Statistics Problem: finding a PDF using the CDF technique

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SUMMARY

The discussion focuses on finding the probability density function (pdf) of the random variable T = X + Y, where X and Y are continuous random variables with a joint pdf defined as f(x,y) = 4xy for 0 < x < 1 and 0 < y < 1. The CDF technique is employed to derive G(t), the cumulative distribution function of T. Participants confirm that separate integrals are necessary for the cases where t < 1 and t > 1, emphasizing that combining these into a single function for G(t) is not feasible when using the CDF method.

PREREQUISITES
  • Understanding of joint probability density functions (pdf)
  • Familiarity with cumulative distribution functions (CDF)
  • Knowledge of double integrals in probability theory
  • Basic concepts of continuous random variables
NEXT STEPS
  • Study the derivation of cumulative distribution functions for sums of random variables
  • Learn about the properties of joint probability density functions
  • Explore alternative methods for finding pdfs, such as moment-generating functions
  • Investigate the implications of changing limits in double integrals
USEFUL FOR

Students studying probability theory, statisticians working with continuous random variables, and educators teaching concepts related to joint distributions and cumulative functions.

_Steve_
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Hey guys, I'm stuck on a question in my homework assignment and I was wondering if you could push me in the right direction? So here's the question:

X and Y are continuous random variables with joint pdf f(x,y)= 4xy (0<x<1, 0<y<1, and otherwise 0). Find the pdf of T=X+Y using the CDF technique.

So this is how I started off, I first let G(t) be the CDF of T, then I look at three different cases:
t<=0: G(t) = P[X+Y<=t] = 0
t>=2: G(t) = P[X+Y<=t] = 1
0<t<2: G(t) = P[X+Y<=t] = ?
So here I'm a little confused, I'm trying to figure out the limits of the double integral I'm supposed to take, I think I might have to take one integral from 0<t<1, then another from 1<=t<2, but then I'm stuck with two "functions" (one from (0,1), one from [1,2)) for my g(t). Are there any possible limits for this double integral that would save me from having to separate 0<t<2 into two double integrals?
 
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_Steve_ said:
Hey guys, I'm stuck on a question in my homework assignment and I was wondering if you could push me in the right direction? So here's the question:

X and Y are continuous random variables with joint pdf f(x,y)= 4xy (0<x<1, 0<y<1, and otherwise 0). Find the pdf of T=X+Y using the CDF technique.

So this is how I started off, I first let G(t) be the CDF of T, then I look at three different cases:
t<=0: G(t) = P[X+Y<=t] = 0
t>=2: G(t) = P[X+Y<=t] = 1
0<t<2: G(t) = P[X+Y<=t] = ?
So here I'm a little confused, I'm trying to figure out the limits of the double integral I'm supposed to take, I think I might have to take one integral from 0<t<1, then another from 1<=t<2, but then I'm stuck with two "functions" (one from (0,1), one from [1,2)) for my g(t). Are there any possible limits for this double integral that would save me from having to separate 0<t<2 into two double integrals?

You are correct in that you do need different integrals for t > 1 and t < 1. If you use the cdf method this is unavoidable. (Other, completely different, methods are easier, but you are told not to use them.)

RGV
 
Ray Vickson said:
You are correct in that you do need different integrals for t > 1 and t < 1. If you use the cdf method this is unavoidable. (Other, completely different, methods are easier, but you are told not to use them.)

RGV

Oh, okay, so how do I get a g(t) with only one function from 0<t<2 instead of having:
g(t)={f(t), t<1
h(t), t>1)
?

Does it make sense to write:
G(t)={0, t<0
F(t), t<1
H(t), t>1
1, t>2}
as:
G(t)={0, t<0
F(t)+H(t), 0<t<2
1, t>2}
?

EDIT: Nevermind, I think I was doing the other part of the question wrong.. thanks for your help!
 
Last edited:

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