# BASIC PROBABILITY : HOMEWORK 2 Exercise 1: where does the

```BASIC PROBABILITY : HOMEWORK 2
Exercise 1: where does the Poisson distribution come from? (corrected)
Fix λ &gt; 0. We denote by Xn has the Binomial distribution with parameters
n and λ/n. Then prove that for any k ∈ N we have
P [Xn = k] −−−→ P [Yλ = k],
n→∞
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 &micro;.
Show that
(1) X + Y is Poisson with parameter λ + &micro;,
(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 integer
values, we have
∞
X
E[X] =
P [X &gt; n].
n=0
An urn contains b blue and r red balls. Balls are removed at random
until the first blue ball is drawn. Show that the expected number drawn is
(b + r + 1)/(b + 1).
1
2
Hint :
Pr
n=0
n+b
b
=
r+b+1
b+1
.
Exercise 6 (homework)
Let X have a distribution function


if x &lt; 0
0
1
F (x) = 2 x if 0 ≤ x ≤ 2

1
if x &gt; 2
and let Y = X 2 . Find
(1) P ( 12 ≤ X ≤ 32 ),
(2) P (1 ≤ X &lt; 2),
(3) P (Y ≤ X),
(4) P (X ≤ 2Y ),
(5) P (X + Y ≤ 34 ),
√
(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 &gt; 0, we have
1
P [X ≥ a] ≤ E[X].
a
Deduce that
1
P [|X − E[X]| ≥ a] ≤ 2 var(X),
a
p
and explain why var(X) is often referred to as the standard deviation.
3
Exercise 1 (solution)
We have
n λ k
λ
P [Xn = k] =
(1 − )n−k
k n
n
n(n − 1) . . . (n − k + 1) λk
λ
(1 − )n−k
=
k
n
k!
n
k
λ
→ e−λ ,
k!
since for any k &gt; 0, we have that (1 − nλ )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 definition
of 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
X
X
h(y),
fX,Y (x, y) = f (x)
fX (x) =
fY (y) =
y
y
X
X
fX,Y (x, y) = h(y)
y
g(x).
y
Now
1=
X
fX (x) =
X
x
x
g(x)
X
h(y),
y
so that
fX (x)fY (y) = g(x)h(y)
X
g(x)
X
h(y) = g(x)h(y) = fX,Y (x, y).
y
x
Exercise 3 (solution)
Let us notice that for X and Y non-negative independent random variables
then
n
X
P [X + Y = n] =
P [X = n − k]P [Y = k],
k=0
so in our specific case of Poisson random variables
n
X
e−λ λn−k e−&micro; &micro;k
P [X + Y = n] =
(n − k)! k!
k=0
4
n e−λ−&micro; X n n−k k e−λ−&micro; (&micro; + λ)n
λ &micro; =
.
=
n! k=0 k
n!
For the second part of the question
P [X = k, X + Y = n]
P [X = k | X + Y = n] =
P [X + Y = n]
k n−k
P [X = k]P [Y = n − k]
n λ &micro;
=
=
.
P [X + Y = n]
k (&micro; + λ)n
Hence the conditional distribution is binomial with parameters n and
λ/(λ + mu).
```