Question 1203801: Let e = (i, j) represent an arbitrary outcome resulting from two rolls of the four-sided die of Example 2.1.1. Tabulate the discrete pdf and sketch the graph of the CDF for the following random variables
(a) Y(e)=i+j.
(b) Z(e)=i-j.
(c) W(e)=(i-j)^2
Answer by asinus(45)   (Show Source):
You can put this solution on YOUR website! To analyze the problem, we first define the sample space for rolling two four-sided dice. Each die has possible outcomes \( \{1, 2, 3, 4\} \), so the sample space \( S \) consists of all ordered pairs \( e = (i, j) \), where \( i, j \in \{1, 2, 3, 4\} \). This gives \( |S| = 16 \) outcomes, each equally likely with probability \( P(e) = \frac{1}{16} \).
### **Part (a): \( Y(e) = i + j \)**
#### 1. Possible Values of \( Y \):
The sum \( Y(i, j) = i + j \) can take values from \( 2 \) (when \( i = 1, j = 1 \)) to \( 8 \) (when \( i = 4, j = 4 \)).
#### 2. Probability Distribution Function (PDF) of \( Y \):
Count the occurrences of each value of \( Y \):
- \( Y = 2 \): \( (1, 1) \) → 1 outcome, \( P(Y = 2) = \frac{1}{16} \)
- \( Y = 3 \): \( (1, 2), (2, 1) \) → 2 outcomes, \( P(Y = 3) = \frac{2}{16} \)
- \( Y = 4 \): \( (1, 3), (2, 2), (3, 1) \) → 3 outcomes, \( P(Y = 4) = \frac{3}{16} \)
- \( Y = 5 \): \( (1, 4), (2, 3), (3, 2), (4, 1) \) → 4 outcomes, \( P(Y = 5) = \frac{4}{16} \)
- \( Y = 6 \): \( (2, 4), (3, 3), (4, 2) \) → 3 outcomes, \( P(Y = 6) = \frac{3}{16} \)
- \( Y = 7 \): \( (3, 4), (4, 3) \) → 2 outcomes, \( P(Y = 7) = \frac{2}{16} \)
- \( Y = 8 \): \( (4, 4) \) → 1 outcome, \( P(Y = 8) = \frac{1}{16} \)
#### 3. CDF of \( Y \):
The cumulative distribution function \( F_Y(y) = P(Y \leq y) \) is:
- \( F_Y(2) = P(Y \leq 2) = \frac{1}{16} \)
- \( F_Y(3) = P(Y \leq 3) = \frac{3}{16} \)
- \( F_Y(4) = P(Y \leq 4) = \frac{6}{16} \)
- \( F_Y(5) = P(Y \leq 5) = \frac{10}{16} \)
- \( F_Y(6) = P(Y \leq 6) = \frac{13}{16} \)
- \( F_Y(7) = P(Y \leq 7) = \frac{15}{16} \)
- \( F_Y(8) = P(Y \leq 8) = 1 \)
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### **Part (b): \( Z(e) = i - j \)**
#### 1. Possible Values of \( Z \):
The difference \( Z(i, j) = i - j \) can take values from \( -3 \) (when \( i = 1, j = 4 \)) to \( 3 \) (when \( i = 4, j = 1 \)).
#### 2. PDF of \( Z \):
Count the occurrences of each value of \( Z \):
- \( Z = -3 \): \( (1, 4) \) → 1 outcome, \( P(Z = -3) = \frac{1}{16} \)
- \( Z = -2 \): \( (1, 3), (2, 4) \) → 2 outcomes, \( P(Z = -2) = \frac{2}{16} \)
- \( Z = -1 \): \( (1, 2), (2, 3), (3, 4) \) → 3 outcomes, \( P(Z = -1) = \frac{3}{16} \)
- \( Z = 0 \): \( (1, 1), (2, 2), (3, 3), (4, 4) \) → 4 outcomes, \( P(Z = 0) = \frac{4}{16} \)
- \( Z = 1 \): \( (2, 1), (3, 2), (4, 3) \) → 3 outcomes, \( P(Z = 1) = \frac{3}{16} \)
- \( Z = 2 \): \( (3, 1), (4, 2) \) → 2 outcomes, \( P(Z = 2) = \frac{2}{16} \)
- \( Z = 3 \): \( (4, 1) \) → 1 outcome, \( P(Z = 3) = \frac{1}{16} \)
#### 3. CDF of \( Z \):
The CDF \( F_Z(z) = P(Z \leq z) \) is:
- \( F_Z(-3) = P(Z \leq -3) = \frac{1}{16} \)
- \( F_Z(-2) = P(Z \leq -2) = \frac{3}{16} \)
- \( F_Z(-1) = P(Z \leq -1) = \frac{6}{16} \)
- \( F_Z(0) = P(Z \leq 0) = \frac{10}{16} \)
- \( F_Z(1) = P(Z \leq 1) = \frac{13}{16} \)
- \( F_Z(2) = P(Z \leq 2) = \frac{15}{16} \)
- \( F_Z(3) = P(Z \leq 3) = 1 \)
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### **Part (c): \( W(e) = (i - j)^2 \)**
#### 1. Possible Values of \( W \):
The squared difference \( W(i, j) = (i - j)^2 \) can take values \( \{0, 1, 4, 9\} \).
#### 2. PDF of \( W \):
Count the occurrences of each value of \( W \):
- \( W = 0 \): \( (1, 1), (2, 2), (3, 3), (4, 4) \) → 4 outcomes, \( P(W = 0) = \frac{4}{16} \)
- \( W = 1 \): \( (1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3) \) → 6 outcomes, \( P(W = 1) = \frac{6}{16} \)
- \( W = 4 \): \( (1, 3), (2, 4), (3, 1), (4, 2) \) → 4 outcomes, \( P(W = 4) = \frac{4}{16} \)
- \( W = 9 \): \( (1, 4), (4, 1) \) → 2 outcomes, \( P(W = 9) = \frac{2}{16} \)
#### 3. CDF of \( W \):
The CDF \( F_W(w) = P(W \leq w) \) is:
- \( F_W(0) = P(W \leq 0) = \frac{4}{16} \)
- \( F_W(1) = P(W \leq 1) = \frac{10}{16} \)
- \( F_W(4) = P(W \leq 4) = \frac{14}{16} \)
- \( F_W(9) = P(W \leq 9) = 1 \)
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### **Graphs of the CDFs**
Let me plot the CDFs for \( Y, Z, \) and \( W \).
Here are the cumulative distribution function (CDF) graphs for each random variable:
1. **\( Y(e) = i + j \)**: The CDF reflects the probabilities for the sum of two dice.
2. **\( Z(e) = i - j \)**: The CDF shows the probabilities for the difference between the dice outcomes.
3. **\( W(e) = (i - j)^2 \)**: The CDF represents the squared differences between the dice outcomes.
Each step in the graphs corresponds to an increase in the cumulative probability based on the respective PDF values.
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