Math Problem Statement

Given two independent random variables X and Y with respective probability density functions p_X(x) and p_Y(y), find the probability density function of their sum Z = X + Y, and compute it explicitly for X ~ N(μ_X, σ_X^2) and Y ~ N(μ_Y, σ_Y^2).

Solution

Problem Explanation:

Part 1:

The first part requires proving that if XX and YY are independent random variables with probability density functions pX(x)p_X(x) and pY(y)p_Y(y), then the probability density function pZ(z)p_Z(z) of their sum Z=X+YZ = X + Y is given by: pZ(z)=RDpX(x)pY(zx)dxp_Z(z) = \int_{\mathbb{R}^D} p_X(x) p_Y(z - x) dx This result comes from the convolution of the probability density functions of independent random variables.

Part 2:

The second part involves explicitly computing pZ(z)p_Z(z) when XN(μX,σX2)X \sim \mathcal{N}(\mu_X, \sigma_X^2) and YN(μY,σY2)Y \sim \mathcal{N}(\mu_Y, \sigma_Y^2), which are one-dimensional independent normal random variables.


Solution:

1. Proof of Convolution Formula:

Let Z=X+YZ = X + Y. The probability density function of ZZ, pZ(z)p_Z(z), is defined as: pZ(z)=P(Z=z)=P(X+Y=z).p_Z(z) = P(Z = z) = P(X + Y = z). By the independence of XX and YY, their joint density is the product of their individual densities. So, we can write: pZ(z)=xRDpX(x)pY(zx)dx.p_Z(z) = \int_{x \in \mathbb{R}^D} p_X(x) p_Y(z - x) dx. This is the convolution formula for the sum of two independent random variables.


2. Sum of Two Normal Distributions:

Let XN(μX,σX2)X \sim \mathcal{N}(\mu_X, \sigma_X^2) and YN(μY,σY2)Y \sim \mathcal{N}(\mu_Y, \sigma_Y^2).

2.1 Mean of ZZ:

The expectation of ZZ is: [ \mathbb{E}[Z] = \mathbb{E}[X + Y] = \mathbb{E}[X] + \mathbb{E}[Y] = \mu_X + \mu_Y. ]

2.2 Variance of ZZ:

Since XX and YY are independent, their variances add: Var(Z)=Var(X)+Var(Y)=σX2+σY2.\text{Var}(Z) = \text{Var}(X) + \text{Var}(Y) = \sigma_X^2 + \sigma_Y^2.

2.3 Distribution of ZZ:

The sum of two independent normal random variables is also normally distributed: ZN(μX+μY,σX2+σY2).Z \sim \mathcal{N}(\mu_X + \mu_Y, \sigma_X^2 + \sigma_Y^2). Thus, the probability density function of ZZ is: pZ(z)=12π(σX2+σY2)exp((z(μX+μY))22(σX2+σY2)).p_Z(z) = \frac{1}{\sqrt{2\pi (\sigma_X^2 + \sigma_Y^2)}} \exp\left(-\frac{(z - (\mu_X + \mu_Y))^2}{2(\sigma_X^2 + \sigma_Y^2)}\right).


Final Results:

  1. Convolution Formula: pZ(z)=RDpX(x)pY(zx)dxp_Z(z) = \int_{\mathbb{R}^D} p_X(x) p_Y(z - x) dx

  2. Probability Density Function for ZZ (Normal Sum): pZ(z)=12π(σX2+σY2)exp((z(μX+μY))22(σX2+σY2)).p_Z(z) = \frac{1}{\sqrt{2\pi (\sigma_X^2 + \sigma_Y^2)}} \exp\left(-\frac{(z - (\mu_X + \mu_Y))^2}{2(\sigma_X^2 + \sigma_Y^2)}\right).

Let me know if you’d like a more detailed step-by-step derivation for either part!


Questions to Explore:

  1. What happens to the convolution formula if XX and YY are not independent?
  2. How would the formula generalize to three independent random variables X+Y+WX + Y + W?
  3. Can you derive the characteristic function approach to sum two independent normal variables?
  4. What is the convolution formula in higher dimensions (D>1D > 1)?
  5. How can the convolution theorem be used in Fourier transforms for finding pZ(z)p_Z(z)?

Tip:

For sums of independent variables, the characteristic function (Fourier transform of the PDF) simplifies convolutions as it converts them into products.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Random Variables
Convolution
Normal Distribution
Sum of Random Variables

Formulas

p_Z(z) = ∫ p_X(x) p_Y(z - x) dx
E[Z] = μ_X + μ_Y
Var(Z) = σ_X^2 + σ_Y^2
Z ~ N(μ_X + μ_Y, σ_X^2 + σ_Y^2)

Theorems

Convolution Theorem for Independent Random Variables
Sum of Independent Normal Distributions

Suitable Grade Level

Undergraduate