Two or more independent random variables can be combined together by either adding or subtracting.
Mean
For the expected value of these sums and differences of the random variables X and Y:
E(X + Y) = E(X) + E(Y) E(X – Y) = E(X) – E(Y) 
In words
The mean of the sum of two random variables is equal to the sum of the individual means.
The mean of the difference of two random variables is equal to the difference of the individual means.
Example
X and Y are random variables with E(X) = 3 and E(Y) = 4.  

E(X + Y) = E(X) + E(Y) = 3 + 4 = 7 

E(X − Y) = E(X) − E(Y) = 3 − 4 = 1 
Variance
For the variance of these sums and differences of the independent random variables X and Y:
VAR(X + Y) = VAR(X) + VAR(Y) VAR(X – Y) = VAR(X) + VAR(Y) 
In words
The variance of either the sum or difference of two independent random variables is equal to thesum of the individual variances.
Example



VAR(X + Y) = VAR(X) + VAR(Y) = 6 + 7 = 13 

VAR(X − Y) = VAR(X) + VAR(Y) = 6 + 7 = 13 
Functions of Combinations of Random Variables
It also follows from our earlier work that:
For independent random variables X and Y with constants a and b: E(aX ± bY) = aE(X) ± bE(Y) and VAR(aX ± bY) = a^{2}VAR(X) + b^{2}(VAR(Y)

Example of an Application of Sums of Random Variables
A survey taken by an airline shows that the distribution of the weights of their passengers has a mean of 68 kg and a standard deviation of 10 kg, while the weight of the luggage carried by each passenger has a mean of 12 kg and a standard deviation of 6 kg.
a. Calculate the mean and standard deviation of the combined weight of a passenger and baggage assuming these are independent.
Let P = weight of a passenger and L = weight of a passenger's baggage.
E(P + L) = E(P) + E(L) = 68 + 12 = 80 kg
VAR(P + L) = VAR(P) + VAR(L) = 10^{2} + 6^{2} = 136 kg
SD(P + L) = √136 = 11.7 kg (to 3 sig.fig.)
b. Calculate the mean and standard deviation of the total weight of two passengers.
Let the two passengers' weights be P_{1} and P_{2.}
E(P_{1} + P_{2}) = E(P_{1}) + E(P_{2}) = 68 + 68 = 176 kg
VAR(P_{1} + P_{2}) = VAR(P_{1}) + VAR(P_{2}) = 10^{2} + 10^{2} = 200 kg
SD(P_{1} + P_{2}) = √200 = 14.1 kg (to 3 sig.fig.)