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Math 5010 § 1. Treibergs
Solutions to Eighth Homework March 13, 2009
151[14] Suppose X is a random variable that is uniform on 1 ≤ x ≤ m. What is P(X = k | a ≤ X ≤ b)? In particular, find P(X > n + k | X > n). The random variable X takes values in the set D = { 1 , 2 , 3 ,... , m}. The pmf of the uniform random variable is fX (x) = 1/m for x ∈ D and fX (x) = 0 if x /∈ D. The answer is simpler if we make the assumption that a, k, b are integers and 1 ≤ a ≤ k ≤ b ≤ m. Under this condition, all the numbers between a and b, inclusive, are in D so that there are b − a + 1 such numbers and P(a ≤ X ≤ b) = (b − a + 1)/m. Similarly a ≤ k ≤ b implies P({X = k} ∩ {a ≤ X ≤ b}) = P(X = k) = 1/m. Thus
P(X = k | a ≤ X ≤ b) =
P({X = k} ∩ {a ≤ X ≤ b}) P(a ≤ X ≤ b)
b − a + 1
Assuming the conditions 0 ≤ n < m and n ≤ n+k ≤ m, the set of numbers in D that satisfy X > n is {n + 1 ≤ X ≤ m} so there are m − n such numbers and P(X > n) = (m − n)/m. Also the set of numbers in D that satisfy {X > n + k} ∩ {X > n} = {X > n + k} is {n + k + 1 ≤ X ≤ m} so there are m − n − k such numbers and P({X > n + k} ∩ {X > n}) = (m − n − k)/m. Thus
P(X > n + k | X > n) =
P({X > n + k} ∩ {X > n}) P(X > n)
m − n − k m − n
We also give a solution in case we make weaker hypotheses on the numbers a, b, k which are not specified in this problem. Since we are conditioning on the event {a ≤ X ≤ b}, which must have positive probability, at least one of the numbers from D have to be included between a and b. In other words, we can assume the weaker conditions a ≤ b, a ≤ m and b ≥ 1. Thus the numbers in D that are between a and b are exactly
max(a, 1), max(a, 1) + 1,... , min(b, m).
For example if m = 6 as in X is the number on one roll of a die, and a = 2, b = 9, then the possible values of X between a and b are max(2, 1) = 2, 3 , 4 , 5 , 6 = min(9, 6). The probability is thus the number of numbers times the probability of any one of them, or
P(a ≤ X ≤ b) =
min(b, m) − max(a, 1) + 1 m
For example on the standard die with a = 2, b = 9 this is (6 − 2 + 1)/6. The intersection event {X = k} ∩ {a ≤ X ≤ b} = {X = k} if max(a, 1) ≤ k ≤ min(b, m) and the empty set if not. Thus the conditional probability
P(X = k | a ≤ X ≤ b) = P({X = k} ∩ {a ≤ X ≤ b}) P(a ≤ X ≤ b)
P(X = k) P(a ≤ X ≤ b)
, if max(a, 1) ≤ k ≤ min(b, m);
0 , otherwise.
min(b, m) − max(a, 1) + 1
, if max(a, 1) ≤ k ≤ min(b, m);
0 , otherwise.
For the second problem, we assume n < m so that {X > n} ∩ D 6 = ∅ so it has positive probability. Using (1),
P(X > n + k | X > n) =
P({X > n + k} ∩ {X > n}) P(X > n)
P(X > n + k) P(X > n)
P(n + k < X ≤ m) P(n < X ≤ m)
, if m > n + k;
0 , otherwise.
m − max(n + k + 1, 1) + 1 m − max(n + 1, 1) + 1
, if m > n + k;
0 , otherwise.
151[42] Prove Chebychev’s Inequality, that for a random variable X with mean μ and variance σ^2 ,
P(|X − μ| ≤ hσ) ≥ 1 −
h^2
for any h > 0. (2)
When an unbiased coin is tossed n times, let the number of heads be m. Show that
P
m n
when n ≥ 100. Given that n = 100, show that the actual probability is
P
m n
2 π
You may assume Stirling’s Formula n! '
2 π nn+^ (^12) e−n. Applying Theorem 4.6.1 to h(x) = (x−E(X))^2 , we get the version of Chebychev’s inequality given in class, P(|X − E(X)| ≥ a) ≤
Var(X) a^2
, for any a > 0. (3)
The desired inequality reverses signs, so we expect to apply it to the complementary event. Furthermore, we replace a = hσ, use the fact that the event {|X−μ| > hσ} ⊂ {|X−μ| ≥ hσ} and (3),
1 − P(|X − μ| ≤ hσ) = P({|X − μ| ≤ hσ}c) = P(|X − μ| > hσ) ≤ P(|X − μ| ≥ hσ)
≤
σ^2 (hσ)^2
h^2
Rearranging gives (2). The second question asks us to apply the inequality to the random variable m, the num- ber of heads in n tosses, whch has the distribution of a binomial random variable m ∼ binomial(n, p = 12 ). From Table 4.1, μ = E(m) = np = 0. 5 n and σ^2 = Var(m) = npq =
- 25 n so σ = 0. 5
n. We may rewrite the event using equivalent inequalities on m, {
- 4 ≤
m n
= { 0. 4 n ≤ m ≤ 0. 6 n} = {− 0. 1 n ≤ m − 0. 5 n ≤ 0. 1 n} = {|m − μ| ≤ 0. 1 n}.
Thus 0. 1 n = hσ = 0. 5 h
n so h = 0. 2
n. Thus applying (2),
P
m n
= P(|m − μ| ≤ 0. 1 n = hσ) ≥ 1 −
h^2
- 04 n
Using the alternative formula for expectation of a nonnegative integer valued random vari- able Theorem 4.3.11,
E(X) =
∑^6
k=
P(X ≥ k) =
∑^6
k=
(7 − k)^5 65
65 + 5^5 + 4^5 + 3 + 2^5 + 1^5
[C.] Suppose the random variable X is distributed according to the Poisson distribution with mean λ. Find fX (x | X is odd ) and E(X | X is odd ). The values of the Poisson variable are taken in D = { 0 , 1 , 2 , 3 ,.. .} and for λ > 0, the pmf for x ∈ D is fX (x) = P(X = x) =
e−λ^ λx x!
We need to consider the event
{X is odd} = {X = 1} ∪ {X = 3} ∪ {X = 5} ∪ · · · =
⋃^ ∞
k=
{X = 2k + 1}.
Its probability of being odd turns out to be less than one half
P(X is odd) =
∑^ ∞
k=
fX (2k+1) =
∑^ ∞
k=
e−λ^ λ^2 k+ (2k + 1)!
= e−λ^ sinh λ = e−λ
eλ^ − e−λ 2
1 − e−^2 λ 2
To compute the conditional mass function, we use the usual formula for conditional proba- bility. If x ∈ D,
fX (x | X is odd) = P(X = x | X is odd) = P({X = x} ∩ {X is odd}) P(X is odd)
P(X = x) P(X is odd)
, if x is odd;
0 , if x is even.
λx x! sinh λ
, if x is odd;
0 , if x is even.
The conditional expectation is given by the formula
E(X | X is odd) =
x∈D
xfX (x | X is odd)
∑^ ∞
k=
(2k + 1) λ^2 k+ (2k + 1)! sinh λ
λ sinh λ
∑^ ∞
k=
λ^2 k (2k)!
λ cosh λ sinh λ
= λ coth λ.
