Solutions to Homework Set No. 4 – Probability Theory (235A), Fall ...

This implies that X + Y ∼ Poisson(λ + µ).
(c) Use the same “discrete convolution” formula to prove directly that if X ∼ Bin(n, p)
and Y ∼ Bin(m, p) are independent then X + Y ∼ Bin(n + m, p). You may make use
of the combinatorial identity (known as the Vandermonde identity or Chu-Vandermonde
identity)
Solution.
k
j=0
P(X + Y = k) =
n m
=
j k − j
n + m
k
P(X = j, Y = k − j)
k
, (n, m ≥ 0, 0 ≤ k ≤ n + m).
j=0
k
n
= p
j
j=0
j (1 − p) n−j
m
p
k − j
k−j (1 − p) m−(k−j)
= p k (1 − p) n+m−k
k
n m
= p
j k − j
j=0
k (1 − p) n+m−k
n + m
6. Prove that if X is a random variable that is independent of itself, then X is a.s.
constant, i.e., there is a constant c ∈ R such that P(X = c) = 1.
Solution. Let FX be the c.d.f. of X. Since X is independent of itself,
P(X ≤ x) = P(X ≤ x, X ≤ x) = P(X ≤ x)P(X ≤ x),
which implies that FX(x) = 0 or 1 for each x ∈ R. Recall that FX(x) is nondecreasing,
limx→−∞ FX(x) = 0 and limx→∞ FX(x) = 1. Thus, there exists c ∈ R such that
lim
x→c− FX(x) = 0 and lim
x→c + FX(x) = 1
Since FX is right continuous, P(X = c) = FX(c) − FX(c−) = 1.
6
k
.