# Probability

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)

1. 👍 2
2. 👎 0
3. 👁 786
1. 1. 1/2
2. 15/64
3. Don't know how to do

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2. 👎 2
2. can you please throw some light on, how is it 15/64 ? I can add for the next step

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3. no. 2) 3/16

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2. 👎 4
4. 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

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5. Correction: Cov(Y1,Y2) = ... = C(X3,X3) = var(X3) = 1/2

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6. Why would the variance of Z1 be 15/64?

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7. How much did you get?

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8. 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

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9. So, so ...and hat happend whit term (n+m) on (n+m)p(1-p)?

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10. Z is an Indicator random variable, It is not direct a substract of tow binomial distribuction

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11. 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

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