VIPSolutionsQUESTION 4. [ 20 Points, LO4.2] Let discrete random variables X and Y have the following joint probability mass function: P(z = 1,y = 1) = 3/8, P(x = —1,y = —1) = 3/8, Plz=1y=-1)=1/8 Plz =-1y=1)=1/8 a. (3 pts, LO4.2) Find E(X). b. (3 pts, LO4.2) Find Var(X). c. (5 pts, LO4.2) Find Cov(X,Y). d. (3 pts, LO4.2) Let Z = X +Y . Find Var(Z). e. (6 pts, LO4.2) Let T = 2X + 2Y — 3. Find Cov(Z, T) where Z = X +Y.
Question:
QUESTION 4. [ 20 Points, LO4.2] Let discrete random variables X and Y have the
following joint probability mass function: P(z = 1,y = 1) = 3/8, P(x = —1,y = —1) = 3/8,
Plz=1y=-1)=1/8 Plz =-1y=1)=1/8
a. (3 pts, LO4.2) Find E(X).
b. (3 pts, LO4.2) Find Var(X).
c. (5 pts, LO4.2) Find Cov(X,Y).
d. (3 pts, LO4.2) Let Z = X +Y . Find Var(Z).
e. (6 pts, LO4.2) Let T = 2X + 2Y — 3. Find Cov(Z, T) where Z = X +Y.
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Asked by: Nadeem
Created at: 2025-12-09 00:41:17
Asked by: Nadeem
Created at: 2025-12-09 00:41:17
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Let's break down each part step by step.
### **Given Joint PMF**
- \( P(x = 1, y = 1) = 3/8 \)
- \( P(x = -1, y = -1) = 3/8 \)
- \( P(x = 1, y = -1 = 1/8 \)
- \( P(x = -1, y = 1) = 1/8 \)
---
## **a. Find \(E(X)\)**
\[
E(X) = \sum_{x,y} x \cdot P(x, y)
\]
\[
= 1 \cdot P(1,1) + 1 \cdot P(1,-1) + (-1) \cdot P(-1,-1) + (-1) \cdot P(-1,1)
\]
\[
= 1 \cdot \frac{3}{8} + 1 \cdot \frac{1}{8} + (-1) \cdot \frac{3}{8} + (-1) \cdot \frac{1}{8}
\]
\[
= \frac{3}{8} + \frac{1}{8} - \frac{3}{8} - \frac{1}{8} =
\]
---
## **b. Find \(Var(X)\)**
\[
Var(X) = E(X^2) - [E(X)]^2
\]
\[
E(X^2) = (1)^2 \cdot [P(1,1) + P(1,-1)] + (-1)^2 \cdot [P(-1,1) + P(-1,-1)]
\]
\[
= 1 \cdot (\frac{3}{8} + \frac{1}{8}) + 1 \cdot (\frac{1}{8} + \frac{3}{8}) = \frac{4}{8} + \frac{4}{8} = 1
\]
\[
Var(X) = 1 - ^2 = 1
\]
---
## **c. Find \(Cov(X, Y)\)**
\[
Cov(X, Y) = E(XY) - E(X)E(Y)
\]
First, compute \(E(Y)\):
\[
E(Y) = 1 \cdot [P(1,1) + P(-1,1)] + (-1) \cdot [P(1,-1) + P(-1,-1)]
\]
\[
= 1 \cdot (\frac{3}{8} + \frac{1}{8}) + (-1) \cdot (\frac{1}{8} + \frac{3}{8}) = \frac{4}{8} - \frac{4}{8} =
\]
Now, compute \(E(XY)\):
\[
E(XY) = (1)(1)P(1,1) + (1)(-1)P(1,-1) + (-1)(1)P(-1,1) + (-1)(-1)P(-1,-1)
\]
\[
= 1 \cdot \frac{3}{8} + (-1) \cdot \frac{1}{8} + (-1) \cdot \frac{1}{8} + 1 \cdot \frac{3}{8}
\]
\[
= \frac{3}{8} - \frac{1}{8} - \frac{1}{8} + \frac{3}{8} = \frac{4}{8} = \frac{1}{2}
\]
So,
\[
Cov(X, Y) = \frac{1}{2} - ()() = \frac{1}{2}
\]
---
## **d. Let \(Z = X + Y\). Find \(Var(Z)\).**
\[
Var(Z) = Var(X + Y) = Var(X) + Var(Y) + 2Cov(X, Y)
\]
We already found \(Var(X) = 1\), \(Cov(X, Y) = \frac{1}{2}\).
Now, \(Var(Y)\):
By symmetry, \(Var(Y) = Var(X) = 1\).
