R.V.S Explaination

Distribution Functions of Independent Discrete Random Variables that are Nonnegative

September 9, 2013– Matthew J. Sobel

1

Independent random variables

Let X and Y be two random variables (r.v.s) defined on the same sample space. Their joint distribution function (d.f.) is FX,Y (a, b) = P (X ≤ a, Y ≤ b) where a and b are any numbers. Similarly, label the d.f.s of X and Y with FX (a) = P (X ≤ a) and FY (b) = P (Y ≤ b).

The r.v.s X and Y are independent if FX,Y (a, b) = FX (a)FY (b) for all pairs of numbers (a, b). If X and Y are discrete r.v.s, then they have probability mass functions pX (a) and pY (b) where a and b are integers and pX (a) = P (X = a) and pY (b) = P (Y = b).

The joint probability mass function of X and Y is pX,Y (a, b) = P (X = a, Y = b). The purpose of this note is to show that nonnegative discrete r.v.s X and Y are independent if and only if pX,Y (a, b) = pX (a)pY (b) for all pairs of nonnegative integers (a, b).

2

If pX,Y (a, b) = pX (a)pY (b) for all pairs of nonnegative integers (a, b), then X and Y are independent.

We need to show that if pX,Y (j, k) = pX (j)pY (k) for all pairs of nonnegative integers (j, k), then FX,Y (a, b) = FX (a)FY (b) for all pairs of integers (a, b): FX,Y (a, b) =

j≤a k≤b

pX,Y (j, k) pX (j)pY (k)

j≤a k≤b

= =

j≤a

pX (j)

k≤b

pY (k)

= FX (a)FY (b) 1

3

If X and Y are independent, then pX,Y (a, b) = pX (a)pY (b) for all pairs of nonnegative integers (a, b).

The r.v.s are nonnegative, so FX , Y (0, 0) = pX,Y (0, 0). Their independence and nonnegativity imply pX,Y (0, 0) = FX,Y (0, 0) = FX (0)FY (0) = pX (0)pY (0). This starts an inductive proof in which the inductive assumption is that for nonnegative integers a and b, pX,Y (a, k) = pX (a)pY (k) for all k < b. That assumption and the definition of independent r.v.s imply pX,Y (a, b) = FX,Y (a, b) − FX,Y (a − 1, b) −

k≤b−1

pX,Y (a, k) pX (a)pY (k).

k≤b−1

= FX (a)FY (b) − FX (a − 1)FY (b) −

The inductive...