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Math  /  Data & Statistics

Question(λ),E(X+7),D(2X−5)(\lambda), E(X+7), D(2 X-5) 3) 21 . Supozojme Cov⁡(X1−X2,X1+X2)=0\operatorname{Cov}\left(X_{1}-X_{2}, X_{1}+X_{2}\right)=0

Studdy Solution

STEP 1

1. X1 X_1 and X2 X_2 are random variables with the same distribution.
2. We need to find the covariance of X1−X2 X_1 - X_2 and X1+X2 X_1 + X_2 .

STEP 2

1. Recall the definition of covariance.
2. Apply the properties of covariance to simplify the expression.
3. Use the assumption that X1 X_1 and X2 X_2 have the same distribution.

STEP 3

Recall the definition of covariance:
Cov⁡(A,B)=E[(A−E[A])(B−E[B])]\operatorname{Cov}(A, B) = E[(A - E[A])(B - E[B])]

STEP 4

Apply the properties of covariance to Cov⁡(X1−X2,X1+X2) \operatorname{Cov}(X_1 - X_2, X_1 + X_2) :
Cov⁡(X1−X2,X1+X2)=Cov⁡(X1,X1)+Cov⁡(X1,X2)−Cov⁡(X2,X1)−Cov⁡(X2,X2)\operatorname{Cov}(X_1 - X_2, X_1 + X_2) = \operatorname{Cov}(X_1, X_1) + \operatorname{Cov}(X_1, X_2) - \operatorname{Cov}(X_2, X_1) - \operatorname{Cov}(X_2, X_2)

STEP 5

Simplify using the properties of covariance:
Cov⁡(X1,X1)=Var⁡(X1)\operatorname{Cov}(X_1, X_1) = \operatorname{Var}(X_1) Cov⁡(X2,X2)=Var⁡(X2)\operatorname{Cov}(X_2, X_2) = \operatorname{Var}(X_2) Cov⁡(X1,X2)=Cov⁡(X2,X1)\operatorname{Cov}(X_1, X_2) = \operatorname{Cov}(X_2, X_1)
Thus,
Cov⁡(X1−X2,X1+X2)=Var⁡(X1)−Var⁡(X2)\operatorname{Cov}(X_1 - X_2, X_1 + X_2) = \operatorname{Var}(X_1) - \operatorname{Var}(X_2)

STEP 6

Since X1 X_1 and X2 X_2 have the same distribution, their variances are equal:
Var⁡(X1)=Var⁡(X2)\operatorname{Var}(X_1) = \operatorname{Var}(X_2)
Therefore,
Cov⁡(X1−X2,X1+X2)=0\operatorname{Cov}(X_1 - X_2, X_1 + X_2) = 0
The covariance is:
0 \boxed{0}

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