Suppose two Poisson processes commence at the same instant, and suppose they are independent of one another. In the first process, on average,
5
events occur per hour. In the other, on average,
2

Question

Suppose two Poisson processes commence at the same instant, and suppose they are independent of one another. In the first process, on average, 
5
 ﻿events occur per hour. In the other, on average, 
2
 ﻿events occur per hour. If the independent random variables 
x
1
 ﻿and 
x
2
 ﻿count the number of events in the first and second processes, respectively, over the course of an hour, what is 
(
i
)
 
E
(
x
1
 
x
2
)
 ﻿and 
(
ii
)
 ﻿Var
(
2
x
1
 
3
x
2
)
 
?
 ﻿Note: The variance of a Poisson distribution with expected value 
λ
 ﻿is also 
λ
.     Provide the complete solution in structural manner from start to end, i need complete detailed solution in the format of - 1st - complete solution with all the calculation, 2nd - explanation of each and every parts, 3rd final conclusion statement of all part at the end. ( Note :- don't use pronoun in the solution)

VIPSolutions · Accepted Answer

Certainly, here is the detailed, step-by-step solution in the requested structure.

---

## 1. **Complete Solution with All Calculation**

### **Given:**
- Two independent Poisson processes.
- First process: Mean rate $ \lambda_1 = 5 $ events/hour.
- Second process: Mean rate $ \lambda_2 = 2 $ events/hour.
- $ x_1 $: Number of events in the first process (in 1 hour).
- $ x_2 $: Number of events in the second process (in 1 hour).
- $ x_1 \sim \text{Poisson}(5) $, $ x_2 \sim \text{Poisson}(2) $.
- $ x_1 $ and $ x_2 $ are independent.

---

### **(i) Find $ E(x_1 x_2) $:**

Since $ x_1 $ and $ x_2 $ are independent,
$$
E(x_1 x_2) = E(x_1) \cdot E(x_2)
$$
For Poisson, $ E(x_1) = \lambda_1 = 5 $, $ E(x_2) = \lambda_2 = 2 $.
$$
E(x_1 x_2) = 5 \times 2 = 10
$$

---

### **(ii) Find $ \mathrm{Var}(2x_1 - 3x_2) $:**

Variance of a linear combination of independent random variables:
$$
\mathrm{Var}(aX + bY) = a^2 \mathrm{Var}(X) + b^2 \mathrm{Var}(Y), \quad \text{if } X, Y \text{ independent}
$$
Given $ a = 2, b = -3 $, $ \mathrm{Var}(x_1) = \lambda_1 = 5 $, $ \mathrm{Var}(x_2) = \lambda_2 = 2 $:

$$
\mathrm{Var}(2x_1 - 3x_2) = 2^2 \mathrm{Var}(x_1) + (-3)^2 \mathrm{Var}(x_2)
$$
$$
= 4 \times 5 + 9 \times 2
$$
$$
= 20 + 18 = 38
$$

---

## 2. **Explanation of Each and Every Part**

### **(i) $ E(x_1 x_2) $:**

- The expectation of the product of two independent random variables is the product of their expectations.
- For Poisson distributed $ x_1 $ and $ x_2 $, the means are $ \lambda_1 $ and $ \lambda_2 $.
- Multiplying these gives the expected value of the product.

### **(ii) $ \mathrm{Var}(2x_1 - 3x_2) $:**

- For independent $ x_1 $, $ x_2 $, the variance of a linear combination is the sum of the variances of the scaled variables.
- Each variance is scaled by the square of its coefficient ($ a^2 $, $ b^2 $).
- For Poisson, the variance equals the mean.
- Compute each term and add.

---

## 3. **Final Conclusion Statement**

- The expected value $ E(x_1 x_2) $ is **10**.
- The variance $ \mathrm{Var}(2x_1 - 3x_2) $ is **38**.

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