Problem 4 Let X₁,...,Xn be independent random variables with values in (1,...,x) and probability mass function Px ifxe (1,...,x) else, respectively. Of course, p, € [0, 1] and p₁ +...+Px = 1. Now, for k = 1,...,x let för is p(x)= Nk:= [1(x,-k), i=1 i.e. N counts how many of the X, assume the value k. (a) Show that for any (n₁,...,nx) € (0, 1,...,n)* with n1 + ... + nx = n n! P{N₁= n₁,..., Nx = nx} = nin! Hint: The number of distinguishable rearrangements of the vector (1,...,1,2,...,2,. n₁ times n₂ times n! ni!…x! n, times ...p. because there are n! possible rearrangements in total, but n₁!--nx! of those rearrangements consist merely of permutations within the blocks of equal entries and thus lead exactly to the same sequence of numbers (i.e. are not distinguishable). (b) Suppose that x =2. Compute EN₁ and EN2.
Problem 4 Let X₁,...,Xn be independent random variables with values in (1,...,x) and probability mass function Px ifxe (1,...,x) else, respectively. Of course, p, € [0, 1] and p₁ +...+Px = 1. Now, for k = 1,...,x let för is p(x)= Nk:= [1(x,-k), i=1 i.e. N counts how many of the X, assume the value k. (a) Show that for any (n₁,...,nx) € (0, 1,...,n)* with n1 + ... + nx = n n! P{N₁= n₁,..., Nx = nx} = nin! Hint: The number of distinguishable rearrangements of the vector (1,...,1,2,...,2,. n₁ times n₂ times n! ni!…x! n, times ...p. because there are n! possible rearrangements in total, but n₁!--nx! of those rearrangements consist merely of permutations within the blocks of equal entries and thus lead exactly to the same sequence of numbers (i.e. are not distinguishable). (b) Suppose that x =2. Compute EN₁ and EN2.
A First Course in Probability (10th Edition)
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![Problem 4
(1,...,x) and probability mass function
Let X₁,...,Xn be independent random variables with values in
p(x)=-
respectively. Of course, p, € [0, 1] and P1 +
is
Px
0
...+Px 1. Now, for k= 1,...,x let
n
if x = (1,...,x)
else,
Nk:= [1(x,-k),
i=1
i.e. N counts how many of the X, assume the value k.
(a) Show that for any (n₁,...,nx) (0, 1,...,n)* with n₁ +...+nx = n
P{N₁ = n₁,..., Nx = nx}=
n!
nink!"
Hint: The number of distinguishable rearrangements of the vector
(1,...,1,2,...,2,
n₁ times n₂ times
n!
ni!…nx!
21
P1
K₂...,K
ng times
ng
PK
because there are n! possible rearrangements in total, but n₁!nk! of those rearrangements
consist merely of permutations within the blocks of equal entries and thus lead exactly to the
same sequence of numbers (i.e. are not distinguishable).
(b) Suppose that x =2. Compute EN₁ and EN₂.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4575c695-56bc-4a6a-843f-ec886ca258f2%2F41aab082-cb63-4249-9b6c-58ef46d08858%2F436yjuo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Problem 4
(1,...,x) and probability mass function
Let X₁,...,Xn be independent random variables with values in
p(x)=-
respectively. Of course, p, € [0, 1] and P1 +
is
Px
0
...+Px 1. Now, for k= 1,...,x let
n
if x = (1,...,x)
else,
Nk:= [1(x,-k),
i=1
i.e. N counts how many of the X, assume the value k.
(a) Show that for any (n₁,...,nx) (0, 1,...,n)* with n₁ +...+nx = n
P{N₁ = n₁,..., Nx = nx}=
n!
nink!"
Hint: The number of distinguishable rearrangements of the vector
(1,...,1,2,...,2,
n₁ times n₂ times
n!
ni!…nx!
21
P1
K₂...,K
ng times
ng
PK
because there are n! possible rearrangements in total, but n₁!nk! of those rearrangements
consist merely of permutations within the blocks of equal entries and thus lead exactly to the
same sequence of numbers (i.e. are not distinguishable).
(b) Suppose that x =2. Compute EN₁ and EN₂.
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