Suppose n = 3 X 1 = 1 {stock price rises Monday} X 2 = 1 {stock price rises on T
ID: 3054329 • Letter: S
Question
Suppose n = 3
X1 = 1 {stock price rises Monday}
X2 = 1 {stock price rises on Tuesday}
X3 = 1 {stock price rises on exactly one day over Monday and Tuesday}
Here, 1{A} denotes an indicator function which takes value 1 if event A happens and 0 otherwise. Assume the stock price movements on Monday and Tuesday are independent, and the probability of stock price rising on any given day is 0.5 ( same for monday and tuesday)
a) show that X1, X2, X3 are pairwise independent
b) show that X1, X2, X3 are not jointly independent
Explanation / Answer
a) Let, A be event that stock price rises then P(X1=1)=0.5=P(X2=1) (given), P(X3=1)=P(Stock price rises on Monday and stock price does not rise on Tuesday)+
P(Stock price rises on Tuesday and stock price does not rise on Monday)=P(X1=1, X2=0)+P(X1=1, X2=0)
=P(X1=1)P(X2=0)+P(X1=1)P(X2=0) [since X1 and X2 are independent]
=(1/2)(1/2)+(1/2)(1/2)=1/2
P(X3=0)=P(stock price rises on both day)+P(stock price rises on none of these days)=P(X1=1, X2=1)+P(X1=0, X2=0)=P(X1=1)P(X2=1)+P(X1=0)P(X2=0)=(1/2)(1/2)+(1/2)(1/2)=1/2
Now, P(X1=1, X3=1)=P(X1=1, X2=0)=P(X1=1)P(X2=0) [since assume the stock price movements on Monday and Tuesday are independent hence X1 and X2 are independent ]
=(1/2)(1/2)=1/4=P(X1=1)P(X3=1)
P(X1=1, X3=0)=P(X1=1, X2=1)=P(X1=1)P(X2=1) [since X1 and X2 are independent]
=(1/2)(1/2)=P(X1=1)P(X3=0)
Similarly P(X1=0, X3=1)=P(X1=0, X2=1)=P(X1=0)P(X2=1)=(1/2)(1/2)=P(X1=0)P(X3=1)
P(X1=0, X3=0)=P(X1=0, X2=0)=P(X1=0)P(X2=0)=(1/2)(1/2)=P(X1=0)P(X3=0)
Hence X1 and X3 are independent since P(X1=x1, X3=x3)=P(X1=x1)P(X3=x3) for all x1, x3=0,1
Now, P(X2=1, X3=1)=P(X1=0, X2=1)=P(X1=0)P(X2=1) [since assume the stock price movements on Monday and Tuesday are independent hence X1 and X2 are independent ]
=(1/2)(1/2)=1/4=P(X2=1)P(X3=1)
P(X2=1, X3=0)=P(X1=1, X2=1)=P(X1=1)P(X2=1) [since X1 and X2 are independent]
=(1/2)(1/2)=P(X2=1)P(X3=0)
Similarly P(X2=0, X3=1)=P(X1=1, X2=0)=P(X1=1)P(X2=0)=(1/2)(1/2)=P(X2=0)P(X3=1)
P(X2=0, X3=0)=P(X1=0, X2=0)=P(X1=0)P(X2=0)=(1/2)(1/2)=P(X2=0)P(X3=0)
Hence X2 and X3 are independent since P(X2=x2, X3=x3)=P(X2=x2)P(X3=x3) for all x2, x3=0,1
Again X1 and X2 are independent.
Hence X1, X2, X3 are pairwise independent.
(b)
P(X1=0, X2=0, X3=0)=P(X1=0, X2=0)=1/4
But P(X1=0)P(X2=0)P(X3=0)=1/8
Hence P(X1=0, X2=0, X3=0) is not equal to P(X1=0)P(X2=0)P(X3=0).
Hence X1, X2, X3 are not jointly independent.