Probability Density Function - Need Help

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SUMMARY

The discussion centers on finding the Probability Density Function (PDF) of W = X + Y, where X and Y have the joint PDF fX,Y(x,y) = 2 for 0 ≤ x ≤ y ≤ 1. The correct PDF derived is fw(w) = w for 0 ≤ w ≤ 1, fw(w) = 2 - w for 1 ≤ w ≤ 2, and fw(w) = 0 otherwise. The initial incorrect solution provided was fw(w) = 2w - 1 for w > 0, which was identified as erroneous due to negative values for w < 1/2 and an integral that did not equal 1.

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  • Understanding of joint probability distributions
  • Knowledge of double integration techniques
  • Familiarity with cumulative distribution functions (CDF)
  • Ability to differentiate functions to find probability density functions (PDF)
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vptran84
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Probability Density Function -- Need Help!

Hi,

Can someone please check my work if i did the problem correctly? thanks in advance.

Here is the problem:

Find the PDF of W = X + Y when X and Y have the joint PDF fx,y (x,y) = 2 for 0<=x<=y<=1, and 0 otherwise.

here is my solution:
<br /> \int_{0}^{1} \int_{0}^{w-y} 2dxdy<br />

I work through the integral and get fw (w) = 2w-1 for w>0, and 0 for w<0.
 
Last edited:
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Your answer is obviously wrong. f(w) is <0 for w<1/2. Moreover, the integral should be 1 - yours is 0.
 
ok, i did a little more thinking :-p and this is what i got now...

For region w>0, the region of integration is outside so CDF Fw (w) is 0

For region 0<=w<=1, i used double integration, and i get w^2/2

For region 1<=w<=2, i get 2w-1-w^2/2

For region w>2, i get 1.

So to find PDF, i take the derivative, and i get the following:

fw(w) = w for 0<=w<=1
fw(w) = 2-w for 1<=w<=2
fw(w) = 0 otherwise.

Please let me know if i did anything wrong.
 
Before I looked at your latest post, I worked it out myself. I got the same result as you did.
 

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