BASIC PROBABILITY : HOMEWORK 2

Exercise 1: where does the Poisson distribution come from? (cor-rected)

Fix λ > 0. We denote by Xn has the Binomial distribution with parametersn and λ/n. Then prove that for any k ∈ N we have

P [Xn = k] −−−→n→∞

P [Yλ = k],

where Yλ has a Poisson distribution of parameter λ.

Exercise 2 (corrected)Prove Theorem 2.2.

Exercise 3 (corrected)Let X and Y be independent Poisson variables with parameters λ and µ.

Show that

(1) X + Y is Poisson with parameter λ+ µ,(2) the conditional distribution of X, given that X + Y = n, is binomial,

and find the parameters.

Exercise 4 (homework)Suppose we have n independent random variables X1, . . . , Xn with common

distribution function F . Find the distribution function of maxi∈[1,n]Xi.

Exercise 5 (homework) Show that if X takes only non-negative integervalues, we have

E[X] =∞∑n=0

P [X > n].

An urn contains b blue and r red balls. Balls are removed at randomuntil the first blue ball is drawn. Show that the expected number drawn is(b+ r + 1)/(b+ 1).

1

2

Hint :∑r

n=0

(n+bb

)=(r+b+1b+1

).

Exercise 6 (homework)Let X have a distribution function

F (x) =

0 if x < 012x if 0 ≤ x ≤ 2

1 if x > 2

and let Y = X2. Find

(1) P (12≤ X ≤ 3

2),

(2) P (1 ≤ X < 2),(3) P (Y ≤ X),(4) P (X ≤ 2Y ),(5) P (X + Y ≤ 3

4),

(6) the distribution function of Z =√X

Exercise 7: Chebyshev’s inequality (homework)Let X be a random variable taking only non-negative values. Then show

that for any a > 0, we have

P [X ≥ a] ≤ 1

aE[X].

Deduce that

P [|X − E[X]| ≥ a] ≤ 1

a2var(X),

and explain why√

var(X) is often referred to as the standard deviation.

3

Exercise 1 (solution)We have

P [Xn = k] =

(n

k

)(λn

)k(1− λ

n)n−k

=n(n− 1) . . . (n− k + 1)

nkλk

k!(1− λ

n)n−k

→ λk

k!e−λ,

since for any k > 0, we have that (1− λn)n−k → e−λ.

Exercise 2 (solution)The condition fX,Y (x, y) = fX(x)fy(y) for all x, y can be rewritten P (X =

x and Y = y) = P (X = x)P (Y = y) for all x, y which is the very definitionof X and Y being independent.

Now for the second part of the theorem. Suppose that fX,Y (x, y) =g(x)h(y) for all x, y, we have

fX(x) =∑y

fX,Y (x, y) = f(x)∑y

h(y),

fY (y) =∑y

fX,Y (x, y) = h(y)∑y

g(x).

Now

1 =∑x

fX(x) =∑x

g(x)∑y

h(y),

so that

fX(x)fY (y) = g(x)h(y)∑x

g(x)∑y

h(y) = g(x)h(y) = fX,Y (x, y).

Exercise 3 (solution)Let us notice that for X and Y non-negative independent random variables

then

P [X + Y = n] =n∑k=0

P [X = n− k]P [Y = k],

so in our specific case of Poisson random variables

P [X + Y = n] =n∑k=0

e−λλn−k

(n− k)!

e−µµk

k!

4

=e−λ−µ

n!

n∑k=0

(n

k

)λn−kµk =

e−λ−µ(µ+ λ)n

n!.

For the second part of the question

P [X = k | X + Y = n] =P [X = k,X + Y = n]

P [X + Y = n]

=P [X = k]P [Y = n− k]

P [X + Y = n]=

(n

k

)λkµn−k

(µ+ λ)n.

Hence the conditional distribution is binomial with parameters n andλ/(λ+ mu).