Let X1 , X2 , X3 be i.i.d. Binomial random variables with parameters n=2 and p=1/2 . Define two new random variables
Y1 =X1−X3,
Y2 =X2−X3.
We further introduce indicator random variables Zi∈{0,1} with Zi=1 if and only if Yi=0 for i=1,2 .
Calculate the covariance of Y1 and Y2 .
(Give an exact answer or a decimal accurate to at least 3 decimal places.)
Find:
1) Cov(Y1,Y2)=
Calculate the variance of Z1 . (Give an exact answer or a decimal accurate to at least 3 decimal places.)
2) Var(Z1)=
Y1 and Y2 have X3 in common, they are not independent.
No it's not. You consider all the outcomes of Y1 to be equally likely, but they are not.
For example;
P(X1 = 0) and P(X3 = 0) = 1/4 * 1/4 = 1/16.
While: P(X1 = 1) and P(X3 = 1) = 1/2 * 1/2 = 1/4.
Cov(y1,y2) = 1/2
Car(z1) =1/4
Wow, that's a good point!
I fixed my calculations and got 0.234375 = 15/64
x1 x3 P(x1*X3) Z p var
0 0 0.0625 0.00 0.375 0.234375
0 1 0.1250 -1.00
0 2 0.0625 -2.00
1 0 0.1250 1.00
1 1 0.2500 0.00
1 2 0.1250 -1.00
2 0 0.0625 2.00
2 1 0.1250 1.00
2 2 0.0625 0.00
How can Z1 take the -1 value? I think it is rather a new binomial variable with a new parameter n and same p. Hint: So new that it changes its type of r.v.
can someone please give an answer to VAR(Z1). thank you in advance
0
2/9
1) Cov(Y1,Y2) = -1/4
2) Var(Z1)= ??????
Z1 =1 only if Yi ==0;
Yi == 0 only when
x1=0 x3=0
x1=1 x3=1
x1=2 x3=3
3 possible scenarios leads to Zi=1
list of possible scenarios are
x1 x3 Y1 Z1
0 0 0 1
0 1 -1 0
0 2 -2 0
1 0 1 0
1 1 0 1
1 2 -1 0
2 0 2 0
2 1 1 0
2 2 0 1
Z1 and Z2 are Binomial R.V. with p=3/9
var(Z1) = p*(1-p) = 3/9*6/9=2/9
right?
Cov(Y1,Y2)=0 , they are independent
Var (Z1)=2/3 (z is a uniform discreet random variable with 3 possible values 0,1,-1)
Cov(Z1,Z2)=0 , they are independent
could someone confirm the approach and answers