Question
Let X, Y, Z be independent discrete random variables with EX]=2 EX² = 20 EY=0 E [Z]=0, EYE [Z3] 16, Var (Y) = Var (Z) = 16. and Var (X) Let AX (Y+Z) and B = XY.
1. Find E [B]
E B =
2. Find Var (B).
Var (B)=
3. Find Var (BX = 2).
Var (BX2)=
4. Find E AB
E [AB] =
(
1. Find E [B]
E B =
2. Find Var (B).
Var (B)=
3. Find Var (BX = 2).
Var (BX2)=
4. Find E AB
E [AB] =
(
Answers
GPT-4o mini
To calculate the requested values, we will utilize the properties of expectation, variance, and the independence of the random variables \(X\), \(Y\), and \(Z\).
### Given Information:
- \(E[X] = 2\)
- \(E[X^2] = 20\)
- \(E[Y] = 0\)
- \(E[Z] = 0\)
- \(E[Y E[Z^3]] = 16\)
- \(\text{Var}(Y) = 16\)
- \(\text{Var}(Z) = 16\)
### 1. Find \(E[B]\) where \(B = XY\)
Using the independence of \(X\) and \(Y\):
\[
E[B] = E[XY] = E[X]E[Y]
\]
Since \(E[Y] = 0\):
\[
E[B] = E[X] \cdot 0 = 0
\]
Thus,
\[
E[B] = 0
\]
### 2. Find \(\text{Var}(B)\)
Using the formula for variance of the product of independent random variables:
\[
\text{Var}(B) = \text{Var}(XY) = E[X^2]E[Y^2] + (E[X])^2 \text{Var}(Y) + (E[Y])^2 \text{Var}(X)
\]
1. Calculate \(E[Y^2]\) using \(\text{Var}(Y)\):
\[
\text{Var}(Y) = E[Y^2] - (E[Y])^2 = E[Y^2] - 0^2 = E[Y^2]
\]
Thus, \(E[Y^2] = \text{Var}(Y) = 16\).
2. Since \(X\) is not directly provided, we can denote \(\text{Var}(X) = \sigma_X^2\).
Substituting into the variance formula:
\[
\text{Var}(B) = E[X^2] \cdot E[Y^2] + (E[X])^2 \text{Var}(Y)
\]
\[
\text{Var}(B) = 20 \cdot 16 + (2^2) \cdot 16 = 320 + 4 \cdot 16
\]
\[
\text{Var}(B) = 320 + 64 = 384
\]
Thus,
\[
\text{Var}(B) = 384
\]
### 3. Find \(\text{Var}(B | X = 2)\)
Since \(X\) is fixed at \(2\):
\[
B | (X = 2) = 2Y
\]
Then:
\[
\text{Var}(B | X = 2) = \text{Var}(2Y) = 4 \cdot \text{Var}(Y) = 4 \cdot 16 = 64
\]
Thus,
\[
\text{Var}(B | X = 2) = 64
\]
### 4. Find \(E[AB]\) where \(A = X(Y + Z)\)
Calculating \(E[AB]\):
\[
E[AB] = E[X(Y + Z)XY] = E[X^2Y] + E[X^2Z]
\]
Since:
\[
E[X^2Y] = E[X^2]E[Y] = E[X^2] \cdot 0 = 0,
\]
and
\[
E[X^2Z] = E[X^2]E[Z] = E[X^2] \cdot 0 = 0,
\]
Thus:
\[
E[AB] = 0 + 0 = 0
\]
Thus,
\[
E[AB] = 0
\]
### Summary of Results:
1. \(E[B] = 0\)
2. \(\text{Var}(B) = 384\)
3. \(\text{Var}(B | X = 2) = 64\)
4. \(E[AB] = 0\)
### Given Information:
- \(E[X] = 2\)
- \(E[X^2] = 20\)
- \(E[Y] = 0\)
- \(E[Z] = 0\)
- \(E[Y E[Z^3]] = 16\)
- \(\text{Var}(Y) = 16\)
- \(\text{Var}(Z) = 16\)
### 1. Find \(E[B]\) where \(B = XY\)
Using the independence of \(X\) and \(Y\):
\[
E[B] = E[XY] = E[X]E[Y]
\]
Since \(E[Y] = 0\):
\[
E[B] = E[X] \cdot 0 = 0
\]
Thus,
\[
E[B] = 0
\]
### 2. Find \(\text{Var}(B)\)
Using the formula for variance of the product of independent random variables:
\[
\text{Var}(B) = \text{Var}(XY) = E[X^2]E[Y^2] + (E[X])^2 \text{Var}(Y) + (E[Y])^2 \text{Var}(X)
\]
1. Calculate \(E[Y^2]\) using \(\text{Var}(Y)\):
\[
\text{Var}(Y) = E[Y^2] - (E[Y])^2 = E[Y^2] - 0^2 = E[Y^2]
\]
Thus, \(E[Y^2] = \text{Var}(Y) = 16\).
2. Since \(X\) is not directly provided, we can denote \(\text{Var}(X) = \sigma_X^2\).
Substituting into the variance formula:
\[
\text{Var}(B) = E[X^2] \cdot E[Y^2] + (E[X])^2 \text{Var}(Y)
\]
\[
\text{Var}(B) = 20 \cdot 16 + (2^2) \cdot 16 = 320 + 4 \cdot 16
\]
\[
\text{Var}(B) = 320 + 64 = 384
\]
Thus,
\[
\text{Var}(B) = 384
\]
### 3. Find \(\text{Var}(B | X = 2)\)
Since \(X\) is fixed at \(2\):
\[
B | (X = 2) = 2Y
\]
Then:
\[
\text{Var}(B | X = 2) = \text{Var}(2Y) = 4 \cdot \text{Var}(Y) = 4 \cdot 16 = 64
\]
Thus,
\[
\text{Var}(B | X = 2) = 64
\]
### 4. Find \(E[AB]\) where \(A = X(Y + Z)\)
Calculating \(E[AB]\):
\[
E[AB] = E[X(Y + Z)XY] = E[X^2Y] + E[X^2Z]
\]
Since:
\[
E[X^2Y] = E[X^2]E[Y] = E[X^2] \cdot 0 = 0,
\]
and
\[
E[X^2Z] = E[X^2]E[Z] = E[X^2] \cdot 0 = 0,
\]
Thus:
\[
E[AB] = 0 + 0 = 0
\]
Thus,
\[
E[AB] = 0
\]
### Summary of Results:
1. \(E[B] = 0\)
2. \(\text{Var}(B) = 384\)
3. \(\text{Var}(B | X = 2) = 64\)
4. \(E[AB] = 0\)