Question 1207013: 4. Discrete independent random variables X and Y are given by the following laws of distribution:
X 0 3 Y - 2 - 1 2
P 0,3 0,7 P 0,2 0,4 0,4
Find M (X + Y) by two ways: 1) composing the law of distribution of X + Y; 2) using the property: M (X + Y) = M (X) + M (Y).
Answer by ElectricPavlov(122)   (Show Source):
You can put this solution on YOUR website! We are tasked with calculating the expected value \( M(X + Y) \) by two methods. Let us proceed step by step.
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### **Step 1: Distribution Laws**
#### Distribution of \( X \):
\[
X: \quad 0, \, 3 \quad \text{with probabilities } \quad P(X = 0) = 0.3, \, P(X = 3) = 0.7.
\]
#### Distribution of \( Y \):
\[
Y: \quad -2, \, -1, \, 2 \quad \text{with probabilities } \quad P(Y = -2) = 0.2, \, P(Y = -1) = 0.4, \, P(Y = 2) = 0.4.
\]
### **Step 2: Method 1 - Composing the Law of Distribution for \( X + Y \)**
Since \( X \) and \( Y \) are independent, the probability of any combination \( (X, Y) \) is given by the product of their probabilities:
\[
P(X + Y = z) = \sum_{(x, y): x + y = z} P(X = x) \cdot P(Y = y).
\]
#### Compute Possible Values of \( X + Y \):
- If \( X = 0 \):
\[
X + Y \in \{-2, -1, 2\} \quad \text{(corresponding to \( Y = -2, -1, 2 \))}.
\]
- If \( X = 3 \):
\[
X + Y \in \{1, 2, 5\} \quad \text{(corresponding to \( Y = -2, -1, 2 \))}.
\]
Thus, the possible values of \( X + Y \) are:
\[
\{-2, -1, 1, 2, 5\}.
\]
#### Compute Probabilities for Each Value of \( X + Y \):
- For \( X + Y = -2 \):
\[
P(X + Y = -2) = P(X = 0) \cdot P(Y = -2) = 0.3 \cdot 0.2 = 0.06.
\]
- For \( X + Y = -1 \):
\[
P(X + Y = -1) = P(X = 0) \cdot P(Y = -1) = 0.3 \cdot 0.4 = 0.12.
\]
- For \( X + Y = 1 \):
\[
P(X + Y = 1) = P(X = 3) \cdot P(Y = -2) = 0.7 \cdot 0.2 = 0.14.
\]
- For \( X + Y = 2 \):
\[
P(X + Y = 2) = P(X = 0) \cdot P(Y = 2) + P(X = 3) \cdot P(Y = -1) = (0.3 \cdot 0.4) + (0.7 \cdot 0.4) = 0.12 + 0.28 = 0.4.
\]
- For \( X + Y = 5 \):
\[
P(X + Y = 5) = P(X = 3) \cdot P(Y = 2) = 0.7 \cdot 0.4 = 0.28.
\]
#### Distribution of \( X + Y \):
\[
X + Y: \quad -2, \, -1, \, 1, \, 2, \, 5 \quad \text{with probabilities } \quad 0.06, \, 0.12, \, 0.14, \, 0.4, \, 0.28.
\]
#### Compute \( M(X + Y) \):
The expected value is:
\[
M(X + Y) = \sum_{z} z \cdot P(X + Y = z).
\]
Substitute:
\[
M(X + Y) = (-2)(0.06) + (-1)(0.12) + (1)(0.14) + (2)(0.4) + (5)(0.28).
\]
\[
M(X + Y) = -0.12 - 0.12 + 0.14 + 0.8 + 1.4 = 2.1.
\]
---
### **Step 3: Method 2 - Using the Property \( M(X + Y) = M(X) + M(Y) \)**
#### Compute \( M(X) \):
\[
M(X) = \sum_{x} x \cdot P(X = x).
\]
\[
M(X) = (0)(0.3) + (3)(0.7) = 2.1.
\]
#### Compute \( M(Y) \):
\[
M(Y) = \sum_{y} y \cdot P(Y = y).
\]
\[
M(Y) = (-2)(0.2) + (-1)(0.4) + (2)(0.4).
\]
\[
M(Y) = -0.4 - 0.4 + 0.8 = 0.
\]
#### Compute \( M(X + Y) \):
\[
M(X + Y) = M(X) + M(Y) = 2.1 + 0 = 2.1.
\]
---
### **Final Answer**:
\[
M(X + Y) = \boxed{2.1}.
\]
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