Assignment 2 Please Answer The 2 Questions Given In The Case Study 5 ✓ Solved

Assignment 2: Please answer the 2 questions given in the case study (5 Marks Each) Consider the following joint distribution of X and Y. For example, P(X=-200, Y=-100) = .3, P(X=-200, Y=100) = .1, etc. XY -....4 a) Calculate the mean and variance of X b) Calculate the mean and variance of Y. c) Calculate the covariance and correlation between X and Y. Is there a positive or negative relationship between X and Y? Is the relationship strong, weak or moderate? Please answer the questions given at the end of the case study (5 Marks Each)

Paper for above instructions

To address the questions posed in Assignment 2 regarding the joint distribution of random variables \(X\) and \(Y\), we need to first extract relevant data from the joint distribution table that has been provided. In this case, the joint distribution gives probabilities for specific values of \(X\) and \(Y\).

Given Joint Probability Distribution

For the purposes of this assignment, let's assume the provided distribution is as follows (this is illustrative):
| X \ Y | -100 | 100 |
|-------------|------|-------|
| -200 | 0.3 | 0.1 |
| 0 | 0.2 | 0.4 |
| 200 | 0.1 | 0.3 |
From the table, we can derive the marginal probabilities of \(X\) and \(Y\).

A) Calculate the Mean and Variance of X

Step 1: Marginal Distribution of X
To find the mean and variance of \(X\), we first need to compute the marginal distribution of \(X\).
- For \(X = -200\):
\[
P(X = -200) = 0.3 + 0.1 = 0.4
\]
- For \(X = 0\):
\[
P(X = 0) = 0.2 + 0.4 = 0.6
\]
- For \(X = 200\):
\[
P(X = 200) = 0.1 + 0.3 = 0.4
\]
Thus, the marginal distribution of \(X\) is:
- \(P(X = -200) = 0.4\)
- \(P(X = 0) = 0.6\)
- \(P(X = 200) = 0.4\)
Step 2: Mean of X
The mean of \(X\) is computed using:
\[
E(X) = \sum (x_i \cdot P(X = x_i))
\]
Calculating:
\[
E(X) = (-200)(0.4) + (0)(0.6) + (200)(0.4) = -80 + 0 + 80 = 0
\]
Step 3: Variance of X
The variance is found using:
\[
Var(X) = E(X^2) - (E(X))^2
\]
We first calculate \(E(X^2)\):
\[
E(X^2) = \sum (x_i^2 \cdot P(X = x_i))
\]
Calculating:
\[
E(X^2) = (-200)^2(0.4) + (0)^2(0.6) + (200)^2(0.4) = 40000(0.4) + 0 + 40000(0.4) = 16000 + 0 + 16000 = 32000
\]
Now compute \(Var(X)\):
\[
Var(X) = 32000 - (0)^2 = 32000
\]

B) Calculate the Mean and Variance of Y

Step 1: Marginal Distribution of Y
Next, we find the marginal distribution of \(Y\).
- For \(Y = -100\):
\[
P(Y = -100) = 0.3 + 0.2 + 0.1 = 0.6
\]
- For \(Y = 100\):
\[
P(Y = 100) = 0.1 + 0.4 + 0.3 = 0.8
\]
Thus, the marginal distribution of \(Y\) is:
- \(P(Y = -100) = 0.6\)
- \(P(Y = 100) = 0.8\)
Step 2: Mean of Y
Calculating \(E(Y)\):
\[
E(Y) = (-100)(0.6) + (100)(0.8) = -60 + 80 = 20
\]
Step 3: Variance of Y
Calculating \(E(Y^2)\):
\[
E(Y^2) = (-100)^2(0.6) + (100)^2(0.8) = 10000(0.6) + 10000(0.8) = 6000 + 8000 = 14000
\]
Now compute \(Var(Y)\):
\[
Var(Y) = 14000 - (20)^2 = 14000 - 400 = 13600
\]

C) Calculate the Covariance and Correlation Between X and Y

Step 1: Covariance
The covariance is calculated using the formula:
\[
Cov(X, Y) = E(XY) - E(X)E(Y)
\]
Step 2: Compute \(E(XY)\)
Calculating \(E(XY)\) with joint probabilities:
\[
E(XY) = \sum (x_i \cdot y_j \cdot P(X=x_i, Y=y_j))
\]
\[
E(XY) = (-200)(-100)(0.3) + (-200)(100)(0.1) + (0)(-100)(0.2) + (0)(100)(0.4) + (200)(-100)(0.1) + (200)(100)(0.3)
\]
Calculating gives:
\[
= 6000 - 2000 + 0 + 0 - 2000 + 6000 = 8000
\]
Step 3: Covariance Calculation
Substituting into the covariance formula:
\[
Cov(X, Y) = 8000 - (0)(20) = 8000
\]
Step 4: Correlation Calculation
The correlation coefficient \( \rho \) is given by:
\[
\rho(X, Y) = \frac{Cov(X, Y)}{\sqrt{Var(X)Var(Y)}}
\]
Calculating:
\[
\rho(X, Y) = \frac{8000}{\sqrt{32000 \cdot 13600}} \approx \frac{8000}{\sqrt{435200000}} \approx \frac{8000}{20856.11} \approx 0.384
\]

Conclusion

Based on the computations:
- The mean of \(X\) is 0, and its variance is \(32000\).
- The mean of \(Y\) is \(20\), and its variance is \(13600\).
- The covariance is \(8000\), and the correlation is approximately \(0.384\), indicating a positive, moderate relationship between \(X\) and \(Y\).

References

1. DeGroot, M.H. & Schervish, M.J. (2012). Probability and Statistics. Pearson.
2. Walpole, R.E., Myers, R.H., & Myers, S.L. (2012). Probability and Statistics. Pearson.
3. Casella, G. & Berger, R.L. (2002). Statistical Inference. Duxbury Press.
4. Johnson, R.A., & Bhattacharyya, G.K. (2010). Statistics: Principles and Methods. Wiley.
5. McClave, J.T. & Sincich, T. (2016). Statistics. Pearson.
6. D'Agostino, R.B. & Stephens, M.A. (1986). Goodness-of-Fit Techniques. Marcel Dekker.
7. Ross, S.M. (2010). A First Course in Probability. Pearson.
8. Freund, J.E. & Perles, B.M. (2006). Statistics. Pearson.
9. Devore, J.L. (2015). Probability and Statistics. Cengage Learning.
10. Mood, A.M., Graybill, F.A., & Boes, D.C. (1974). Introduction to the Theory of Statistics. McGraw-Hill.