Sum of Two Independent Random Variables

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SUMMARY

The discussion focuses on finding the density function of the sum of two independent and identically distributed random variables, X and Y, both uniformly distributed over the interval (-1, 1). The density function fZ of Z = X + Y is derived using the convolution of the individual densities, leading to the integral fZ = ∫ fX(z-y)fY(y) dy. The correct interpretation of the integrand indicates that it equals 1/2 within the specified bounds and 0 outside, clarifying the earlier confusion in the calculations.

PREREQUISITES
  • Understanding of Uniform Distribution, specifically Uniform(-1, 1)
  • Knowledge of convolution of probability density functions
  • Familiarity with integral calculus
  • Basic concepts of independent random variables
NEXT STEPS
  • Study the properties of convolution in probability theory
  • Learn about the Central Limit Theorem and its implications for sums of random variables
  • Explore the derivation of density functions for other distributions
  • Investigate the application of integral calculus in probability density functions
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Statisticians, data scientists, and students studying probability theory who are interested in understanding the behavior of sums of independent random variables.

Shannon Young
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Suppose X and Y are Uniform(-1, 1) such that X and Y are independent and identically distributed. What is the density of Z = X + Y?

Here is what I have done so far (I am new to this forum, so, my formatting is very bad). I know that
fX(x) = fY(x) = 1/2 if -1<x<1 and 0 otherwise

The density of Z will be given by
fZ= \intfX(z-y)fY(y)dy

fY(y) = 1/2 if -1<y<1 and 0 otherwise
So,

fZ= \int(1/2)fX(z-y)dy (bounds of integration -1 to 1)

The integrand = 0 if -1<z-y<1 or if z-1<y<z+1

That is where I get stuck, and need help to complete. Your assistance is appreciated, thanks
 
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Shannon Young said:
Suppose X and Y are Uniform(-1, 1) such that X and Y are independent and identically distributed. What is the density of Z = X + Y?

Here is what I have done so far (I am new to this forum, so, my formatting is very bad). I know that
fX(x) = fY(x) = 1/2 if -1<x<1 and 0 otherwise

The density of Z will be given by
fZ= \intfX(z-y)fY(y)dy

fY(y) = 1/2 if -1<y<1 and 0 otherwise
So,

fZ= \int(1/2)fX(z-y)dy (bounds of integration -1 to 1)

The integrand = 0 if -1<z-y<1 or if z-1<y<z+1

That is where I get stuck, and need help to complete. Your assistance is appreciated, thanks
The integrand = 0 if -1<z-y<1 or if z-1<y<z+1
This is incorrect, it should read =1/2 within the interval and =0 outside.
 
mathman said:
This is incorrect, it should read =1/2 within the interval and =0 outside.

I figured the problem out, thanks
 

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