Question

Let X, Y, and Z be three independent, identically distributed random variables, each with a density function f(x). Determine E[X+Y+Z].

a) 3E[f(x)]
b) E[f(x)]/3
c) E[f(x)] + Var[f(x)]
d) 3E[f(x)] + 3Var[f(x)]

Answer 1

The correct answer is 3E[X], which represents the sum of the expectations of the three independent identically distributed random variables.

Step-by-step explanation:

The expectation, E[X+Y+Z], of the sum of three independent, identically distributed random variables X, Y, and Z is found by using the linearity of expectation. Since these variables are identically distributed with the same density function f(x), they also have the same expectation, denoted as E[X], E[Y], and E[Z]. Therefore, the expected value of the sum is E[X] + E[Y] + E[Z]. Given the independence and identical distribution, the expected value of each variable is just E[X].

The direct answer in two lines is:

E[X+Y+Z] = E[X] + E[Y] + E[Z]
E[X+Y+Z] = 3E[X]

Note that E[f(x)] pertains to the expectation of the probability density function, which is not the same as the expectation of the random variables themselves. Thus, the correct answer to the student's question is neither of the given options, but rather the sum of each variable's expectation, resulting in 3E[X].