance is estimated via X = X1 +X2. She does this again the next day, getting another estimate, Y , and her final estimate is the average over the two days: XB = (X + Y )/2.

Which method gives a more accurate answer?

Answer 1

Let D be the true distrance between Emily house and point P.

Method A:

Mean of XA = E[XA] = D

Variance of XA = Var[XA] =

Standard deviation of XA = 0.05

Method B:

E[X] = E[X1 + X2] = E[X1] + E[X2] = D/2 + D/2 = D

Var[X] = Var[X1 + X2] = Var[X1] + Var[X2] = + = (X1 and X2 are independent)

Similarly,

E[Y] = D

Var[Y] =

E[XB] = E[(X + Y)/2] = (E[X] + E[Y])/2 = (D + D)/2 = D

Var[XB] = Var[(X + Y)/2] = (Var[X] + Var[Y])/ = ( + ) /4 (X and Y are independent)

=

Standard deviation of XB = 0.05

Since Var[XA] = Var[XB] ,  both method will give the answer with the same accuracy.