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BASIC PROBABILITY : HOMEWORK 2

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

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

P[Xn = k] n

P[Y = k],

where Y has a Poisson distribution of parameter .

Exercise 2 (corrected)Prove Theorem 2.2.

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

Show that

(1) X+ Y is Poisson with parameter + ,(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, . . . , X n 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

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).

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Hint :r

n=0

n+bb

=r+b+1b+1

.

Exercise 6 (homework)Let X have a distribution function

F(x) =

0 if x < 012

x if 0 x 21 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 34),(6) the distribution function of Z =

X

Exercise 7: Chebyshevs 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] 1a

E[X].

Deduce thatP[|XE[X]| a] 1

a2var(X),

and explain why

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

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Exercise 1 (solution)We have

P[Xn = k] =

nk

n

k(1

n)nk

=n(n 1) . . . (n k + 1)

nkk

k!(1

n)nk

k

k!e,

since for any k > 0, we have that (1 n

)nk e.

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).

Now1 =

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

enk

(n k)!ek

k!

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=e

n!

nk=0

n

k

nkk =

e( + )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

knk

( + )n.

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