Question 1199680: Let X and Y be independent random variables with distributions :X∼N(−1,2),Y∼N(1,4).Find the pdf of Z=X+Y.
Answer by textot(100)   (Show Source):
You can put this solution on YOUR website! **1. Identify the Distributions**
* **X:** X ~ N(-1, 2)
* Mean (μ_X) = -1
* Standard Deviation (σ_X) = √2
* **Y:** Y ~ N(1, 4)
* Mean (μ_Y) = 1
* Standard Deviation (σ_Y) = 2
**2. Determine the Distribution of Z = X + Y**
* **Sum of Independent Normal Random Variables:**
* If X and Y are independent and normally distributed, then their sum Z = X + Y is also normally distributed.
* **Mean of Z:**
* μ_Z = μ_X + μ_Y = -1 + 1 = 0
* **Variance of Z:**
* Var(Z) = Var(X) + Var(Y) = σ_X² + σ_Y² = 2 + 4 = 6
* **Standard Deviation of Z:**
* σ_Z = √Var(Z) = √6
**3. Conclusion**
* **Distribution of Z:** Z ~ N(0, √6)
**In summary:**
The sum of the independent normal random variables X and Y, denoted as Z, follows a normal distribution with a mean of 0 and a standard deviation of √6.
**Note:**
* The PDF (probability density function) of a normal distribution with mean μ and standard deviation σ is given by:
f(z) = (1 / (σ√(2π))) * exp(-(z - μ)² / (2σ²))
* You can use this formula to calculate the probability density of Z at any specific value of z.
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