SOLUTION: 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 fol

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 $
---
### **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 $
---
### **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 $
---
### **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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