Let X,Y,Z be independent discrete random variables with
E[X]=2, E[Y]=0, E[Z]=0,
E[X^2]=20
E[Y^2]= E[Z^2]=16,
and
Var(X)=Var(Y)=Var(Z)=16.
Let A=X(Y+Z) and B=XY.
1. Find E[B].
E[B]= ?
2. Find Var(B).
Var(B)= ?
3. Find E[AB].
E[AB]= ?
4. Find Var (B|X = 2)
To find the expected value E[B], we can use the linearity of expectation. We know that A, X, Y, and Z are all independent random variables.
1. E[B] = E[XY]
Since X and Y are independent, the expected value of the product is the product of the expected values.
E[B] = E[X] * E[Y]
Given that E[X]=2 and E[Y]=0:
E[B] = 2 * 0
E[B] = 0
So, E[B] = 0.
To find the variance Var(B), we can use the properties of variance.
2. Var(B) = Var(XY)
Since X and Y are independent, we can use the property Var(aX) = a^2 * Var(X).
Var(B) = Var(X) * E[Y]^2 + E[X]^2 * Var(Y) + Var(X) * Var(Y)
Given that Var(X) = Var(Y) = 16, and E[Y] = 0:
Var(B) = 16 * 0^2 + 2^2 * 16 + 16 * 16
Var(B) = 4 * 16 + 16 * 16
Var(B) = 64 + 256
Var(B) = 320
So, Var(B) = 320.
To find E[AB], we can again use the linearity of expectation.
3. E[AB] = E[X(Y+Z)]
Since X, Y, and Z are independent, we can use the property E[aX] = a * E[X].
E[AB] = E[X] * E[Y+Z]
Given that E[X] = 2 and E[Y] = E[Z] = 0:
E[AB] = 2 * (E[Y] + E[Z])
E[AB] = 2 * (0 + 0)
E[AB] = 2 * 0
E[AB] = 0
So, E[AB] = 0.
To find Var(B|X = 2), we need to find the conditional variance of B given that X = 2.
4. Var(B|X = 2)
Since X and Y are independent, we know that B = XY.
Var(B|X = 2) = Var(XY|X = 2)
Using the property Var(aX) = a^2 * Var(X), we can rewrite this as
Var(B|X = 2) = 2^2 * Var(Y|X = 2)
Since Y and Z are independent, Var(Y+Z|X = 2) = Var(Y|X = 2) + Var(Z|X = 2)
Given that Var(Y) = Var(Z) = 16, we have
Var(B|X = 2) = 4 * (Var(Y) + Var(Z))
Var(B|X = 2) = 4 * (16 + 16)
Var(B|X = 2) = 4 * 32
Var(B|X = 2) = 128
So, Var(B|X = 2) = 128.