An introduction to random variables (part 2)

This article follows on from An Introduction to Random Variables (Part 1). Here, fundamental results regarding the expectation and variance of random variables (discrete or continuous) are stated and proved.

Property 1

For some constant c ∈ ℜ and random variable X,

E(cX) = cE(X). (1)

Proof:
From Part 1-(1), for X discrete

while from Part 1-(8), for X continuous

Property 2

For discrete or continuous random variables X and Y,

E(X + Y) = E(X) + E(Y). (2)

Proof:
Suppose X and Y have joint mass function f_X,Y : ℜ² → [0, 1] given by f_X,Y(x,y) = P(X = x and Y = y). Then, for X and Y discrete, an extension of Part 1-(1) gives

Noting Part 1-(8), for X and Y continuous the proof begins

where f_X,Y(x, y) : ℜ² → [0, ∞) is the joint density function of X and Y. The proof then proceeds in a similar way to the discrete case with summations replaced by integrations.

Property 3

For discrete or continuous independent random variables X and Y,

E(XY) = E(X)⋅E(Y) (3)

Proof:
The proof of (3) is first presented for discrete random variables X and Y. Let X and Y have joint mass function f_X,Y(x, y) : ℜ² → [0, 1] given by f_X,Y = P(X = x and Y = y). If X and Y are independent, then (by definition) the probability of Y occurring is not affected by the occurrence or non-occurrence of X. For X and Y independent,

P(X = x and Y = y) = P((X = x) ∩ (Y = y)) = P(X = x)⋅P(Y = y),

so that f_X,Y(x, y) = f_X(x)⋅f_Y(y). Therefore,

For X and Y continuous, the proof begins

where f_X,Y : ℜ² → [0, ∞) is the joint density function of X and Y. The proof then proceeds in a similar way to the discrete case with summations replaced by integrations.

Property 4

For the discrete or continuous random variable X,

(4)

Proof:
The proof of (4) holds for X discrete or continuous. As a shorthand notation, let μ = E(X). Then,

Property 5

For the discrete or continuous random variable X and the constant c ∈ ℜ,

Var(cX) = c²⋅Var(X). (5)

Proof:
The proof of (5) holds for X discrete or continuous. Again, let μ = E(X). Then,

Var(cX) = E((cX - cμ)²) = E(c²⋅(X - μ)²) = c²⋅E((X - μ)²) = c²⋅Var(X).

Property 6

For discrete or continuous independent random variables X and Y,

Var(X + Y) = Var(X) + Var(Y).(6)

Proof:
The proof of (6) holds for X and Y discrete or continuous. As a shorthand notation, let μ_X = E(X) and μ_Y = E(Y). Then,

However, from (3), if X and Y are independent random variables, then