Question

A shipping company handles containers in three different sizes: (1) 27 ft3 (3 × 3 × 3), (2) 125 ft3, and (3) 512 ft3. Let Xi (i = 1, 2, 3) denote the number of type i containers shipped during a given week. With μi = E(Xi) and σi2 = V(Xi), suppose that the mean values and standard deviations are as follows: μ1 = 220 μ2 = 250 μ3 = 120 σ1 = 9 σ2 = 13 σ3 = 8 (a) Assuming that X1, X2, X3 are independent, calculate the expected value and variance of the total volume shipped. [Hint: Volume = 27X1 + 125X2 + 512X3.] expected value ft3 variance ft6 (b) Would your calculations necessarily be correct if the Xi's were not independent? Explain. Both the expected value and the variance would be correct. The expected value would be correct, but the variance would not be correct. Neither the expected value nor the variance would be correct. The expected value would not be correct, but the variance would be correct.

Answer 1

a. Expectation is linear, so that

$E[27X_1+125X_2+512X_3]=27E[X_1]+125E[X_2]+512E[X_3]=98,630$

The variance of a sum of independent random variables is equal to a weighted sum of the variances, with weights equal to the squares of the coefficients of the $X_i$:

$V[27X_1+125X_2+512X_3]=27^2V[X_1]+125^2V[X_2]+512^2V[X_3]=2,306,838$

b. If the $X_i$ were dependent on one another, we would have the same expectation, but now the variance of a sum of random variables becomes the sum of their covariances:

$\displaystyle\sum_{i=1}^3V[\alpha_iX_i]=\sum_{i=1}^3\sum_{j=1}^3\mathrm{Cov}[X_i,X_j]=\sum_{i=1}^3{\alpha_i}^2V[X_i]+2\sum_{i\neq j}\alpha_i\alpha_j\mathrm{Cov}[X_i,X_j]$

where

$\mathrm{Cov}[X_i,X_j]=E[(X_i-E[X_i])(X_j-E[X_j])]=E[X_iX_j]-E[X_i]E[X_j]$

When we assumed independence, we were granted $E[X_iX_j]=E[X_i]E[X_j]$, but this may not be the case if the variables are dependent.