SOLUTION: Find random variable X+2Y and Y median, dispersion and correlation coefficient, if X and Y are independent variables, X~P(4)(Poisson variable) and Y~T(1, 6)(continuous distribution

Question 1195038: Find random variable X+2Y and Y median, dispersion and correlation coefficient, if X and Y are independent variables, X~P(4)(Poisson variable) and Y~T(1, 6)(continuous distribution)
Answer by proyaop(69) (Show Source): You can put this solution on YOUR website!
Certainly, let's analyze the random variable X + 2Y and find the requested statistics.
**1. Define the Distributions**
* **X:** Poisson distribution with mean (λ) = 4.
* Probability Mass Function (PMF): P(X = k) = (e^(-λ) * λ^k) / k!
* **Y:** Continuous uniform distribution over the interval [1, 6].
* Probability Density Function (PDF): f(y) = 1/5 for 1 ≤ y ≤ 6, and 0 otherwise.
**2. X + 2Y**
* Since X and Y are independent, the distribution of X + 2Y will not have a simple closed-form expression.
* However, we can simulate values of X and Y and then calculate values of X + 2Y to analyze its properties.
**3. Median, Dispersion, and Correlation Coefficient of Y**
* **Median of Y:**
* For a continuous uniform distribution, the median is the midpoint of the interval: (1 + 6) / 2 = 3.5
* **Dispersion of Y (Standard Deviation):**
* For a continuous uniform distribution over the interval [a, b], the standard deviation is:
* σ_Y = (b - a) / √12 = (6 - 1) / √12 ≈ 1.44
* **Correlation Coefficient between X and Y:**
* Since X and Y are independent, their correlation coefficient is 0.
**Simulation in Python**
```python
import numpy as np
from scipy.stats import poisson, uniform
# Define the distributions of X and Y
X = poisson.rvs(mu=4, size=10000) # Poisson distribution with mean 4
Y = uniform.rvs(loc=1, scale=5, size=10000) # Uniform distribution between 1 and 6
# Create the new random variable Z = X + 2Y
Z = X + 2*Y
# Calculate the median of Y
median_y = np.median(Y)
# Calculate the dispersion of Y (standard deviation)
dispersion_y = np.std(Y)
# Since X and Y are independent, their correlation coefficient is 0
correlation_coefficient = 0
# Print the results
print(f"Median of Y: {median_y}")
print(f"Dispersion of Y (Standard Deviation): {dispersion_y}")
print(f"Correlation Coefficient between X and Y: {correlation_coefficient}")
```
This code will generate a large number of samples from the distributions of X and Y, and then calculate the median, dispersion, and correlation coefficient of Y.
**Key Points:**
* The distribution of X + 2Y will be more complex than either X or Y individually due to the combination of a discrete (Poisson) and continuous (uniform) distribution.
* The median of Y for a uniform distribution is easily calculated as the midpoint of the interval.
* The independence of X and Y ensures a correlation coefficient of 0.
I hope this explanation is helpful! Let me know if you have any further questions.

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