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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,...Asked by ram121
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 .
1. cov(Y1,Y2)
2. var(Z1)
3. cov(Z1,Z2)
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 .
1. cov(Y1,Y2)
2. var(Z1)
3. cov(Z1,Z2)
Answers
Answered by
Anonymous
1. 1/2
2. 15/64
3. Don't know how to do
2. 15/64
3. Don't know how to do
Answered by
Anonymous
can you please throw some light on, how is it 15/64 ? I can add for the next step
Answered by
Anon
no. 2) 3/16
Answered by
noname
for 1: I am getting
Cov(Y1,Y2) = ... = E(X3^2) = var(X3)+(E(X3))^2 = 1/2 + (1)^2 = 3/2 ?
E[X3]=n.p = (2.1/2)=1, var(x3) = n.p(1-p)=(X3)=n.2.1/2.1/2 = 1/2
Cov(Y1,Y2) = ... = E(X3^2) = var(X3)+(E(X3))^2 = 1/2 + (1)^2 = 3/2 ?
E[X3]=n.p = (2.1/2)=1, var(x3) = n.p(1-p)=(X3)=n.2.1/2.1/2 = 1/2
Answered by
noname
Correction: Cov(Y1,Y2) = ... = C(X3,X3) = var(X3) = 1/2
Answered by
Amana
Why would the variance of Z1 be 15/64?
Answered by
Anonymous
How much did you get?
Answered by
newmew
I got 1/4. I found this from google search:
"Your Z=X−Y will not be a "shifted binomial" unless p=1/2, or the trivial cases where at least one of n and m is zero. For the case p=1/2, m−Y has the same distribution as Y so X+Y and X−Y+m have the same distribution, which is indeed binomial.
In general consider the means and variances of the distributions:
X has mean np and variance np(1−p)
Y has mean mp and variance mp(1−p)
X+Y has mean (n+m)p and variance (n+m)p(1−p)
Z=X−Y has mean (n−m)p and variance (n+m)p(1−p)"
which makes probability of success of y1 =1/2. So, so, var(z1) = p(1-p)=1/4
"Your Z=X−Y will not be a "shifted binomial" unless p=1/2, or the trivial cases where at least one of n and m is zero. For the case p=1/2, m−Y has the same distribution as Y so X+Y and X−Y+m have the same distribution, which is indeed binomial.
In general consider the means and variances of the distributions:
X has mean np and variance np(1−p)
Y has mean mp and variance mp(1−p)
X+Y has mean (n+m)p and variance (n+m)p(1−p)
Z=X−Y has mean (n−m)p and variance (n+m)p(1−p)"
which makes probability of success of y1 =1/2. So, so, var(z1) = p(1-p)=1/4
Answered by
Anonymous
So, so ...and hat happend whit term (n+m) on (n+m)p(1-p)?
Answered by
Anonymous
Z is an Indicator random variable, It is not direct a substract of tow binomial distribuction
Answered by
just saying
for 3:
to me it looks that knowing Z1 doesn't inform you for Z2 and vice versa, so they look independent variables to me -> Cov(z1,z2)= 0. (couldn't figure anything else and made up this nice explanation to comfort my self
to me it looks that knowing Z1 doesn't inform you for Z2 and vice versa, so they look independent variables to me -> Cov(z1,z2)= 0. (couldn't figure anything else and made up this nice explanation to comfort my self
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