So,
\[
Var(Z) = 1 + 1 + 2 \cdot \frac{1}{2} = 1 + 1 + 1 = 3
\]
---
## **e. Let \(T = 2X + 2Y - 3\). Find \(Cov(Z, T)\) where \(Z = X + Y\).**
\[
Cov(Z, T) = Cov(X + Y, 2X + 2Y - 3)
\]
Constant terms do not contribute to covariance, so:
\[
Cov(X + Y, 2X + 2Y - 3) = Cov(X + Y, 2X + 2Y)
\]
\[
= 2Cov(X+Y, X) + 2Cov(X+Y, Y)
\]
Let's calculate \(Cov(X+Y, X)\):
\[
Cov(X+Y, X) = Cov(X, X) + Cov(Y, X) = Var(X) + Cov(Y, X) = 1 + \frac{1}{2} = \frac{3}{2}
\]
Similarly,
\[
Cov(X+Y, Y) = Cov(X, Y) + Cov(Y, Y) = \frac{1}{2} + 1 = \frac{3}{2}
\]
So,
\[
Cov(Z, T) = 2 \cdot \frac{3}{2} + 2 \cdot \frac{3}{2} = 3 + 3 = 6
\]
---
## **Summary Table**
| Part | Answer |
|------|--------|
| a | \(E(X) = \) |
| b | \(Var(X) = 1\) |
| c | \(Cov(X, Y) = \frac{1}{2}\) |
| d | \(Var(Z) = 3\) |
| e | \(Cov(Z, T) = 6\) |
Question:
whAT A PART ANSWR
This is AI generated Response.
Asked by: Nadeem
Created at: 2025-12-09 00:42:30
Asked by: Nadeem
Created at: 2025-12-09 00:42:30
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# Discrete Random Variables Analysis
## Given Joint PMF
- \( P(x = 1, y = 1) = \frac{3}{8} \)
- \( P(x = -1, y = -1) = \frac{3}{8} \)
- \( P(x = 1, y = -1) = \frac{1}{8} \)
- \( P(x = -1, y = 1) = \frac{1}{8} \)
---
## a. Find \(E(X)\)
\[
E(X) = \sum_{x,y} x \cdot P(x, y)
\]
\[
= 1 \cdot \frac{3}{8} + 1 \cdot \frac{1}{8} - 1 \cdot \frac{3}{8} - 1 \cdot \frac{1}{8}
\]
\[
= \frac{3}{8} + \frac{1}{8} - \frac{3}{8} - \frac{1}{8} = 0
\]
---
## b. Find \(Var(X)\)
\[
Var(X) = E(X^2) - [E(X)]^2
\]
\[
E(X^2) = 1 \cdot (\frac{3}{8} + \frac{1}{8}) + 1 \cdot (\frac{1}{8} + \frac{3}{8}) = 1
\]
\[
Var(X) = 1 - 0^2 = 1
\]
---
## c. Find \(Cov(X, Y)\)
\[
Cov(X, Y) = E(XY) - E(X)E(Y)
\]
Calculate \(E(Y)\):
\[
E(Y) = 1 \cdot (\frac{3}{8} + \frac{1}{8}) - 1 \cdot (\frac{1}{8} + \frac{3}{8}) = 0
\]
Calculate \(E(XY)\):
\[
E(XY) = 1 \cdot \frac{3}{8} - 1 \cdot \frac{1}{8} - 1 \cdot \frac{1}{8} + 1 \cdot \frac{3}{8} = \frac{1}{2}
\]
Thus,
\[
Cov(X, Y) = \frac{1}{2} - (0)(0) = \frac{1}{2}
\]
---
## d. Let \(Z = X + Y\). Find \(Var(Z)\)
\[
Var(Z) = Var(X + Y) = Var(X) + Var(Y) + 2Cov(X, Y)
\]
Since \(Var(Y) = Var(X) = 1\):
\[
Var(Z) = 1 + 1 + 2 \cdot \frac{1}{2} = 3
\]
---
## e. Let \(T = 2X + 2Y - 3\). Find \(Cov(Z, T)\)
\[
Cov(Z, T) = Cov(X + Y, 2X + 2Y - 3)
\]
\[
= Cov(X + Y, 2X + 2Y) = 2Cov(X + Y, X) + 2Cov(X + Y, Y)
\]
Calculate \(Cov(X + Y, X)\):
\[
Cov(X + Y, X) = Var(X) + Cov(Y, X) = 1 + \frac{1}{2} = \frac{3}{2}
\]
Calculate \(Cov(X + Y, Y)\):
\[
Cov(X + Y, Y) = Cov(X, Y) + Var(Y) = \frac{1}{2} + 1 = \frac{3}{2}
\]
Thus,
\[
Cov(Z, T) = 2 \cdot \frac{3}{2} + 2 \cdot \frac{3}{2} = 6
\]
---
## Summary of Results
| Part | Answer |
|------|--------------------|
| a | \(E(X) = 0\) |
| b | \(Var(X) = 1\) |
| c | \(Cov(X, Y) = \frac{1}{2}\) |
| d | \(Var(Z) = 3\) |
| e | \(Cov(Z, T) = 6\) |
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